Did you know? Two draws from the same bag can each have a 40% unconditional chance of being red, yet the colour of the first draw changes the probability of red on the second. If there are four red and six blue counters, the chance of red on the second draw is 4/10 before anything is revealed. But if the first was red, the chance becomes 3/9; if the first was blue, it becomes 4/9. Equal unconditional probabilities do not make the draws independent.
This JC2 H2 Mathematics Tutor Bukit Timah article answers a targeted A-Level H2 Mathematics 9758 question: how do you calculate sequential and conditional probabilities when sampling is without replacement, and why does the denominator change? We begin with a small urn, show the two-stage tree, verify ordered and unordered counting, and then work through 36 original cases. The maths is the same whether the objects are counters, cards or fictional test items.
The official SEAB 2027 H2 Mathematics (9758) syllabus includes conditional probabilities, multiplication and addition of probabilities, and tables, trees and counting techniques in §6.1. This lesson uses those tools without importing the binomial model where trial independence is absent. The exercises are invented for learning and do not represent real school data or actual examination questions.
Quick answer: the second draw has a smaller bag
Suppose a bag contains r red and b blue counters, and two are drawn in order without replacement. Let n=r+b. The first red probability is r/n. If the first counter was red, only r−1 red counters remain out of n−1, so P(R₂|R₁)=(r−1)/(n−1). If the first counter was blue, all r red counters remain out of n−1, so P(R₂|B₁)=r/(n−1).
- Two red in order: P(R₁∩R₂)=(r/n)((r−1)/(n−1)).
- Red then blue: P(R₁∩B₂)=(r/n)(b/(n−1)).
- Blue then red: P(B₁∩R₂)=(b/n)(r/(n−1)).
- Exactly one red: add the two orders, giving 2rb/[n(n−1)].
- At least one red: subtract the all-blue event, giving 1−b(b−1)/[n(n−1)].
- Red on second draw alone: P(R₂)=r/n, even though the conditional chances depend on the first colour.
Full worked example: four red, six blue
There are four red and six blue counters, ten in all. Draw two without replacement. The probability that both are red is (4/10)(3/9)=12/90=2/15. The key is that there are only three red counters among nine remaining after a first red. Using (4/10)² would describe independent draws with replacement, not this experiment.
For red then blue, the probability is (4/10)(6/9)=24/90=4/15. For blue then red, it is (6/10)(4/9)=24/90=4/15. Therefore exactly one red in two draws has probability 8/15. The two orders are mutually exclusive, so their probabilities are added. Do not multiply them together; they represent different paths through the sample space.
At least one red is the complement of two blue. The probability of two blue is (6/10)(5/9)=30/90=1/3. Hence P(at least one red)=2/3. Check by adding the disjoint outcomes: two red (2/15) and exactly one red (8/15) total 10/15=2/3. Both paths to the answer agree, giving an independent arithmetic audit.
The conditional denominator is not always ten
Suppose someone reveals that the first counter was red. Your sample space for the second draw is now nine remaining counters, three of them red. The relevant denominator is nine, not ten. This is conditional probability in everyday form: information about the first outcome changes which objects remain possible on the second.
If the first counter was blue, the second draw still sees nine counters, but all four reds remain. Thus P(R₂|B₁)=4/9 rather than 3/9. The first event changes the composition of the bag. A learner should describe the updated inventory in words before writing a conditional fraction; that habit prevents a correct-looking but wrong use of r/n.
Why the second draw is red with probability 4/10 before any information
It may seem surprising that the unconditional second-draw probability remains 4/10. Use the two possible first-draw paths. Either first red then second red, probability (4/10)(3/9), or first blue then second red, probability (6/10)(4/9). Add them: 12/90+24/90=36/90=4/10. The first-draw paths partition the event “second is red”.
The equal marginals P(R₁)=P(R₂)=4/10 arise from the symmetry of the two random draw positions, not independence. In fact, the conditional probability P(R₂|R₁)=3/9 differs from P(R₂)=4/10. A student who uses equality of these two unconditional probabilities to declare independence has confused two separate concepts.
Why dependence matters for the binomial model
A standard binomial model counts successes over a fixed number of independent trials with a constant success chance. Without-replacement draws from a finite bag are not independent in the ordinary two-colour experiment: the chance of red changes after each draw. Therefore the count of reds in two such draws should not simply be labelled B(2,4/10).
With replacement, the first counter goes back into the bag before the second draw, restoring four red and six blue each time. Then the draws are independent and the probability of two reds is (4/10)²=0.16, compared with 2/15≈0.1333 without replacement. The comparison explains precisely which modelling assumption changed. For a larger population sampled in a particular way, dependence may sometimes be small, but an exact H2 question should use the stated procedure, not an approximation by habit.
