JC2 H2 Maths Bukit Timah | Binomial Thresholds

Did you know? “At least four successes” includes exactly four, while “more than four” does not. In a binomial question this one-word difference changes the tail probability. For an integer-valued count X, P(X≥4)=1−P(X≤3), whereas P(X>4)=1−P(X≤4). Subtracting the wrong boundary can leave an answer that looks plausible but excludes a whole probability mass.

This JC2 H2 Mathematics Tutor Bukit Timah guide explains how to translate exactly k, fewer than k, at most k, at least k and more than k into correct binomial events. The quick answer comes first, then a complete five-trial example, boundary cases and 36 original probability clinics with exact fractions and checked complementary tails. The purpose is precise event language, not merely recognising the binomial formula on a calculator screen.

The 2027 SEAB H2 Mathematics syllabus 9758 includes using B(n,p) as a discrete probability model and calculating relevant probabilities. It explicitly excludes the normal approximation to the binomial distribution and finding the general cumulative distribution function of a discrete random variable. This guide computes individual requested binomial event probabilities and their sums, without deriving an entire cumulative distribution function or using an excluded normal approximation.

Quick answer: translate the comparison symbol before calculating

Let X~B(n,p), with n a nonnegative integer and 0≤p≤1. For 0≤k≤n, the single-outcome formula is P(X=k)=C(n,k)pᵏ(1−p)ⁿ⁻ᵏ. Since X takes integer values only, changing a strict inequality to a non-strict one can add or remove exactly one probability mass. The correct complement uses the event just below or at the stated integer boundary, depending on the wording.

Wording	Mathematical event	Exact working
Exactly k	X=k	P(X=k)
At most k	X≤k	Σ from j=0 to k P(X=j)
Fewer than k	X<k	P(X≤k−1)
At least k	X≥k	1−P(X≤k−1)
More than k	X>k	1−P(X≤k)
Between a and b inclusive	a≤X≤b	P(X≤b)−P(X≤a−1)

First define the success: X counts that one chosen event across n trials.
Check binomial assumptions: a fixed number of independent binary trials with a common success probability p.
Translate English literally: “at most” means ≤, “fewer than” means <, “at least” means ≥ and “more than” means >.
Keep integer boundaries: X<k is the same as X≤k−1.
Use a complement if shorter: X≥k is the complement of X≤k−1, not X≤k.
Use a probability mass for exactly: calculate just P(X=k), not a tail.
Check against nearby tails: P(X≥k)=P(X=k)+P(X>k).

Fully worked example: five trials with p=0.3

Assume five independent trials, each with success probability 0.3, and let X count successes. Then X~B(5,0.3). For exactly two successes, the binomial formula gives P(X=2)=C(5,2)(0.3)²(0.7)³=10×0.09×0.343=0.3087. There is no need to add other possible counts because the event is exactly two.

For at most two successes, sum the three possible counts: P(X=0)=0.7⁵=0.16807, P(X=1)=5(0.3)(0.7)⁴=0.36015, and P(X=2)=0.3087. Their total is P(X≤2)=0.83692. Fewer than two successes means X=0 or 1 only, so P(X<2)=0.16807+0.36015=0.52822.

Now for at least two, take the complement of fewer than two: P(X≥2)=1−0.52822=0.47178. More than two is the complement of at most two: P(X>2)=1−0.83692=0.16308. Check the difference: 0.47178−0.16308=0.3087, which is exactly the probability mass at X=2. The five pieces agree.

The one-mass difference explains the off-by-one trap

The two events X≥k and X>k differ only by the outcome X=k. Therefore P(X≥k)−P(X>k)=P(X=k). Similarly, X≤k and X<k differ by the same outcome: P(X≤k)−P(X<k)=P(X=k). These identities are more reliable than trying to memorise a calculator menu for each English phrase.

In the five-trial example, the probability of exactly two is 0.3087, large enough to change a tail answer substantially. The mistake is not necessarily a tiny numerical rounding issue. It is a wrong event definition. Draw a quick number line labelled 0,1,2,3,4,5, and circle the included counts before calculating.

A symmetric example: eight fair trials

Suppose X~B(8,0.5). The probability of exactly four successes is C(8,4)/2⁸=70/256=0.2734375. The distribution is symmetric around four. That gives P(X<4)=P(X>4)=0.36328125, because the two strict tails are mirror images. But P(X≤4)=P(X≥4)=0.63671875, because each inclusive tail contains the central mass at four.

These values show why an inclusive upper tail and an inclusive lower tail can both be greater than one-half. They overlap at X=4, so their probabilities do not sum to one. Instead, P(X≤4)+P(X>4)=1, and P(X<4)+P(X≥4)=1. The complementary event must actually partition the possible counts.

A binomial variable counts integers, not arbitrary decimals

If an H2 question states “fewer than 4.5 successes”, then because X is integer-valued, the event is X≤4. But “at most 4.5” is also X≤4. By contrast, “more than 4.5” is X≥5. It is often easiest to convert a non-integer threshold into the corresponding set of legal success counts before using a discrete probability formula.

Do not apply a continuity correction or a normal approximation to justify this change. The integer nature of X already settles the event exactly. The 2027 syllabus excludes the normal approximation to binomial, and this problem does not need it.

The boundary k=0

For any valid binomial count X, at least zero successes is certain: P(X≥0)=1. Fewer than zero successes is impossible: P(X<0)=0. Exactly zero successes has probability P(X=0)=(1−p)ⁿ. More than zero successes has probability 1−(1−p)ⁿ, the familiar complement calculation for at least one success.

The earlier Binomial “At Least One” guide owns that particular zero-boundary problem. This article places it inside the general comparison-language system with arbitrary k, where the change between strict and non-strict inequalities is the key decision.

The boundary k=n

A binomial count cannot exceed its fixed number of trials n. Thus P(X≤n)=1 and P(X>n)=0. Exactly n successes has probability pⁿ, so P(X≥n)=pⁿ. Fewer than n successes has probability 1−pⁿ. These identities are immediate model checks, and a calculator result that gives a nonzero P(X>n) signals a wrongly entered event or distribution.

