High School Maths · Year 2 · Unit 13 · Lesson 4 of 5

Dependent Events: Find the Hidden Number of Counters

About 15 minutes · dependent events and changing branches · trees for two and three picks without replacement · conditional probability from a tree · "given that" questions about the first pick · reverse problems that lead to a quadratic equation · no calculator · exit quiz

Hook

2 black socks · 2 white socks

A drawer holds 2 black socks and 2 white socks. You take two socks in the dark. Is the probability of a matching pair one half?

No. After the first sock is taken, only 3 socks are left and only 1 of them matches it. So P(match) = 13, whatever the first sock is. With a tree: 24 × 13 + 24 × 13 = 13. The second pick depends on the first. In this lesson you also work backwards: from a probability to the number of socks or counters in the bag.

Learn

Master the essentials

R = red, G = green. First counter, then second counter.

Dependent events

The first result changes the probabilities for the second.

Taking counters out one at a time without putting them back

Changing branches

On the next branches there is one item fewer, and one fewer of the colour already taken.

8 red, 3 blue: P(R, R) = 811 × 710

Branches are conditional

The second-stage branches are conditional probabilities.

The branch after "red first" is P(second red | first red)

Three picks

Keep going: each pick has one item fewer.

7 red, 3 blue: 710 × 69 × 58

Conditional from a tree

P(A | B) = P(A and B) ÷ P(B). Add all the paths for B first.

P(first red | second red)

Reverse problems

Call the unknown total n. The second pick has n − 1 items.

P(both red) = 5n × 4n − 1

A quadratic

Clear the fractions, expand, and solve the quadratic equation.

n(n − 1) = 30, so n² − n − 30 = 0

Check your answer

Reject a root that is negative, or smaller than a number of counters you already know.

(n − 6)(n + 5) = 0: n = 6

Worked examples

See it done

1. A box holds 4 faulty and 8 good batteries. Three are taken at random, without replacement. Find the probability that exactly one of them is faulty.

  1. Exactly one faulty: FGG, GFG or GGF. That is 3 paths.
  2. FGG: 412 × 811 × 710 = 2241320.
  3. The other two paths have the same three tops and bottoms, in a different order, so they are equal.
  4. 3 × 2241320 = 6721320 = 2855.

2855

2. A bag holds 5 red and 3 green counters. Two are taken at random, without replacement. Given that the second counter is green, find the probability that the first was green.

  1. P(G, G) = 38 × 27 = 656.
  2. P(second green) = P(R, G) + P(G, G) = 1556 + 656 = 2156.
  3. P(first green | second green) = 656 ÷ 2156 = 621 = 27.
R = red, G = green. First counter, then second counter.

27

3. A bag holds n sweets. 4 are orange and the rest are yellow. Two are taken at random, without replacement. P(both orange) = 27. Show that n² − n − 42 = 0 and find n.

  1. 4n × 3n − 1 = 27, so 12n(n − 1) = 27.
  2. Cross-multiply: 2n(n − 1) = 84, so n² − n = 42.
  3. n² − n − 42 = 0 ✓. Factorise: (n − 7)(n + 6) = 0.
  4. n cannot be negative, so n = 7. Check: 47 × 36 = 27 ✓.

n = 7

Common mistake: Keeping the same bottoms on the second branch. Without replacement, the total goes down by 1, and so does the colour that was taken. In a reverse problem the second fraction is r − 1n − 1, not rn.

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Sketch the tree: each pick has one item fewer. For "given that", divide by the probability of what you are told. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.A bag holds 6 red and 4 white beads. Two are taken at random, without replacement. Find the probability that the second bead is white, given that the first is red.

Not quite. This puts the first bead back. After a red bead is taken, 9 beads are left.

Solution

  1. After a red bead is taken: 9 beads are left, and 4 of them are white.
  2. P(second white | first red) = 49.

Answer: C. 49.

Not quite. This takes a white bead away. The first bead was red.

Solution

  1. After a red bead is taken: 9 beads are left, and 4 of them are white.
  2. P(second white | first red) = 49.

Answer: C. 49.

Correct.