A counting approach gives the same two-red answer
If the order of two selected counters is irrelevant, the probability of obtaining two reds can be calculated by combinations: choose two reds from the four available, divided by choosing two counters from ten. This is C(4,2)/C(10,2)=6/45=2/15. The value matches the sequential calculation. The counting method describes unordered two-item selections; the tree method describes ordered selections, but the event “both red” is the same.
The agreement is a good independent check, not an invitation to mix incompatible sample spaces midway through one expression. If you count unordered pairs in the numerator, use unordered pairs in the denominator. If you count sequential ordered paths, use consistent ordered path probabilities. Different counting strategies can lead to the same event probability when used coherently.
What if the first draw is blue?
In the ten-counter example, the probability that the first draw is blue and the second is red is (6/10)(4/9)=4/15. But if the question says “given the first draw was blue”, the conditional probability is 4/9, not 4/15. The difference is that 4/15 describes a joint event from the original ten-counter experiment, while 4/9 describes a probability within the already-known blue-first group.
A useful language check is whether the question says “and” or “given”. “Blue and then red” describes a path, so multiply the first probability by the conditional second probability. “Red second given blue first” begins after the first outcome has been revealed, so use only the updated conditional fraction. This distinction applies beyond urns to cards, inventory selection and many tree-diagram problems.
A two-way table of all ordered colour patterns
| First draw | Second draw | Probability, r=4, b=6 |
|---|---|---|
| Red | Red | (4/10)(3/9)=2/15 |
| Red | Blue | (4/10)(6/9)=4/15 |
| Blue | Red | (6/10)(4/9)=4/15 |
| Blue | Blue | (6/10)(5/9)=1/3 |
All four paths are mutually exclusive and exhaustive. Their probabilities sum to 2/15+4/15+4/15+1/3=1. Use the sum-to-one audit whenever a probability tree has a small number of possible outcomes. If it does not sum to one, at least one branch probability or a denominator has been misplaced.
A one-minute tutoring diagnostic
Ask a learner to write P(R₂|R₁) and P(R₂|B₁) for four red and six blue without replacement. The answers are 3/9 and 4/9. Then ask for P(R₂) before any first-draw information: 4/10. Finally ask whether the two draws are independent. They are not, because P(R₂|R₁)≠P(R₂). Each question tests a distinct mathematical statement.
If the student instead gives 4/10 for both conditionals, they have ignored the updated inventory. If they give 2/15 for P(R₂|R₁), they have mixed a joint event with a conditional one. If they call the trials independent merely because P(R₁)=P(R₂), they have confused equal marginals with independence. Each error can be repaired with a brief changed-case drawing rather than repeating an entire textbook.
Thirty-six original sequential-draw clinics
The next 36 cases vary the number of red and blue counters. Each assumes two ordered draws without replacement. The exact fractions are calculated using the evolving contents of the bag, and the tree probabilities are checked to sum to one. Cover the worked lines, state the updated numbers in words, and calculate the probability before viewing the answer. These are fictional mathematical experiments rather than collected data.
1. 2 red and 5 blue: update after the first draw
Problem. A fictional bag holds 2 red and 5 blue counters, n=7. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 2/7.
Sequential calculation. After a red first draw, 1 reds remain out of 6 counters. Therefore P(R₂|R₁)=1/6, and the two-red path is (2/7)((1)/(6))=1/21. The red-blue path is 5/21, and the blue-red path is 5/21. Adding those last two disjoint paths gives P(exactly one red)=10/21. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 10/21, so at least one red has probability 11/21. Unconditionally the second draw is red with probability 2/7, while after a red first draw it is 1/6; those differ when both colours are present. The second draw is conditional on the first when the objects are not replaced. The four ordered path probabilities sum to one: 2+10+10+20=42 over a common denominator 42.
2. 3 red and 10 blue: update after the first draw
Problem. A fictional bag holds 3 red and 10 blue counters, n=13. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 3/13.
Sequential calculation. After a red first draw, 2 reds remain out of 12 counters. Therefore P(R₂|R₁)=1/6, and the two-red path is (3/13)((2)/(12))=1/26. The red-blue path is 5/26, and the blue-red path is 5/26. Adding those last two disjoint paths gives P(exactly one red)=5/13. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 15/26, so at least one red has probability 11/26. Unconditionally the second draw is red with probability 3/13, while after a red first draw it is 1/6; those differ when both colours are present. A smaller denominator is not enough; the numerator must reflect what was removed as well. The four ordered path probabilities sum to one: 6+30+30+90=156 over a common denominator 156.
3. 4 red and 4 blue: update after the first draw
Problem. A fictional bag holds 4 red and 4 blue counters, n=8. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 4/8.