If k is outside the possible count range, use the support of X to classify the event before computing. For example, when n=8, “at least nine successes” is impossible and “at most nine” is certain. A careful answer should not invent negative factorials or unsupported binomial terms for impossible counts.

An interval event: at least two but no more than five

Suppose X~B(8,0.5) and the event is “at least two but at most five”. It includes X=2,3,4,5, so P(2≤X≤5)=[C(8,2)+C(8,3)+C(8,4)+C(8,5)]/256. The numerator is 28+56+70+56=210, giving 210/256=105/128≈0.8203125.

A complementary method subtracts the excluded lower counts X=0,1 and upper counts X=6,7,8. Those categories do not overlap, so the complements can be added. An equivalent cumulative expression is P(X≤5)−P(X≤1). The lower boundary is one less than the smallest included value; that is the same off-by-one rule applied to an interval.

A different interval: more than two but at most five

For the same X~B(8,0.5), “more than two but at most five” includes X=3,4,5, not X=2. Its probability is [C(8,3)+C(8,4)+C(8,5)]/256=(56+70+56)/256=182/256=91/128. This differs from the previous interval by exactly P(X=2)=28/256=7/64.

The two example phrases sound similar but specify different sets of counts. In any interval question, write the smallest included integer and largest included integer. Then calculate the sum over that range. This is faster and more robust than attempting to interpret four words while entering a cumulative calculator command.

Why the sum of disjoint events is safe

The events X=0, X=1, …, X=n are mutually exclusive, because a particular experiment produces one count. Therefore adding their individual probabilities gives the probability of a union of those exact counts. For X≤k, sum the masses 0 through k. For X≥k, sum k through n or take an appropriate complement. You may not add overlapping tails without accounting for their overlap.

For example, X≤4 and X≥4 overlap at exactly X=4. Adding their probabilities counts P(X=4) twice. The correct complementary pair is X≤4 and X>4, or X<4 and X≥4. Set language comes before arithmetic, even when a graphing calculator handles the probability sums automatically.

When the binomial model is suitable

Before using B(n,p), check the stated mechanism: a fixed number n of trials, binary success/failure outcomes, a constant success probability p and independent trials. If the probability changes after each draw without replacement from a finite bag, a simple binomial model may not apply exactly. The earlier Without Replacement guide owns that dependence issue.

Event translation and model choice are separate skills. You can correctly interpret “at least four” and still apply the wrong distribution. A complete H2 solution first defines X and its model, then identifies which integer counts satisfy the requested event, and only then evaluates the corresponding probability.

The graphing calculator is a check, not an interpreter

A suitable graphing calculator can calculate individual binomial masses and probabilities summed over a range. But the user must supply the correct event boundary. If a cumulative feature returns P(X≤k), it must be complemented directly for “more than k” and shifted by one for “at least k”. A calculator cannot determine whether the word “at least” in a sentence was read correctly.

Write the event in symbols before entering a number. This can be one short line: “At least four means P(X≥4)=1−P(X≤3).” It gives a mathematical audit trail, and it makes a result easier to review if it disagrees with the expected magnitude. The 2027 syllabus expects students to understand graphing-calculator limitations and present working when required.

A one-minute probability diagnostic

Let X~B(5,0.3). Ask five questions in order: exactly two (0.3087), fewer than two (0.52822), at most two (0.83692), at least two (0.47178), and more than two (0.16308). Then ask which differences equal the single mass P(X=2). The learner should identify both inclusive-minus-strict differences without recalculating the entire binomial distribution.

Next change the threshold to zero: P(X≥0)=1 but P(X>0)=1−0.7⁵. Change it to n=5: P(X≤5)=1 and P(X>5)=0. These boundary questions help repair a learner’s event understanding faster than repeated numerical calculation with the same incorrect inequality.

Thirty-six original binomial event-translation clinics

The next 36 exercises use binomial counts with different integer trial totals, rational success probabilities and thresholds k. Each supplies the exact probability mass, the two lower tails and the two upper tails, with all complements verified using exact integer arithmetic. Several cases have k=0 or k=n to reinforce the support boundaries. These are original hypothetical probability models, not actual examination outcomes or official past-year questions.

1. n=5, p=1/5, threshold k=5

Try the word-to-symbol exercise. A fictional experiment consists of 5 independent success/failure trials, each with success probability p=1/5. Let X count successes, so X~B(5,1/5). At the integer threshold k=5, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=5)=C(5,5)(1/5)^5(1−1/5)^0=1/3125≈0.00032. Summing masses for values strictly below 5 gives P(X<5)=3124/3125≈0.99968. Including the threshold adds its mass: P(X≤5)=1≈1. The difference between these two lower-tail values is precisely the single mass at 5, not a rounding artefact.

2. n=6, p=3/4, threshold k=0

Try the word-to-symbol exercise. A fictional experiment consists of 6 independent success/failure trials, each with success probability p=3/4. Let X count successes, so X~B(6,3/4). At the integer threshold k=0, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=0)=C(6,0)(3/4)^0(1−3/4)^6=1/4096≈0.000244. Summing masses for values strictly below 0 gives P(X<0)=0≈0. Including the threshold adds its mass: P(X≤0)=1/4096≈0.000244. The difference between these two lower-tail values is precisely the single mass at 0, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥0)=1−P(X≤-1)=1≈1 and P(X>0)=1−P(X≤0)=4095/4096≈0.999756. The exact sums P(X<0)+P(X≥0)=1 and P(X≤0)+P(X>0)=1 both hold. The difference between P(X≥k) and P(X>k) is also exactly P(X=k). At this threshold the mass added by changing a strict sign to an inclusive sign is 0.0244% of all model outcomes. Explain that difference in words before calculating another case.