Solution

  1. After a red bead is taken: 9 beads are left, and 4 of them are white.
  2. P(second white | first red) = 49.

Not quite. This is P(red, then white). "Given that the first is red" means the first pick has already happened.

Solution

  1. After a red bead is taken: 9 beads are left, and 4 of them are white.
  2. P(second white | first red) = 49.

Answer: C. 49.

2.A class has 11 boys and 9 girls. Two students are chosen at random to be leaders. Find the probability that both are girls.

Correct.

Solution

  1. First girl: 920. Then 19 students are left and 8 are girls: 819.
  2. 920 × 819 = 72380 = 1895.

Not quite. This puts the first girl back. The same student cannot be chosen twice.

Solution

  1. First girl: 920. Then 19 students are left and 8 are girls: 819.
  2. 920 × 819 = 72380 = 1895.

Answer: A. 1895.

Not quite. This takes 1 off the top of the second fraction but not off the bottom.

Solution

  1. First girl: 920. Then 19 students are left and 8 are girls: 819.
  2. 920 × 819 = 72380 = 1895.

Answer: A. 1895.

Not quite. This adds the fractions. For "and", multiply.

Solution

  1. First girl: 920. Then 19 students are left and 8 are girls: 819.
  2. 920 × 819 = 72380 = 1895.

Answer: A. 1895.

3.A bag holds 5 blue and 2 yellow counters. Three counters are taken at random, without replacement. Find the probability that all three are blue.

Not quite. This puts each counter back.

Solution

  1. 57 × 46 × 35.
  2. = 60210 = 27.

Answer: B. 27.

Correct.

Solution

  1. 57 × 46 × 35.
  2. = 60210 = 27.

Not quite. This takes 1 off the tops but not off the bottoms.

Solution

  1. 57 × 46 × 35.
  2. = 60210 = 27.

Answer: B. 27.

Not quite. This stops after two picks.

Solution

  1. 57 × 46 × 35.
  2. = 60210 = 27.

Answer: B. 27.

Exam standard (Several steps, like real exam questions.)

4.The table shows the counters in a bag. Two counters are taken at random, without replacement. Find the probability that they are the same colour.

ColourRedGreenYellow
Number of counters435

The counters in a bag.

Not quite. This puts the first counter back.

Solution

  1. There are 12 counters, then 11.
  2. RR: 412 × 311 = 12132. GG: 312 × 211 = 6132. YY: 512 × 411 = 20132.
  3. Same colour: 38132 = 1966.

Answer: D. 1966.

Not quite. This leaves out green, green.

Solution

  1. There are 12 counters, then 11.
  2. RR: 412 × 311 = 12132. GG: 312 × 211 = 6132. YY: 512 × 411 = 20132.
  3. Same colour: 38132 = 1966.

Answer: D. 1966.

Not quite. This is the probability that they are different colours.

Solution

  1. There are 12 counters, then 11.
  2. RR: 412 × 311 = 12132. GG: 312 × 211 = 6132. YY: 512 × 411 = 20132.
  3. Same colour: 38132 = 1966.

Answer: D. 1966.

Correct.

Solution

  1. There are 12 counters, then 11.
  2. RR: 412 × 311 = 12132. GG: 312 × 211 = 6132. YY: 512 × 411 = 20132.
  3. Same colour: 38132 = 1966.

5.A bag holds 8 black and 4 white counters. Two are taken at random, without replacement. Given that at least one of them is white, find the probability that both are white.

Not quite. This is P(both white). Divide by P(at least one white).

Solution

  1. P(both white) = 412 × 311 = 12132.
  2. P(no white) = 812 × 711 = 56132, so P(at least one) = 76132.
  3. P(both | at least one) = 12132 ÷ 76132 = 1276 = 319.

Answer: D. 319.

Not quite. This is P(at least one white), the condition itself.

Solution

  1. P(both white) = 412 × 311 = 12132.
  2. P(no white) = 812 × 711 = 56132, so P(at least one) = 76132.
  3. P(both | at least one) = 12132 ÷ 76132 = 1276 = 319.

Answer: D. 319.

Not quite. This is P(second white | first white). The condition is "at least one white".

Solution

  1. P(both white) = 412 × 311 = 12132.
  2. P(no white) = 812 × 711 = 56132, so P(at least one) = 76132.
  3. P(both | at least one) = 12132 ÷ 76132 = 1276 = 319.