Sequential calculation. After a red first draw, 3 reds remain out of 7 counters. Therefore P(R₂|R₁)=3/7, and the two-red path is (4/8)((3)/(7))=3/14. The red-blue path is 2/7, and the blue-red path is 2/7. Adding those last two disjoint paths gives P(exactly one red)=4/7. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 3/14, so at least one red has probability 11/14. Unconditionally the second draw is red with probability 1/2, while after a red first draw it is 3/7; those differ when both colours are present. Two different ordered paths may have the same joint probability even though their conditions differ. The four ordered path probabilities sum to one: 12+16+16+12=56 over a common denominator 56.
4. 5 red and 9 blue: update after the first draw
Problem. A fictional bag holds 5 red and 9 blue counters, n=14. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 5/14.
Sequential calculation. After a red first draw, 4 reds remain out of 13 counters. Therefore P(R₂|R₁)=4/13, and the two-red path is (5/14)((4)/(13))=10/91. The red-blue path is 45/182, and the blue-red path is 45/182. Adding those last two disjoint paths gives P(exactly one red)=45/91. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 36/91, so at least one red has probability 55/91. Unconditionally the second draw is red with probability 5/14, while after a red first draw it is 4/13; those differ when both colours are present. Exactly one red means red-then-blue or blue-then-red; add those disjoint paths. The four ordered path probabilities sum to one: 20+45+45+72=182 over a common denominator 182.
5. 6 red and 3 blue: update after the first draw
Problem. A fictional bag holds 6 red and 3 blue counters, n=9. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 6/9.
Sequential calculation. After a red first draw, 5 reds remain out of 8 counters. Therefore P(R₂|R₁)=5/8, and the two-red path is (6/9)((5)/(8))=5/12. The red-blue path is 1/4, and the blue-red path is 1/4. Adding those last two disjoint paths gives P(exactly one red)=1/2. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 1/12, so at least one red has probability 11/12. Unconditionally the second draw is red with probability 2/3, while after a red first draw it is 5/8; those differ when both colours are present. At least one red is often easier by subtracting the two-blue probability. The four ordered path probabilities sum to one: 30+18+18+6=72 over a common denominator 72.
6. 7 red and 8 blue: update after the first draw
Problem. A fictional bag holds 7 red and 8 blue counters, n=15. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 7/15.
Sequential calculation. After a red first draw, 6 reds remain out of 14 counters. Therefore P(R₂|R₁)=3/7, and the two-red path is (7/15)((6)/(14))=1/5. The red-blue path is 4/15, and the blue-red path is 4/15. Adding those last two disjoint paths gives P(exactly one red)=8/15. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 4/15, so at least one red has probability 11/15. Unconditionally the second draw is red with probability 7/15, while after a red first draw it is 3/7; those differ when both colours are present. Equal first- and second-position marginal probabilities do not imply independence. The four ordered path probabilities sum to one: 42+56+56+56=210 over a common denominator 210.
7. 8 red and 2 blue: update after the first draw
Problem. A fictional bag holds 8 red and 2 blue counters, n=10. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 8/10.
Sequential calculation. After a red first draw, 7 reds remain out of 9 counters. Therefore P(R₂|R₁)=7/9, and the two-red path is (8/10)((7)/(9))=28/45. The red-blue path is 8/45, and the blue-red path is 8/45. Adding those last two disjoint paths gives P(exactly one red)=16/45. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 1/45, so at least one red has probability 44/45. Unconditionally the second draw is red with probability 4/5, while after a red first draw it is 7/9; those differ when both colours are present. The combination check C(r,2)/C(n,2) should agree with the ordered two-red result. The four ordered path probabilities sum to one: 56+16+16+2=90 over a common denominator 90.
8. 9 red and 7 blue: update after the first draw
Problem. A fictional bag holds 9 red and 7 blue counters, n=16. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 9/16.
Sequential calculation. After a red first draw, 8 reds remain out of 15 counters. Therefore P(R₂|R₁)=8/15, and the two-red path is (9/16)((8)/(15))=3/10. The red-blue path is 21/80, and the blue-red path is 21/80. Adding those last two disjoint paths gives P(exactly one red)=21/40. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 7/40, so at least one red has probability 33/40. Unconditionally the second draw is red with probability 9/16, while after a red first draw it is 8/15; those differ when both colours are present. With replacement, the conditional red chance would stay r/n after a red first draw. The four ordered path probabilities sum to one: 72+63+63+42=240 over a common denominator 240.
9. 10 red and 12 blue: update after the first draw
Problem. A fictional bag holds 10 red and 12 blue counters, n=22. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 10/22.
Sequential calculation. After a red first draw, 9 reds remain out of 21 counters. Therefore P(R₂|R₁)=3/7, and the two-red path is (10/22)((9)/(21))=15/77. The red-blue path is 20/77, and the blue-red path is 20/77. Adding those last two disjoint paths gives P(exactly one red)=40/77. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 2/7, so at least one red has probability 5/7. Unconditionally the second draw is red with probability 5/11, while after a red first draw it is 3/7; those differ when both colours are present. Always state n=r+b before the first draw and n−1 after one counter has been removed. The four ordered path probabilities sum to one: 90+120+120+132=462 over a common denominator 462.