3. n=7, p=1/4, threshold k=6

Try the word-to-symbol exercise. A fictional experiment consists of 7 independent success/failure trials, each with success probability p=1/4. Let X count successes, so X~B(7,1/4). At the integer threshold k=6, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=6)=C(7,6)(1/4)^6(1−1/4)^1=21/16384≈0.001282. Summing masses for values strictly below 6 gives P(X<6)=8181/8192≈0.998657. Including the threshold adds its mass: P(X≤6)=16383/16384≈0.999939. The difference between these two lower-tail values is precisely the single mass at 6, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥6)=1−P(X≤5)=11/8192≈0.001343 and P(X>6)=1−P(X≤6)=1/16384≈0.000061. The exact sums P(X<6)+P(X≥6)=1 and P(X≤6)+P(X>6)=1 both hold. “At least” includes the threshold; “more than” excludes it. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.1282% of all model outcomes. Explain that difference in words before calculating another case.

4. n=8, p=4/5, threshold k=3

Try the word-to-symbol exercise. A fictional experiment consists of 8 independent success/failure trials, each with success probability p=4/5. Let X count successes, so X~B(8,4/5). At the integer threshold k=3, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=3)=C(8,3)(4/5)^3(1−4/5)^5=3584/390625≈0.009175. Summing masses for values strictly below 3 gives P(X<3)=481/390625≈0.001231. Including the threshold adds its mass: P(X≤3)=813/78125≈0.010406. The difference between these two lower-tail values is precisely the single mass at 3, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥3)=1−P(X≤2)=390144/390625≈0.998769 and P(X>3)=1−P(X≤3)=77312/78125≈0.989594. The exact sums P(X<3)+P(X≥3)=1 and P(X≤3)+P(X>3)=1 both hold. “At most” includes the threshold; “fewer than” excludes it. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.9175% of all model outcomes. Explain that difference in words before calculating another case.

5. n=9, p=1/3, threshold k=8

Try the word-to-symbol exercise. A fictional experiment consists of 9 independent success/failure trials, each with success probability p=1/3. Let X count successes, so X~B(9,1/3). At the integer threshold k=8, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=8)=C(9,8)(1/3)^8(1−1/3)^1=2/2187≈0.000914. Summing masses for values strictly below 8 gives P(X<8)=19664/19683≈0.999035. Including the threshold adds its mass: P(X≤8)=19682/19683≈0.999949. The difference between these two lower-tail values is precisely the single mass at 8, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥8)=1−P(X≤7)=19/19683≈0.000965 and P(X>8)=1−P(X≤8)=1/19683≈0.000051. The exact sums P(X<8)+P(X≥8)=1 and P(X≤8)+P(X>8)=1 both hold. The complement of X≥k is X≤k−1, not X≤k. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.0914% of all model outcomes. Explain that difference in words before calculating another case.

6. n=10, p=2/5, threshold k=2

Try the word-to-symbol exercise. A fictional experiment consists of 10 independent success/failure trials, each with success probability p=2/5. Let X count successes, so X~B(10,2/5). At the integer threshold k=2, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=2)=C(10,2)(2/5)^2(1−2/5)^8=236196/1953125≈0.120932. Summing masses for values strictly below 2 gives P(X<2)=452709/9765625≈0.046357. Including the threshold adds its mass: P(X≤2)=1633689/9765625≈0.16729. The difference between these two lower-tail values is precisely the single mass at 2, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥2)=1−P(X≤1)=9312916/9765625≈0.953643 and P(X>2)=1−P(X≤2)=8131936/9765625≈0.83271. The exact sums P(X<2)+P(X≥2)=1 and P(X≤2)+P(X>2)=1 both hold. The complement of X>k is X≤k, with no minus-one adjustment. At this threshold the mass added by changing a strict sign to an inclusive sign is 12.0932% of all model outcomes. Explain that difference in words before calculating another case.

7. n=11, p=1/2, threshold k=7

Try the word-to-symbol exercise. A fictional experiment consists of 11 independent success/failure trials, each with success probability p=1/2. Let X count successes, so X~B(11,1/2). At the integer threshold k=7, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=7)=C(11,7)(1/2)^7(1−1/2)^4=165/1024≈0.161133. Summing masses for values strictly below 7 gives P(X<7)=743/1024≈0.725586. Including the threshold adds its mass: P(X≤7)=227/256≈0.886719. The difference between these two lower-tail values is precisely the single mass at 7, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥7)=1−P(X≤6)=281/1024≈0.274414 and P(X>7)=1−P(X≤7)=29/256≈0.113281. The exact sums P(X<7)+P(X≥7)=1 and P(X≤7)+P(X>7)=1 both hold. A binomial count is integer-valued, so write the actual supported values before calculating. At this threshold the mass added by changing a strict sign to an inclusive sign is 16.1133% of all model outcomes. Explain that difference in words before calculating another case.

8. n=12, p=3/5, threshold k=11

Try the word-to-symbol exercise. A fictional experiment consists of 12 independent success/failure trials, each with success probability p=3/5. Let X count successes, so X~B(12,3/5). At the integer threshold k=11, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=11)=C(12,11)(3/5)^11(1−3/5)^1=4251528/244140625≈0.017414. Summing masses for values strictly below 11 gives P(X<11)=239357656/244140625≈0.980409. Including the threshold adds its mass: P(X≤11)=243609184/244140625≈0.997823. The difference between these two lower-tail values is precisely the single mass at 11, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥11)=1−P(X≤10)=4782969/244140625≈0.019591 and P(X>11)=1−P(X≤11)=531441/244140625≈0.002177. The exact sums P(X<11)+P(X≥11)=1 and P(X≤11)+P(X>11)=1 both hold. The probabilities of X≤k and X>k sum to one exactly. At this threshold the mass added by changing a strict sign to an inclusive sign is 1.7414% of all model outcomes. Explain that difference in words before calculating another case.