Answer: D. 319.

Correct.

Solution

  1. P(both white) = 412 × 311 = 12132.
  2. P(no white) = 812 × 711 = 56132, so P(at least one) = 76132.
  3. P(both | at least one) = 12132 ÷ 76132 = 1276 = 319.

6.A bag holds n counters, and 5 of them are red. Two are taken at random, without replacement. P(both red) = 514. Which equation does n satisfy?

Not quite. This uses n for both bottoms, as if the first counter were put back.

Solution

  1. 5n × 4n − 1 = 514, so 20n(n − 1) = 514.
  2. 5n(n − 1) = 280, so n² − n = 56.
  3. n² − n − 56 = 0.

Answer: D. n² − n − 56 = 0.

Not quite. This sets n(n − 1) equal to the 20 on top and forgets the 14.

Solution

  1. 5n × 4n − 1 = 514, so 20n(n − 1) = 514.
  2. 5n(n − 1) = 280, so n² − n = 56.
  3. n² − n − 56 = 0.

Answer: D. n² − n − 56 = 0.

Not quite. This uses n + 1 for the second bottom. One counter has gone, so it is n − 1.

Solution

  1. 5n × 4n − 1 = 514, so 20n(n − 1) = 514.
  2. 5n(n − 1) = 280, so n² − n = 56.
  3. n² − n − 56 = 0.

Answer: D. n² − n − 56 = 0.

Correct.

Solution

  1. 5n × 4n − 1 = 514, so 20n(n − 1) = 514.
  2. 5n(n − 1) = 280, so n² − n = 56.
  3. n² − n − 56 = 0.

7.A bag holds 3 green counters and some yellow counters. Two are taken at random, without replacement. P(both green) = 126. How many yellow counters are in the bag?

Not quite. This is n, the total number of counters. Take away the 3 green ones.

Solution

  1. 3n × 2n − 1 = 126, so n(n − 1) = 156.
  2. n² − n − 156 = 0, so (n − 13)(n + 12) = 0.
  3. n = 13 counters, so 13 − 3 = 10 are yellow.

Answer: B. 10.

Correct.

Solution

  1. 3n × 2n − 1 = 126, so n(n − 1) = 156.
  2. n² − n − 156 = 0, so (n − 13)(n + 12) = 0.
  3. n = 13 counters, so 13 − 3 = 10 are yellow.

Not quite. This is n − 1, the total after one counter is taken.

Solution

  1. 3n × 2n − 1 = 126, so n(n − 1) = 156.
  2. n² − n − 156 = 0, so (n − 13)(n + 12) = 0.
  3. n = 13 counters, so 13 − 3 = 10 are yellow.

Answer: B. 10.

Not quite. This is n(n − 1). Solve the quadratic to find n.

Solution

  1. 3n × 2n − 1 = 126, so n(n − 1) = 156.
  2. n² − n − 156 = 0, so (n − 13)(n + 12) = 0.
  3. n = 13 counters, so 13 − 3 = 10 are yellow.

Answer: B. 10.

Challenge (Harder problems that mix ideas or need reasoning.)

8.A committee of 3 is chosen at random from 6 men and 3 women. Find the probability that at least one woman is chosen.

Correct.

Solution

  1. P(no women) = 69 × 58 × 47 = 120504 = 521.
  2. P(at least one woman) = 1 − 521 = 1621.

Not quite. This is P(no women). Take it away from 1.

Solution

  1. P(no women) = 69 × 58 × 47 = 120504 = 521.
  2. P(at least one woman) = 1 − 521 = 1621.

Answer: A. 1621.

Not quite. This puts each person back. A person cannot be chosen twice.

Solution

  1. P(no women) = 69 × 58 × 47 = 120504 = 521.
  2. P(at least one woman) = 1 − 521 = 1621.

Answer: A. 1621.

Not quite. This is exactly one woman. Two or three women also count.

Solution

  1. P(no women) = 69 × 58 × 47 = 120504 = 521.
  2. P(at least one woman) = 1 − 521 = 1621.

Answer: A. 1621.