10. 2 red and 6 blue: update after the first draw
Problem. A fictional bag holds 2 red and 6 blue counters, n=8. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 2/8.
Sequential calculation. After a red first draw, 1 reds remain out of 7 counters. Therefore P(R₂|R₁)=1/7, and the two-red path is (2/8)((1)/(7))=1/28. The red-blue path is 3/14, and the blue-red path is 3/14. Adding those last two disjoint paths gives P(exactly one red)=3/7. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 15/28, so at least one red has probability 13/28. Unconditionally the second draw is red with probability 1/4, while after a red first draw it is 1/7; those differ when both colours are present. If a computed probability exceeds one, examine whether you have divided or multiplied the wrong events. The four ordered path probabilities sum to one: 2+12+12+30=56 over a common denominator 56.
11. 3 red and 11 blue: update after the first draw
Problem. A fictional bag holds 3 red and 11 blue counters, n=14. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 3/14.
Sequential calculation. After a red first draw, 2 reds remain out of 13 counters. Therefore P(R₂|R₁)=2/13, and the two-red path is (3/14)((2)/(13))=3/91. The red-blue path is 33/182, and the blue-red path is 33/182. Adding those last two disjoint paths gives P(exactly one red)=33/91. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 55/91, so at least one red has probability 36/91. Unconditionally the second draw is red with probability 3/14, while after a red first draw it is 2/13; those differ when both colours are present. The second draw is conditional on the first when the objects are not replaced. The four ordered path probabilities sum to one: 6+33+33+110=182 over a common denominator 182.
12. 4 red and 5 blue: update after the first draw
Problem. A fictional bag holds 4 red and 5 blue counters, n=9. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 4/9.
Sequential calculation. After a red first draw, 3 reds remain out of 8 counters. Therefore P(R₂|R₁)=3/8, and the two-red path is (4/9)((3)/(8))=1/6. The red-blue path is 5/18, and the blue-red path is 5/18. Adding those last two disjoint paths gives P(exactly one red)=5/9. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 5/18, so at least one red has probability 13/18. Unconditionally the second draw is red with probability 4/9, while after a red first draw it is 3/8; those differ when both colours are present. A smaller denominator is not enough; the numerator must reflect what was removed as well. The four ordered path probabilities sum to one: 12+20+20+20=72 over a common denominator 72.
13. 5 red and 10 blue: update after the first draw
Problem. A fictional bag holds 5 red and 10 blue counters, n=15. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 5/15.
Sequential calculation. After a red first draw, 4 reds remain out of 14 counters. Therefore P(R₂|R₁)=2/7, and the two-red path is (5/15)((4)/(14))=2/21. The red-blue path is 5/21, and the blue-red path is 5/21. Adding those last two disjoint paths gives P(exactly one red)=10/21. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 3/7, so at least one red has probability 4/7. Unconditionally the second draw is red with probability 1/3, while after a red first draw it is 2/7; those differ when both colours are present. Two different ordered paths may have the same joint probability even though their conditions differ. The four ordered path probabilities sum to one: 20+50+50+90=210 over a common denominator 210.
14. 6 red and 4 blue: update after the first draw
Problem. A fictional bag holds 6 red and 4 blue counters, n=10. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 6/10.
Sequential calculation. After a red first draw, 5 reds remain out of 9 counters. Therefore P(R₂|R₁)=5/9, and the two-red path is (6/10)((5)/(9))=1/3. The red-blue path is 4/15, and the blue-red path is 4/15. Adding those last two disjoint paths gives P(exactly one red)=8/15. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 2/15, so at least one red has probability 13/15. Unconditionally the second draw is red with probability 3/5, while after a red first draw it is 5/9; those differ when both colours are present. Exactly one red means red-then-blue or blue-then-red; add those disjoint paths. The four ordered path probabilities sum to one: 30+24+24+12=90 over a common denominator 90.
15. 7 red and 9 blue: update after the first draw
Problem. A fictional bag holds 7 red and 9 blue counters, n=16. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 7/16.
Sequential calculation. After a red first draw, 6 reds remain out of 15 counters. Therefore P(R₂|R₁)=2/5, and the two-red path is (7/16)((6)/(15))=7/40. The red-blue path is 21/80, and the blue-red path is 21/80. Adding those last two disjoint paths gives P(exactly one red)=21/40. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 3/10, so at least one red has probability 7/10. Unconditionally the second draw is red with probability 7/16, while after a red first draw it is 2/5; those differ when both colours are present. At least one red is often easier by subtracting the two-blue probability. The four ordered path probabilities sum to one: 42+63+63+72=240 over a common denominator 240.