9. n=13, p=2/3, threshold k=1

Try the word-to-symbol exercise. A fictional experiment consists of 13 independent success/failure trials, each with success probability p=2/3. Let X count successes, so X~B(13,2/3). At the integer threshold k=1, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=1)=C(13,1)(2/3)^1(1−2/3)^12=26/1594323≈0.000016. Summing masses for values strictly below 1 gives P(X<1)=1/1594323≈0.000001. Including the threshold adds its mass: P(X≤1)=1/59049≈0.000017. The difference between these two lower-tail values is precisely the single mass at 1, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥1)=1−P(X≤0)=1594322/1594323≈0.999999 and P(X>1)=1−P(X≤1)=59048/59049≈0.999983. The exact sums P(X<1)+P(X≥1)=1 and P(X≤1)+P(X>1)=1 both hold. The probabilities of X

10. n=14, p=1/5, threshold k=4

Try the word-to-symbol exercise. A fictional experiment consists of 14 independent success/failure trials, each with success probability p=1/5. Let X count successes, so X~B(14,1/5). At the integer threshold k=4, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=4)=C(14,4)(1/5)^4(1−1/5)^10=1049624576/6103515625≈0.17197. Summing masses for values strictly below 4 gives P(X<4)=4261412864/6103515625≈0.69819. Including the threshold adds its mass: P(X≤4)=1062207488/1220703125≈0.87016. The difference between these two lower-tail values is precisely the single mass at 4, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥4)=1−P(X≤3)=1842102761/6103515625≈0.30181 and P(X>4)=1−P(X≤4)=158495637/1220703125≈0.12984. The exact sums P(X<4)+P(X≥4)=1 and P(X≤4)+P(X>4)=1 both hold. For k=0, every possible binomial count is at least zero. At this threshold the mass added by changing a strict sign to an inclusive sign is 17.1970% of all model outcomes. Explain that difference in words before calculating another case.

11. n=5, p=4/5, threshold k=5

Try the word-to-symbol exercise. A fictional experiment consists of 5 independent success/failure trials, each with success probability p=4/5. Let X count successes, so X~B(5,4/5). At the integer threshold k=5, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=5)=C(5,5)(4/5)^5(1−4/5)^0=1024/3125≈0.32768. Summing masses for values strictly below 5 gives P(X<5)=2101/3125≈0.67232. Including the threshold adds its mass: P(X≤5)=1≈1. The difference between these two lower-tail values is precisely the single mass at 5, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥5)=1−P(X≤4)=1024/3125≈0.32768 and P(X>5)=1−P(X≤5)=0≈0. The exact sums P(X<5)+P(X≥5)=1 and P(X≤5)+P(X>5)=1 both hold. For k=n, a count more than n is impossible, and at most n is certain. At this threshold the mass added by changing a strict sign to an inclusive sign is 32.7680% of all model outcomes. Explain that difference in words before calculating another case.

12. n=6, p=1/3, threshold k=1

Try the word-to-symbol exercise. A fictional experiment consists of 6 independent success/failure trials, each with success probability p=1/3. Let X count successes, so X~B(6,1/3). At the integer threshold k=1, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=1)=C(6,1)(1/3)^1(1−1/3)^5=64/243≈0.263374. Summing masses for values strictly below 1 gives P(X<1)=64/729≈0.087791. Including the threshold adds its mass: P(X≤1)=256/729≈0.351166. The difference between these two lower-tail values is precisely the single mass at 1, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥1)=1−P(X≤0)=665/729≈0.912209 and P(X>1)=1−P(X≤1)=473/729≈0.648834. The exact sums P(X<1)+P(X≥1)=1 and P(X≤1)+P(X>1)=1 both hold. If the probability mass at k is large, the strict versus inclusive choice can change the answer substantially. At this threshold the mass added by changing a strict sign to an inclusive sign is 26.3374% of all model outcomes. Explain that difference in words before calculating another case.

13. n=7, p=2/5, threshold k=6

Try the word-to-symbol exercise. A fictional experiment consists of 7 independent success/failure trials, each with success probability p=2/5. Let X count successes, so X~B(7,2/5). At the integer threshold k=6, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=6)=C(7,6)(2/5)^6(1−2/5)^1=1344/78125≈0.017203. Summing masses for values strictly below 6 gives P(X<6)=76653/78125≈0.981158. Including the threshold adds its mass: P(X≤6)=77997/78125≈0.998362. The difference between these two lower-tail values is precisely the single mass at 6, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥6)=1−P(X≤5)=1472/78125≈0.018842 and P(X>6)=1−P(X≤6)=128/78125≈0.001638. The exact sums P(X<6)+P(X≥6)=1 and P(X≤6)+P(X>6)=1 both hold. The difference between P(X≤k) and P(X

14. n=8, p=1/2, threshold k=0

Try the word-to-symbol exercise. A fictional experiment consists of 8 independent success/failure trials, each with success probability p=1/2. Let X count successes, so X~B(8,1/2). At the integer threshold k=0, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=0)=C(8,0)(1/2)^0(1−1/2)^8=1/256≈0.003906. Summing masses for values strictly below 0 gives P(X<0)=0≈0. Including the threshold adds its mass: P(X≤0)=1/256≈0.003906. The difference between these two lower-tail values is precisely the single mass at 0, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥0)=1−P(X≤-1)=1≈1 and P(X>0)=1−P(X≤0)=255/256≈0.996094. The exact sums P(X<0)+P(X≥0)=1 and P(X≤0)+P(X>0)=1 both hold. The difference between P(X≥k) and P(X>k) is also exactly P(X=k). At this threshold the mass added by changing a strict sign to an inclusive sign is 0.3906% of all model outcomes. Explain that difference in words before calculating another case.

15. n=9, p=3/5, threshold k=0

Try the word-to-symbol exercise. A fictional experiment consists of 9 independent success/failure trials, each with success probability p=3/5. Let X count successes, so X~B(9,3/5). At the integer threshold k=0, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=0)=C(9,0)(3/5)^0(1−3/5)^9=512/1953125≈0.000262. Summing masses for values strictly below 0 gives P(X<0)=0≈0. Including the threshold adds its mass: P(X≤0)=512/1953125≈0.000262. The difference between these two lower-tail values is precisely the single mass at 0, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥0)=1−P(X≤-1)=1≈1 and P(X>0)=1−P(X≤0)=1952613/1953125≈0.999738. The exact sums P(X<0)+P(X≥0)=1 and P(X≤0)+P(X>0)=1 both hold. “At least” includes the threshold; “more than” excludes it. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.0262% of all model outcomes. Explain that difference in words before calculating another case.