9.A box holds 6 good and 4 faulty bulbs. Two are taken at random, without replacement. Given that exactly one of them is faulty, find the probability that it was the first bulb taken.

Not quite. This is P(first faulty) without the condition.

Solution

  1. FG: 410 × 69 = 2490. GF: 610 × 49 = 2490.
  2. P(exactly one faulty) = 4890.
  3. P(FG | exactly one) = 2490 ÷ 4890 = 12.

Answer: C. 12.

Not quite. This is P(faulty, then good). Divide by P(exactly one faulty).

Solution

  1. FG: 410 × 69 = 2490. GF: 610 × 49 = 2490.
  2. P(exactly one faulty) = 4890.
  3. P(FG | exactly one) = 2490 ÷ 4890 = 12.

Answer: C. 12.

Correct.

Solution

  1. FG: 410 × 69 = 2490. GF: 610 × 49 = 2490.
  2. P(exactly one faulty) = 4890.
  3. P(FG | exactly one) = 2490 ÷ 4890 = 12.

Not quite. This is P(exactly one faulty), the condition itself.

Solution

  1. FG: 410 × 69 = 2490. GF: 610 × 49 = 2490.
  2. P(exactly one faulty) = 4890.
  3. P(FG | exactly one) = 2490 ÷ 4890 = 12.

Answer: C. 12.

10.A bag holds x red and 4 blue counters. Two are taken at random, without replacement. P(one counter of each colour) = 815. Find every possible value of x.

Not quite. x = 2 also works: 2 × 26 × 45 = 1630 = 815.

Solution

  1. 2 × xx + 4 × 4x + 3 = 815 (red then blue, or blue then red).
  2. 120x = 8(x + 4)(x + 3), so 15x = x² + 7x + 12.
  3. x² − 8x + 12 = 0, so (x − 2)(x − 6) = 0.
  4. Both roots are positive whole numbers, and both give 815 when checked.

Answer: C. x = 2 or x = 6.

Not quite. x = 6 also works: 2 × 610 × 49 = 4890 = 815.

Solution

  1. 2 × xx + 4 × 4x + 3 = 815 (red then blue, or blue then red).
  2. 120x = 8(x + 4)(x + 3), so 15x = x² + 7x + 12.
  3. x² − 8x + 12 = 0, so (x − 2)(x − 6) = 0.
  4. Both roots are positive whole numbers, and both give 815 when checked.

Answer: C. x = 2 or x = 6.

Correct.

Solution

  1. 2 × xx + 4 × 4x + 3 = 815 (red then blue, or blue then red).
  2. 120x = 8(x + 4)(x + 3), so 15x = x² + 7x + 12.
  3. x² − 8x + 12 = 0, so (x − 2)(x − 6) = 0.
  4. Both roots are positive whole numbers, and both give 815 when checked.

Not quite. The signs are wrong: (x − 2)(x − 6) = 0 gives x = 2 or x = 6.

Solution

  1. 2 × xx + 4 × 4x + 3 = 815 (red then blue, or blue then red).
  2. 120x = 8(x + 4)(x + 3), so 15x = x² + 7x + 12.
  3. x² − 8x + 12 = 0, so (x − 2)(x − 6) = 0.
  4. Both roots are positive whole numbers, and both give 815 when checked.

Answer: C. x = 2 or x = 6.

Apply

Find the hidden number

A bag holds n counters. 4 are green and the rest are yellow. Two counters are taken at random, without replacement. P(both yellow) = 13.

1. Write P(both yellow) in terms of n.

2. Show that n² − 13n + 30 = 0.

3. Solve the equation. Explain why one of the solutions must be rejected.

4. Find the probability that both counters are green.

  1. n − 4n × n − 5n − 1, because n − 4 counters are yellow.
  2. 3(n − 4)(n − 5) = n(n − 1), so 3n² − 27n + 60 = n² − n. 2n² − 26n + 60 = 0; divide by 2: n² − 13n + 30 = 0 ✓.
  3. (n − 3)(n − 10) = 0, so n = 3 or n = 10. The bag already holds 4 green counters, so n cannot be 3: n = 10.
  4. 410 × 39 = 1290 = 215.

Next: the Lesson 4 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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