16. 8 red and 3 blue: update after the first draw
Problem. A fictional bag holds 8 red and 3 blue counters, n=11. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 8/11.
Sequential calculation. After a red first draw, 7 reds remain out of 10 counters. Therefore P(R₂|R₁)=7/10, and the two-red path is (8/11)((7)/(10))=28/55. The red-blue path is 12/55, and the blue-red path is 12/55. Adding those last two disjoint paths gives P(exactly one red)=24/55. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 3/55, so at least one red has probability 52/55. Unconditionally the second draw is red with probability 8/11, while after a red first draw it is 7/10; those differ when both colours are present. Equal first- and second-position marginal probabilities do not imply independence. The four ordered path probabilities sum to one: 56+24+24+6=110 over a common denominator 110.
17. 9 red and 8 blue: update after the first draw
Problem. A fictional bag holds 9 red and 8 blue counters, n=17. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 9/17.
Sequential calculation. After a red first draw, 8 reds remain out of 16 counters. Therefore P(R₂|R₁)=1/2, and the two-red path is (9/17)((8)/(16))=9/34. The red-blue path is 9/34, and the blue-red path is 9/34. Adding those last two disjoint paths gives P(exactly one red)=9/17. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 7/34, so at least one red has probability 27/34. Unconditionally the second draw is red with probability 9/17, while after a red first draw it is 1/2; those differ when both colours are present. The combination check C(r,2)/C(n,2) should agree with the ordered two-red result. The four ordered path probabilities sum to one: 72+72+72+56=272 over a common denominator 272.
18. 10 red and 2 blue: update after the first draw
Problem. A fictional bag holds 10 red and 2 blue counters, n=12. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 10/12.
Sequential calculation. After a red first draw, 9 reds remain out of 11 counters. Therefore P(R₂|R₁)=9/11, and the two-red path is (10/12)((9)/(11))=15/22. The red-blue path is 5/33, and the blue-red path is 5/33. Adding those last two disjoint paths gives P(exactly one red)=10/33. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 1/66, so at least one red has probability 65/66. Unconditionally the second draw is red with probability 5/6, while after a red first draw it is 9/11; those differ when both colours are present. With replacement, the conditional red chance would stay r/n after a red first draw. The four ordered path probabilities sum to one: 90+20+20+2=132 over a common denominator 132.
19. 2 red and 7 blue: update after the first draw
Problem. A fictional bag holds 2 red and 7 blue counters, n=9. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 2/9.
Sequential calculation. After a red first draw, 1 reds remain out of 8 counters. Therefore P(R₂|R₁)=1/8, and the two-red path is (2/9)((1)/(8))=1/36. The red-blue path is 7/36, and the blue-red path is 7/36. Adding those last two disjoint paths gives P(exactly one red)=7/18. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 7/12, so at least one red has probability 5/12. Unconditionally the second draw is red with probability 2/9, while after a red first draw it is 1/8; those differ when both colours are present. Always state n=r+b before the first draw and n−1 after one counter has been removed. The four ordered path probabilities sum to one: 2+14+14+42=72 over a common denominator 72.
20. 3 red and 12 blue: update after the first draw
Problem. A fictional bag holds 3 red and 12 blue counters, n=15. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 3/15.
Sequential calculation. After a red first draw, 2 reds remain out of 14 counters. Therefore P(R₂|R₁)=1/7, and the two-red path is (3/15)((2)/(14))=1/35. The red-blue path is 6/35, and the blue-red path is 6/35. Adding those last two disjoint paths gives P(exactly one red)=12/35. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 22/35, so at least one red has probability 13/35. Unconditionally the second draw is red with probability 1/5, while after a red first draw it is 1/7; those differ when both colours are present. If a computed probability exceeds one, examine whether you have divided or multiplied the wrong events. The four ordered path probabilities sum to one: 6+36+36+132=210 over a common denominator 210.
21. 4 red and 6 blue: update after the first draw
Problem. A fictional bag holds 4 red and 6 blue counters, n=10. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 4/10.
Sequential calculation. After a red first draw, 3 reds remain out of 9 counters. Therefore P(R₂|R₁)=1/3, and the two-red path is (4/10)((3)/(9))=2/15. The red-blue path is 4/15, and the blue-red path is 4/15. Adding those last two disjoint paths gives P(exactly one red)=8/15. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 1/3, so at least one red has probability 2/3. Unconditionally the second draw is red with probability 2/5, while after a red first draw it is 1/3; those differ when both colours are present. The second draw is conditional on the first when the objects are not replaced. The four ordered path probabilities sum to one: 12+24+24+30=90 over a common denominator 90.
22. 5 red and 11 blue: update after the first draw
Problem. A fictional bag holds 5 red and 11 blue counters, n=16. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 5/16.