16. n=10, p=2/3, threshold k=8

Try the word-to-symbol exercise. A fictional experiment consists of 10 independent success/failure trials, each with success probability p=2/3. Let X count successes, so X~B(10,2/3). At the integer threshold k=8, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=8)=C(10,8)(2/3)^8(1−2/3)^2=1280/6561≈0.195092. Summing masses for values strictly below 8 gives P(X<8)=13795/19683≈0.700859. Including the threshold adds its mass: P(X≤8)=17635/19683≈0.895951. The difference between these two lower-tail values is precisely the single mass at 8, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥8)=1−P(X≤7)=5888/19683≈0.299141 and P(X>8)=1−P(X≤8)=2048/19683≈0.104049. The exact sums P(X<8)+P(X≥8)=1 and P(X≤8)+P(X>8)=1 both hold. “At most” includes the threshold; “fewer than” excludes it. At this threshold the mass added by changing a strict sign to an inclusive sign is 19.5092% of all model outcomes. Explain that difference in words before calculating another case.

17. n=11, p=1/5, threshold k=6

Try the word-to-symbol exercise. A fictional experiment consists of 11 independent success/failure trials, each with success probability p=1/5. Let X count successes, so X~B(11,1/5). At the integer threshold k=6, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=6)=C(11,6)(1/5)^6(1−1/5)^5=473088/48828125≈0.009689. Summing masses for values strictly below 6 gives P(X<6)=48259072/48828125≈0.988346. Including the threshold adds its mass: P(X≤6)=9746432/9765625≈0.998035. The difference between these two lower-tail values is precisely the single mass at 6, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥6)=1−P(X≤5)=569053/48828125≈0.011654 and P(X>6)=1−P(X≤6)=19193/9765625≈0.001965. The exact sums P(X<6)+P(X≥6)=1 and P(X≤6)+P(X>6)=1 both hold. The complement of X≥k is X≤k−1, not X≤k. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.9689% of all model outcomes. Explain that difference in words before calculating another case.

18. n=12, p=3/4, threshold k=12

Try the word-to-symbol exercise. A fictional experiment consists of 12 independent success/failure trials, each with success probability p=3/4. Let X count successes, so X~B(12,3/4). At the integer threshold k=12, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=12)=C(12,12)(3/4)^12(1−3/4)^0=531441/16777216≈0.031676. Summing masses for values strictly below 12 gives P(X<12)=16245775/16777216≈0.968324. Including the threshold adds its mass: P(X≤12)=1≈1. The difference between these two lower-tail values is precisely the single mass at 12, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥12)=1−P(X≤11)=531441/16777216≈0.031676 and P(X>12)=1−P(X≤12)=0≈0. The exact sums P(X<12)+P(X≥12)=1 and P(X≤12)+P(X>12)=1 both hold. The complement of X>k is X≤k, with no minus-one adjustment. At this threshold the mass added by changing a strict sign to an inclusive sign is 3.1676% of all model outcomes. Explain that difference in words before calculating another case.

19. n=13, p=1/4, threshold k=3

Try the word-to-symbol exercise. A fictional experiment consists of 13 independent success/failure trials, each with success probability p=1/4. Let X count successes, so X~B(13,1/4). At the integer threshold k=3, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=3)=C(13,3)(1/4)^3(1−1/4)^10=8444007/33554432≈0.251651. Summing masses for values strictly below 3 gives P(X<3)=11160261/33554432≈0.332602. Including the threshold adds its mass: P(X≤3)=4901067/8388608≈0.584253. The difference between these two lower-tail values is precisely the single mass at 3, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥3)=1−P(X≤2)=22394171/33554432≈0.667398 and P(X>3)=1−P(X≤3)=3487541/8388608≈0.415747. The exact sums P(X<3)+P(X≥3)=1 and P(X≤3)+P(X>3)=1 both hold. A binomial count is integer-valued, so write the actual supported values before calculating. At this threshold the mass added by changing a strict sign to an inclusive sign is 25.1651% of all model outcomes. Explain that difference in words before calculating another case.

20. n=14, p=4/5, threshold k=1

Try the word-to-symbol exercise. A fictional experiment consists of 14 independent success/failure trials, each with success probability p=4/5. Let X count successes, so X~B(14,4/5). At the integer threshold k=1, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=1)=C(14,1)(4/5)^1(1−4/5)^13=56/6103515625≈0. Summing masses for values strictly below 1 gives P(X<1)=1/6103515625≈0. Including the threshold adds its mass: P(X≤1)=57/6103515625≈0. The difference between these two lower-tail values is precisely the single mass at 1, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥1)=1−P(X≤0)=6103515624/6103515625≈1 and P(X>1)=1−P(X≤1)=6103515568/6103515625≈1. The exact sums P(X<1)+P(X≥1)=1 and P(X≤1)+P(X>1)=1 both hold. The probabilities of X≤k and X>k sum to one exactly. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.0000% of all model outcomes. Explain that difference in words before calculating another case.

