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

Tree Diagrams with Three Stages: The Smart Way to Find "At Least One"

About 15 minutes · tree diagrams with three stages · multiplying along a path · counting the paths for "exactly two" · "at least one" = 1 − P(none) · different probabilities at each stage · the smallest number of trials · no calculator · exit quiz

Hook

0.8 · 0.8 · 0.8

A basketball player scores each free throw with probability 0.8, independently. She takes three. Is the probability that she misses at least one of them more or less than a half?

Most people guess "much less". But P(scores all three) = 0.8 × 0.8 × 0.8 = 0.512, so P(misses at least one) = 1 − 0.512 = 0.488: almost a half! In Year 1 you drew trees with two stages. With three stages the tree has 8 paths, and the quick way to "at least one" is 1 − P(none).

Learn

Master the essentials

O = on time, L = late. Day 1, day 2, then day 3.

Three stages

Draw one set of branches for each event. Two outcomes each time gives 2 × 2 × 2 = 8 paths.

Three days, each late or on time

Multiply along a path

The probability of one path is the product of its three branches.

0.6 × 0.6 × 0.4 = 0.144

Exactly two of three

Three paths fit: ✓✓✗, ✓✗✓ and ✗✓✓. Work out each one and add.

Three paths, not one

Equal paths

When every stage has the same branches, those paths are equal: work out one and multiply by the number of paths.

3 × 0.144 = 0.432

At least one

P(at least one) = 1 − P(none). One short sum instead of seven paths.

1 − 0.5³ = 0.875

At least two

Add "exactly two" and "exactly three", or work out 1 − P(none) − P(exactly one).

1 − P(0) − P(1)

Changing branches

The probabilities can be different at each stage: three different people, or three different days.

0.9 × 0.8 × 0.5 = 0.36

n trials

P(none in n independent trials) = P(fail) to the power n. Use it to find how many trials you need.

P(no head in n throws) = (12)ⁿ

Worked examples

See it done

1. The probability that a bus is on time is 0.7 on each day, independently. Find the probability that it is on time on exactly two of three days.

  1. Exactly two days on time: OOL, OLO or LOO. That is 3 paths.
  2. Each path has two 0.7s and one 0.3: 0.7 × 0.7 × 0.3 = 0.147.
  3. Add the three equal paths: 3 × 0.147 = 0.441.
O = on time, L = late. Day 1, day 2, then day 3.

0.441

2. Ana, Ben and Cara each take a driving test. Their probabilities of passing are 34, 23 and 12, independently. Find the probability that (a) at least one of them passes, (b) exactly one of them passes.

  1. (a) P(nobody passes) = 14 × 13 × 12 = 124.
  2. P(at least one passes) = 1 − 124 = 2324.
  3. (b) Only Ana: 34 × 13 × 12 = 324. Only Ben: 14 × 23 × 12 = 224.
  4. Only Cara: 14 × 13 × 12 = 124.
  5. Exactly one: 324 + 224 + 124 = 624 = 14.

(a) 2324 (b) 14

3. A fair coin is thrown n times. What is the smallest value of n for which the probability of at least one head is more than 0.95?

  1. P(at least one head) = 1 − (12)ⁿ. This must be more than 0.95.
  2. So (12)ⁿ must be less than 0.05 = 120.
  3. n = 4: (12)⁴ = 116, which is more than 120. Not enough.
  4. n = 5: (12)⁵ = 132, which is less than 120 ✓.

n = 5

Common mistake: Working out only ONE path for "exactly two out of three". There are three paths (✓✓✗, ✓✗✓, ✗✓✓). And for "at least one", do not add up seven paths and risk missing one: work out 1 − P(none).

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Decide which paths you need, multiply along each one, then add. For "at least one", use 1 − P(none). 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 spinner lands on red with probability 25. It is spun three times. Find the probability that it lands on red all three times.

Not quite. This adds the three probabilities. For "and", multiply.