Sequential calculation. After a red first draw, 4 reds remain out of 15 counters. Therefore P(R₂|R₁)=4/15, and the two-red path is (5/16)((4)/(15))=1/12. The red-blue path is 11/48, and the blue-red path is 11/48. Adding those last two disjoint paths gives P(exactly one red)=11/24. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 11/24, so at least one red has probability 13/24. Unconditionally the second draw is red with probability 5/16, while after a red first draw it is 4/15; those differ when both colours are present. A smaller denominator is not enough; the numerator must reflect what was removed as well. The four ordered path probabilities sum to one: 20+55+55+110=240 over a common denominator 240.
23. 6 red and 5 blue: update after the first draw
Problem. A fictional bag holds 6 red and 5 blue counters, n=11. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 6/11.
Sequential calculation. After a red first draw, 5 reds remain out of 10 counters. Therefore P(R₂|R₁)=1/2, and the two-red path is (6/11)((5)/(10))=3/11. The red-blue path is 3/11, and the blue-red path is 3/11. Adding those last two disjoint paths gives P(exactly one red)=6/11. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 2/11, so at least one red has probability 9/11. Unconditionally the second draw is red with probability 6/11, while after a red first draw it is 1/2; those differ when both colours are present. Two different ordered paths may have the same joint probability even though their conditions differ. The four ordered path probabilities sum to one: 30+30+30+20=110 over a common denominator 110.
24. 7 red and 10 blue: update after the first draw
Problem. A fictional bag holds 7 red and 10 blue counters, n=17. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 7/17.
Sequential calculation. After a red first draw, 6 reds remain out of 16 counters. Therefore P(R₂|R₁)=3/8, and the two-red path is (7/17)((6)/(16))=21/136. The red-blue path is 35/136, and the blue-red path is 35/136. Adding those last two disjoint paths gives P(exactly one red)=35/68. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 45/136, so at least one red has probability 91/136. Unconditionally the second draw is red with probability 7/17, while after a red first draw it is 3/8; those differ when both colours are present. Exactly one red means red-then-blue or blue-then-red; add those disjoint paths. The four ordered path probabilities sum to one: 42+70+70+90=272 over a common denominator 272.
25. 8 red and 4 blue: update after the first draw
Problem. A fictional bag holds 8 red and 4 blue counters, n=12. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 8/12.
Sequential calculation. After a red first draw, 7 reds remain out of 11 counters. Therefore P(R₂|R₁)=7/11, and the two-red path is (8/12)((7)/(11))=14/33. The red-blue path is 8/33, and the blue-red path is 8/33. Adding those last two disjoint paths gives P(exactly one red)=16/33. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 1/11, so at least one red has probability 10/11. Unconditionally the second draw is red with probability 2/3, while after a red first draw it is 7/11; those differ when both colours are present. At least one red is often easier by subtracting the two-blue probability. The four ordered path probabilities sum to one: 56+32+32+12=132 over a common denominator 132.
26. 9 red and 9 blue: update after the first draw
Problem. A fictional bag holds 9 red and 9 blue counters, n=18. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 9/18.
Sequential calculation. After a red first draw, 8 reds remain out of 17 counters. Therefore P(R₂|R₁)=8/17, and the two-red path is (9/18)((8)/(17))=4/17. The red-blue path is 9/34, and the blue-red path is 9/34. Adding those last two disjoint paths gives P(exactly one red)=9/17. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 4/17, so at least one red has probability 13/17. Unconditionally the second draw is red with probability 1/2, while after a red first draw it is 8/17; those differ when both colours are present. Equal first- and second-position marginal probabilities do not imply independence. The four ordered path probabilities sum to one: 72+81+81+72=306 over a common denominator 306.
27. 10 red and 3 blue: update after the first draw
Problem. A fictional bag holds 10 red and 3 blue counters, n=13. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 10/13.
Sequential calculation. After a red first draw, 9 reds remain out of 12 counters. Therefore P(R₂|R₁)=3/4, and the two-red path is (10/13)((9)/(12))=15/26. The red-blue path is 5/26, and the blue-red path is 5/26. Adding those last two disjoint paths gives P(exactly one red)=5/13. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 1/26, so at least one red has probability 25/26. Unconditionally the second draw is red with probability 10/13, while after a red first draw it is 3/4; those differ when both colours are present. The combination check C(r,2)/C(n,2) should agree with the ordered two-red result. The four ordered path probabilities sum to one: 90+30+30+6=156 over a common denominator 156.
28. 2 red and 8 blue: update after the first draw
Problem. A fictional bag holds 2 red and 8 blue counters, n=10. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 2/10.
Sequential calculation. After a red first draw, 1 reds remain out of 9 counters. Therefore P(R₂|R₁)=1/9, and the two-red path is (2/10)((1)/(9))=1/45. The red-blue path is 8/45, and the blue-red path is 8/45. Adding those last two disjoint paths gives P(exactly one red)=16/45. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 28/45, so at least one red has probability 17/45. Unconditionally the second draw is red with probability 1/5, while after a red first draw it is 1/9; those differ when both colours are present. With replacement, the conditional red chance would stay r/n after a red first draw. The four ordered path probabilities sum to one: 2+16+16+56=90 over a common denominator 90.