21. n=5, p=1/2, threshold k=5

Try the word-to-symbol exercise. A fictional experiment consists of 5 independent success/failure trials, each with success probability p=1/2. Let X count successes, so X~B(5,1/2). At the integer threshold k=5, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=5)=C(5,5)(1/2)^5(1−1/2)^0=1/32≈0.03125. Summing masses for values strictly below 5 gives P(X<5)=31/32≈0.96875. Including the threshold adds its mass: P(X≤5)=1≈1. The difference between these two lower-tail values is precisely the single mass at 5, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥5)=1−P(X≤4)=1/32≈0.03125 and P(X>5)=1−P(X≤5)=0≈0. The exact sums P(X<5)+P(X≥5)=1 and P(X≤5)+P(X>5)=1 both hold. The probabilities of X

22. n=6, p=3/5, threshold k=3

Try the word-to-symbol exercise. A fictional experiment consists of 6 independent success/failure trials, each with success probability p=3/5. Let X count successes, so X~B(6,3/5). At the integer threshold k=3, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=3)=C(6,3)(3/5)^3(1−3/5)^3=864/3125≈0.27648. Summing masses for values strictly below 3 gives P(X<3)=112/625≈0.1792. Including the threshold adds its mass: P(X≤3)=1424/3125≈0.45568. The difference between these two lower-tail values is precisely the single mass at 3, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥3)=1−P(X≤2)=513/625≈0.8208 and P(X>3)=1−P(X≤3)=1701/3125≈0.54432. The exact sums P(X<3)+P(X≥3)=1 and P(X≤3)+P(X>3)=1 both hold. For k=0, every possible binomial count is at least zero. At this threshold the mass added by changing a strict sign to an inclusive sign is 27.6480% of all model outcomes. Explain that difference in words before calculating another case.

23. n=7, p=2/3, threshold k=5

Try the word-to-symbol exercise. A fictional experiment consists of 7 independent success/failure trials, each with success probability p=2/3. Let X count successes, so X~B(7,2/3). At the integer threshold k=5, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=5)=C(7,5)(2/3)^5(1−2/3)^2=224/729≈0.30727. Summing masses for values strictly below 5 gives P(X<5)=313/729≈0.429355. Including the threshold adds its mass: P(X≤5)=179/243≈0.736626. The difference between these two lower-tail values is precisely the single mass at 5, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥5)=1−P(X≤4)=416/729≈0.570645 and P(X>5)=1−P(X≤5)=64/243≈0.263374. The exact sums P(X<5)+P(X≥5)=1 and P(X≤5)+P(X>5)=1 both hold. For k=n, a count more than n is impossible, and at most n is certain. At this threshold the mass added by changing a strict sign to an inclusive sign is 30.7270% of all model outcomes. Explain that difference in words before calculating another case.

24. n=8, p=1/5, threshold k=2

Try the word-to-symbol exercise. A fictional experiment consists of 8 independent success/failure trials, each with success probability p=1/5. Let X count successes, so X~B(8,1/5). At the integer threshold k=2, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=2)=C(8,2)(1/5)^2(1−1/5)^6=114688/390625≈0.293601. Summing masses for values strictly below 2 gives P(X<2)=196608/390625≈0.503316. Including the threshold adds its mass: P(X≤2)=311296/390625≈0.796918. The difference between these two lower-tail values is precisely the single mass at 2, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥2)=1−P(X≤1)=194017/390625≈0.496684 and P(X>2)=1−P(X≤2)=79329/390625≈0.203082. The exact sums P(X<2)+P(X≥2)=1 and P(X≤2)+P(X>2)=1 both hold. If the probability mass at k is large, the strict versus inclusive choice can change the answer substantially. At this threshold the mass added by changing a strict sign to an inclusive sign is 29.3601% of all model outcomes. Explain that difference in words before calculating another case.

25. n=9, p=3/4, threshold k=2

Try the word-to-symbol exercise. A fictional experiment consists of 9 independent success/failure trials, each with success probability p=3/4. Let X count successes, so X~B(9,3/4). At the integer threshold k=2, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=2)=C(9,2)(3/4)^2(1−3/4)^7=81/65536≈0.001236. Summing masses for values strictly below 2 gives P(X<2)=7/65536≈0.000107. Including the threshold adds its mass: P(X≤2)=11/8192≈0.001343. The difference between these two lower-tail values is precisely the single mass at 2, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥2)=1−P(X≤1)=65529/65536≈0.999893 and P(X>2)=1−P(X≤2)=8181/8192≈0.998657. The exact sums P(X<2)+P(X≥2)=1 and P(X≤2)+P(X>2)=1 both hold. The difference between P(X≤k) and P(X

26. n=10, p=1/4, threshold k=3

Try the word-to-symbol exercise. A fictional experiment consists of 10 independent success/failure trials, each with success probability p=1/4. Let X count successes, so X~B(10,1/4). At the integer threshold k=3, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=3)=C(10,3)(1/4)^3(1−1/4)^7=32805/131072≈0.250282. Summing masses for values strictly below 3 gives P(X<3)=137781/262144≈0.525593. Including the threshold adds its mass: P(X≤3)=203391/262144≈0.775875. The difference between these two lower-tail values is precisely the single mass at 3, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥3)=1−P(X≤2)=124363/262144≈0.474407 and P(X>3)=1−P(X≤3)=58753/262144≈0.224125. The exact sums P(X<3)+P(X≥3)=1 and P(X≤3)+P(X>3)=1 both hold. The difference between P(X≥k) and P(X>k) is also exactly P(X=k). At this threshold the mass added by changing a strict sign to an inclusive sign is 25.0282% of all model outcomes. Explain that difference in words before calculating another case.

27. n=11, p=4/5, threshold k=0

Try the word-to-symbol exercise. A fictional experiment consists of 11 independent success/failure trials, each with success probability p=4/5. Let X count successes, so X~B(11,4/5). At the integer threshold k=0, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=0)=C(11,0)(4/5)^0(1−4/5)^11=1/48828125≈0. Summing masses for values strictly below 0 gives P(X<0)=0≈0. Including the threshold adds its mass: P(X≤0)=1/48828125≈0. The difference between these two lower-tail values is precisely the single mass at 0, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥0)=1−P(X≤-1)=1≈1 and P(X>0)=1−P(X≤0)=48828124/48828125≈1. The exact sums P(X<0)+P(X≥0)=1 and P(X≤0)+P(X>0)=1 both hold. “At least” includes the threshold; “more than” excludes it. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.0000% of all model outcomes. Explain that difference in words before calculating another case.