29. 3 red and 2 blue: update after the first draw
Problem. A fictional bag holds 3 red and 2 blue counters, n=5. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 3/5.
Sequential calculation. After a red first draw, 2 reds remain out of 4 counters. Therefore P(R₂|R₁)=1/2, and the two-red path is (3/5)((2)/(4))=3/10. The red-blue path is 3/10, and the blue-red path is 3/10. Adding those last two disjoint paths gives P(exactly one red)=3/5. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 1/10, so at least one red has probability 9/10. Unconditionally the second draw is red with probability 3/5, while after a red first draw it is 1/2; those differ when both colours are present. Always state n=r+b before the first draw and n−1 after one counter has been removed. The four ordered path probabilities sum to one: 6+6+6+2=20 over a common denominator 20.
30. 4 red and 7 blue: update after the first draw
Problem. A fictional bag holds 4 red and 7 blue counters, n=11. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 4/11.
Sequential calculation. After a red first draw, 3 reds remain out of 10 counters. Therefore P(R₂|R₁)=3/10, and the two-red path is (4/11)((3)/(10))=6/55. The red-blue path is 14/55, and the blue-red path is 14/55. Adding those last two disjoint paths gives P(exactly one red)=28/55. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 21/55, so at least one red has probability 34/55. Unconditionally the second draw is red with probability 4/11, while after a red first draw it is 3/10; those differ when both colours are present. If a computed probability exceeds one, examine whether you have divided or multiplied the wrong events. The four ordered path probabilities sum to one: 12+28+28+42=110 over a common denominator 110.
31. 5 red and 12 blue: update after the first draw
Problem. A fictional bag holds 5 red and 12 blue counters, n=17. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 5/17.
Sequential calculation. After a red first draw, 4 reds remain out of 16 counters. Therefore P(R₂|R₁)=1/4, and the two-red path is (5/17)((4)/(16))=5/68. The red-blue path is 15/68, and the blue-red path is 15/68. Adding those last two disjoint paths gives P(exactly one red)=15/34. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 33/68, so at least one red has probability 35/68. Unconditionally the second draw is red with probability 5/17, while after a red first draw it is 1/4; those differ when both colours are present. The second draw is conditional on the first when the objects are not replaced. The four ordered path probabilities sum to one: 20+60+60+132=272 over a common denominator 272.
32. 6 red and 6 blue: update after the first draw
Problem. A fictional bag holds 6 red and 6 blue counters, n=12. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 6/12.
Sequential calculation. After a red first draw, 5 reds remain out of 11 counters. Therefore P(R₂|R₁)=5/11, and the two-red path is (6/12)((5)/(11))=5/22. The red-blue path is 3/11, and the blue-red path is 3/11. Adding those last two disjoint paths gives P(exactly one red)=6/11. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 5/22, so at least one red has probability 17/22. Unconditionally the second draw is red with probability 1/2, while after a red first draw it is 5/11; those differ when both colours are present. A smaller denominator is not enough; the numerator must reflect what was removed as well. The four ordered path probabilities sum to one: 30+36+36+30=132 over a common denominator 132.
33. 7 red and 11 blue: update after the first draw
Problem. A fictional bag holds 7 red and 11 blue counters, n=18. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 7/18.
Sequential calculation. After a red first draw, 6 reds remain out of 17 counters. Therefore P(R₂|R₁)=6/17, and the two-red path is (7/18)((6)/(17))=7/51. The red-blue path is 77/306, and the blue-red path is 77/306. Adding those last two disjoint paths gives P(exactly one red)=77/153. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 55/153, so at least one red has probability 98/153. Unconditionally the second draw is red with probability 7/18, while after a red first draw it is 6/17; those differ when both colours are present. Two different ordered paths may have the same joint probability even though their conditions differ. The four ordered path probabilities sum to one: 42+77+77+110=306 over a common denominator 306.
34. 8 red and 5 blue: update after the first draw
Problem. A fictional bag holds 8 red and 5 blue counters, n=13. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 8/13.
Sequential calculation. After a red first draw, 7 reds remain out of 12 counters. Therefore P(R₂|R₁)=7/12, and the two-red path is (8/13)((7)/(12))=14/39. The red-blue path is 10/39, and the blue-red path is 10/39. Adding those last two disjoint paths gives P(exactly one red)=20/39. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 5/39, so at least one red has probability 34/39. Unconditionally the second draw is red with probability 8/13, while after a red first draw it is 7/12; those differ when both colours are present. Exactly one red means red-then-blue or blue-then-red; add those disjoint paths. The four ordered path probabilities sum to one: 56+40+40+20=156 over a common denominator 156.
35. 9 red and 10 blue: update after the first draw
Problem. A fictional bag holds 9 red and 10 blue counters, n=19. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 9/19.
Sequential calculation. After a red first draw, 8 reds remain out of 18 counters. Therefore P(R₂|R₁)=4/9, and the two-red path is (9/19)((8)/(18))=4/19. The red-blue path is 5/19, and the blue-red path is 5/19. Adding those last two disjoint paths gives P(exactly one red)=10/19. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 5/19, so at least one red has probability 14/19. Unconditionally the second draw is red with probability 9/19, while after a red first draw it is 4/9; those differ when both colours are present. At least one red is often easier by subtracting the two-blue probability. The four ordered path probabilities sum to one: 72+90+90+90=342 over a common denominator 342.
36. 10 red and 4 blue: update after the first draw
Problem. A fictional bag holds 10 red and 4 blue counters, n=14. Two counters are selected one after another without replacement. Find the probability of both being red, the probability of exactly one red, and the conditional probability of a red second given a red first. Explain why the second fraction is not simply 10/14.
Sequential calculation. After a red first draw, 9 reds remain out of 13 counters. Therefore P(R₂|R₁)=9/13, and the two-red path is (10/14)((9)/(13))=45/91. The red-blue path is 20/91, and the blue-red path is 20/91. Adding those last two disjoint paths gives P(exactly one red)=40/91. Each numerator reflects the objects that remain after the first outcome.
Complement and independence audit. Two blue counters have probability 6/91, so at least one red has probability 85/91. Unconditionally the second draw is red with probability 5/7, while after a red first draw it is 9/13; those differ when both colours are present. Equal first- and second-position marginal probabilities do not imply independence. The four ordered path probabilities sum to one: 90+40+40+12=182 over a common denominator 182.
Eight quick checks with answers
- Four red and six blue, two draws without replacement: P(RR)=2/15.
- P(RB)=4/15 and P(BR)=4/15.
- P(exactly one red)=8/15.
- P(at least one red)=2/3.
- P(R₂|R₁)=3/9=1/3.
- P(R₂|B₁)=4/9.
- P(R₂)=4/10=2/5 before knowing the first colour.
- With replacement, two red would instead have probability (4/10)²=4/25.
A five-day learning plan
Day one: draw a two-stage tree and update the bag after each branch. Day two: practise joint events such as red-then-blue, explicitly writing the conditional second fraction. Day three: calculate exactly one red and at least one red using disjoint paths and complements. Day four: verify two-red probabilities by combinations and reconcile ordered with unordered counting. Day five: compare the same experiment with and without replacement, and explain which procedure can support independent trials.
In a three-student tutorial, one learner can draw the tree, another can calculate conditional fractions and a third can check that all path probabilities sum to one. Rotate responsibilities and change the starting inventory. The desired outcome is reasoning about the changed sample space rather than relying on a remembered multiplication pattern.
Frequently asked questions
Why does the denominator change?
The first counter is not returned, so the second draw occurs from n−1 objects rather than n.
Are the two draws independent?
Not in a mixed finite bag without replacement; the first colour changes the probability of the second colour.
Why can P(R₁) equal P(R₂) without independence?
The positions are symmetric, so they have equal marginal red probabilities, but their conditional relationship still reflects the changed bag.
Does a binomial model apply to two draws without replacement?
Not exactly in the ordinary mixed-bag experiment because the trial outcomes are dependent and the success chance changes.
Can combinations and a probability tree agree?
Yes. They describe the same event using unordered selections or ordered sequences, provided each method is applied consistently.
Does “given first red” require multiplying by P(first red)?
No. The conditional P(second red | first red) is simply the updated fraction. Multiply by P(first red) only when you need the joint event of both reds.
Is this assessed in H2 Mathematics 9758?
Yes. It uses basic conditional probability, tree diagrams, addition and multiplication rules and counting methods from §6.1.
What if the first drawn counter is replaced?
Then the bag returns to its original composition before the next draw, so the red chance remains r/n and the two draws are independent under a thoroughly randomised replacement procedure.
Related learning and source
- Composite Functions: Hidden Domain After a Square Root
- Implicit Circle Differentiation: Tangent and Normal
- Without Replacement: Why the Second Draw Changes Probability: current guide
- Zero Correlation Does Not Prove Independence
Related resources include Why P(A|B) Is Not P(B|A), JC H2 Conditional Probability and Independence, and the Bukit Timah JC2 Tuition hub. The immutable Clementi three-pupil Mathematics tutoring reference illustrates the small-group teaching format at another level.
Official curriculum source: SEAB Singapore–Cambridge H2 Mathematics 9758 (2027), Probability §6.1. All counters and events are invented instructional examples, not actual school results or official past examination papers.