28. n=12, p=1/3, threshold k=11

Try the word-to-symbol exercise. A fictional experiment consists of 12 independent success/failure trials, each with success probability p=1/3. Let X count successes, so X~B(12,1/3). At the integer threshold k=11, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=11)=C(12,11)(1/3)^11(1−1/3)^1=8/177147≈0.000045. Summing masses for values strictly below 11 gives P(X<11)=531416/531441≈0.999953. Including the threshold adds its mass: P(X≤11)=531440/531441≈0.999998. The difference between these two lower-tail values is precisely the single mass at 11, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥11)=1−P(X≤10)=25/531441≈0.000047 and P(X>11)=1−P(X≤11)=1/531441≈0.000002. The exact sums P(X<11)+P(X≥11)=1 and P(X≤11)+P(X>11)=1 both hold. “At most” includes the threshold; “fewer than” excludes it. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.0045% of all model outcomes. Explain that difference in words before calculating another case.

29. n=13, p=2/5, threshold k=4

Try the word-to-symbol exercise. A fictional experiment consists of 13 independent success/failure trials, each with success probability p=2/5. Let X count successes, so X~B(13,2/5). At the integer threshold k=4, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=4)=C(13,4)(2/5)^4(1−2/5)^9=45034704/244140625≈0.184462. Summing masses for values strictly below 4 gives P(X<4)=41157153/244140625≈0.16858. Including the threshold adds its mass: P(X≤4)=86191857/244140625≈0.353042. The difference between these two lower-tail values is precisely the single mass at 4, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥4)=1−P(X≤3)=202983472/244140625≈0.83142 and P(X>4)=1−P(X≤4)=157948768/244140625≈0.646958. The exact sums P(X<4)+P(X≥4)=1 and P(X≤4)+P(X>4)=1 both hold. The complement of X≥k is X≤k−1, not X≤k. At this threshold the mass added by changing a strict sign to an inclusive sign is 18.4462% of all model outcomes. Explain that difference in words before calculating another case.

30. n=14, p=1/2, threshold k=12

Try the word-to-symbol exercise. A fictional experiment consists of 14 independent success/failure trials, each with success probability p=1/2. Let X count successes, so X~B(14,1/2). At the integer threshold k=12, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=12)=C(14,12)(1/2)^12(1−1/2)^2=91/16384≈0.005554. Summing masses for values strictly below 12 gives P(X<12)=8139/8192≈0.99353. Including the threshold adds its mass: P(X≤12)=16369/16384≈0.999084. The difference between these two lower-tail values is precisely the single mass at 12, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥12)=1−P(X≤11)=53/8192≈0.00647 and P(X>12)=1−P(X≤12)=15/16384≈0.000916. The exact sums P(X<12)+P(X≥12)=1 and P(X≤12)+P(X>12)=1 both hold. The complement of X>k is X≤k, with no minus-one adjustment. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.5554% of all model outcomes. Explain that difference in words before calculating another case.

31. n=5, p=1/5, threshold k=5

Upper tails and independent checks. Using the correct complementary events gives P(X≥5)=1−P(X≤4)=1/3125≈0.00032 and P(X>5)=1−P(X≤5)=0≈0. The exact sums P(X<5)+P(X≥5)=1 and P(X≤5)+P(X>5)=1 both hold. A binomial count is integer-valued, so write the actual supported values before calculating. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.0320% of all model outcomes. Explain that difference in words before calculating another case.

32. n=6, p=3/4, threshold k=5

Try the word-to-symbol exercise. A fictional experiment consists of 6 independent success/failure trials, each with success probability p=3/4. Let X count successes, so X~B(6,3/4). At the integer threshold k=5, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=5)=C(6,5)(3/4)^5(1−3/4)^1=729/2048≈0.355957. Summing masses for values strictly below 5 gives P(X<5)=1909/4096≈0.466064. Including the threshold adds its mass: P(X≤5)=3367/4096≈0.822021. The difference between these two lower-tail values is precisely the single mass at 5, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥5)=1−P(X≤4)=2187/4096≈0.533936 and P(X>5)=1−P(X≤5)=729/4096≈0.177979. The exact sums P(X<5)+P(X≥5)=1 and P(X≤5)+P(X>5)=1 both hold. The probabilities of X≤k and X>k sum to one exactly. At this threshold the mass added by changing a strict sign to an inclusive sign is 35.5957% of all model outcomes. Explain that difference in words before calculating another case.

33. n=7, p=1/4, threshold k=5

Try the word-to-symbol exercise. A fictional experiment consists of 7 independent success/failure trials, each with success probability p=1/4. Let X count successes, so X~B(7,1/4). At the integer threshold k=5, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=5)=C(7,5)(1/4)^5(1−1/4)^2=189/16384≈0.011536. Summing masses for values strictly below 5 gives P(X<5)=16173/16384≈0.987122. Including the threshold adds its mass: P(X≤5)=8181/8192≈0.998657. The difference between these two lower-tail values is precisely the single mass at 5, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥5)=1−P(X≤4)=211/16384≈0.012878 and P(X>5)=1−P(X≤5)=11/8192≈0.001343. The exact sums P(X<5)+P(X≥5)=1 and P(X≤5)+P(X>5)=1 both hold. The probabilities of X

34. n=8, p=4/5, threshold k=2

Try the word-to-symbol exercise. A fictional experiment consists of 8 independent success/failure trials, each with success probability p=4/5. Let X count successes, so X~B(8,4/5). At the integer threshold k=2, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=2)=C(8,2)(4/5)^2(1−4/5)^6=448/390625≈0.001147. Summing masses for values strictly below 2 gives P(X<2)=33/390625≈0.000084. Including the threshold adds its mass: P(X≤2)=481/390625≈0.001231. The difference between these two lower-tail values is precisely the single mass at 2, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥2)=1−P(X≤1)=390592/390625≈0.999916 and P(X>2)=1−P(X≤2)=390144/390625≈0.998769. The exact sums P(X<2)+P(X≥2)=1 and P(X≤2)+P(X>2)=1 both hold. For k=0, every possible binomial count is at least zero. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.1147% of all model outcomes. Explain that difference in words before calculating another case.

35. n=9, p=1/3, threshold k=9

Try the word-to-symbol exercise. A fictional experiment consists of 9 independent success/failure trials, each with success probability p=1/3. Let X count successes, so X~B(9,1/3). At the integer threshold k=9, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=9)=C(9,9)(1/3)^9(1−1/3)^0=1/19683≈0.000051. Summing masses for values strictly below 9 gives P(X<9)=19682/19683≈0.999949. Including the threshold adds its mass: P(X≤9)=1≈1. The difference between these two lower-tail values is precisely the single mass at 9, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥9)=1−P(X≤8)=1/19683≈0.000051 and P(X>9)=1−P(X≤9)=0≈0. The exact sums P(X<9)+P(X≥9)=1 and P(X≤9)+P(X>9)=1 both hold. For k=n, a count more than n is impossible, and at most n is certain. At this threshold the mass added by changing a strict sign to an inclusive sign is 0.0051% of all model outcomes. Explain that difference in words before calculating another case.

36. n=10, p=2/5, threshold k=8

Try the word-to-symbol exercise. A fictional experiment consists of 10 independent success/failure trials, each with success probability p=2/5. Let X count successes, so X~B(10,2/5). At the integer threshold k=8, calculate exactly k, fewer than k, at most k, at least k and more than k. Before entering numbers, list which integer counts each phrase includes, especially if the threshold is at the edge of the support.

Exact probability-mass and lower-tail calculations. The single mass is P(X=8)=C(10,8)(2/5)^8(1−2/5)^2=20736/1953125≈0.010617. Summing masses for values strictly below 8 gives P(X<8)=9645561/9765625≈0.987705. Including the threshold adds its mass: P(X≤8)=9749241/9765625≈0.998322. The difference between these two lower-tail values is precisely the single mass at 8, not a rounding artefact.

Upper tails and independent checks. Using the correct complementary events gives P(X≥8)=1−P(X≤7)=120064/9765625≈0.012295 and P(X>8)=1−P(X≤8)=16384/9765625≈0.001678. The exact sums P(X<8)+P(X≥8)=1 and P(X≤8)+P(X>8)=1 both hold. If the probability mass at k is large, the strict versus inclusive choice can change the answer substantially. At this threshold the mass added by changing a strict sign to an inclusive sign is 1.0617% of all model outcomes. Explain that difference in words before calculating another case.

Ten short exact answers

If X~B(5,0.3), P(X=2)=0.3087.
For that model, P(X≤2)=0.83692 and P(X<2)=0.52822.
P(X≥2)=0.47178 and P(X>2)=0.16308.
P(X≥2)−P(X>2)=P(X=2)=0.3087.
For X~B(8,0.5), P(X=4)=70/256=0.2734375.
For that fair model, P(X≤4)=P(X≥4)=0.63671875.
P(X>4)=P(X<4)=0.36328125.
For X~B(8,0.5), P(2≤X≤5)=105/128.
For X~B(8,0.5), P(2
P(X≥0)=1 and P(X>n)=0 for every binomial count with n trials.

Five-day probability-event revision plan

Day one: translate the five most common English comparison phrases into the exact inequality symbols, without calculating any probabilities. Day two: apply the binomial mass formula only to “exactly k” questions, keeping track of n, p and k. Day three: compute at-most and fewer-than tails and compare their difference with one mass. Day four: use complements for at-least and more-than questions, checking the one-unit shift in the threshold. Day five: combine interval events and boundary cases k=0, k=n and non-integer thresholds, explaining why discrete support determines the valid counts.

In a three-learner tutorial, one pupil can translate the English statement into an event, another can calculate the binomial mass or tail and the third can audit the complementary event and support boundaries. Rotate roles after changing only “at least” to “more than”. The aim is to stop off-by-one errors at the event-definition stage, before they become calculator errors with plausible decimals.

Frequently asked questions

Is “at least k” the same as “more than k”?

No. At least k includes X=k; more than k excludes it. Their probabilities differ by P(X=k).

Is “at most k” the same as “fewer than k”?

No. At most k includes X=k; fewer than k stops at k−1 for an integer count.

What is the complement of X≥k?

It is X≤k−1 for integer-valued X, so P(X≥k)=1−P(X≤k−1).

What is the complement of X>k?

It is X≤k, so P(X>k)=1−P(X≤k).

What does “exactly k” require?

One binomial probability mass, C(n,k)pᵏ(1−p)ⁿ⁻ᵏ, without adding outcomes at other counts.

How do I handle k=0 or k=n?

Use the support 0≤X≤n. At least zero is certain, more than n impossible, and exactly n has probability pⁿ.

Should I approximate binomial with normal?

No for this 2027 H2 syllabus task. Normal approximation to the binomial distribution is expressly excluded, and direct binomial probabilities are available.

Is computing P(X≤k) the same as deriving an entire CDF?

No. This guide calculates specific requested binomial event probabilities, not a general discrete cumulative distribution function. The latter is excluded from the current syllabus.

Related JC2 H2 guides and official syllabus

Graphing f(|x|) Versus |f(x)|
Foot of a Perpendicular From a Point to a 3D Line
E(aX+b) and Var(aX+b): Why the Square Matters
Binomial Exactly k, At Most k or More Than k?: current guide

For the simpler zero-success complement, read Binomial At Least One. For recovering the model itself, see Recovering n and p from Mean and Variance. The JC2 Bukit Timah tuition hub provides further learning context, and the immutable Clementi small-group mathematics reference illustrates the teaching format at another level.

Official syllabus: SEAB 2027 H2 Mathematics 9758, §6.2 Discrete Random Variables. All distributions and events are original fictional teaching models, not observed student outcomes or official past-year examination questions. Binomial probabilities are computed directly without a normal approximation.

JC2 H2 Mathematics Tutor Bukit Timah | Binomial Exactly k, At Most k or More Than k?