Solution

  1. Multiply along the path red, red, red.
  2. 25 × 25 × 25 = 8125.

Answer: C. 8125.

Not quite. This cubes the top but multiplies the bottom by 3.

Solution

  1. Multiply along the path red, red, red.
  2. 25 × 25 × 25 = 8125.

Answer: C. 8125.

Correct.

Solution

  1. Multiply along the path red, red, red.
  2. 25 × 25 × 25 = 8125.

Not quite. This cubes the bottom but not the top.

Solution

  1. Multiply along the path red, red, red.
  2. 25 × 25 × 25 = 8125.

Answer: C. 8125.

2.The probability of rain on any day is 0.3, independently of other days. Find the probability that it does not rain on any of three days.

Not quite. This takes 0.3 away from 1 three times. Multiply the probabilities of no rain instead.

Solution

  1. P(no rain on one day) = 1 − 0.3 = 0.7.
  2. Three dry days: 0.7 × 0.7 × 0.7.
  3. = 0.343.

Answer: C. 0.343.

Not quite. This is P(rain on all three days).

Solution

  1. P(no rain on one day) = 1 − 0.3 = 0.7.
  2. Three dry days: 0.7 × 0.7 × 0.7.
  3. = 0.343.

Answer: C. 0.343.

Correct.

Solution

  1. P(no rain on one day) = 1 − 0.3 = 0.7.
  2. Three dry days: 0.7 × 0.7 × 0.7.
  3. = 0.343.

Not quite. This is P(no rain) on one day only.

Solution

  1. P(no rain on one day) = 1 − 0.3 = 0.7.
  2. Three dry days: 0.7 × 0.7 × 0.7.
  3. = 0.343.

Answer: C. 0.343.

3.A fair dice is rolled three times. Which calculation gives the probability of exactly one six?

Correct.

Solution

  1. One six and two non-sixes: 6NN, N6N or NN6. That is 3 paths.
  2. Each path is 16 × 56 × 56.
  3. Multiply one path by 3.

Not quite. This is one path only. The six could come on the first, second or third roll.

Solution

  1. One six and two non-sixes: 6NN, N6N or NN6. That is 3 paths.
  2. Each path is 16 × 56 × 56.
  3. Multiply one path by 3.

Answer: A. 3 × 16 × (56)².

Not quite. This is P(at least one six). It includes two and three sixes.

Solution

  1. One six and two non-sixes: 6NN, N6N or NN6. That is 3 paths.
  2. Each path is 16 × 56 × 56.
  3. Multiply one path by 3.

Answer: A. 3 × 16 × (56)².

Not quite. This adds 1/6 three times and forgets the rolls that are not sixes.

Solution

  1. One six and two non-sixes: 6NN, N6N or NN6. That is 3 paths.
  2. Each path is 16 × 56 × 56.
  3. Multiply one path by 3.

Answer: A. 3 × 16 × (56)².

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

4.The table shows the probability that each of three machines breaks down on a given day. Find the probability that exactly one machine breaks down.

MachineP(breaks down)
X0.1
Y0.2
Z0.25

The three machines break down independently.

Not quite. This adds the three probabilities. Each path needs the other two machines to keep working.

Solution

  1. Only X: 0.1 × 0.8 × 0.75 = 0.06.
  2. Only Y: 0.9 × 0.2 × 0.75 = 0.135.
  3. Only Z: 0.9 × 0.8 × 0.25 = 0.18.
  4. Exactly one: 0.06 + 0.135 + 0.18 = 0.375.

Answer: B. 0.375.

Correct.

Solution

  1. Only X: 0.1 × 0.8 × 0.75 = 0.06.
  2. Only Y: 0.9 × 0.2 × 0.75 = 0.135.
  3. Only Z: 0.9 × 0.8 × 0.25 = 0.18.
  4. Exactly one: 0.06 + 0.135 + 0.18 = 0.375.

Not quite. This is P(at least one breaks down). It includes two or three breaking down.

Solution

  1. Only X: 0.1 × 0.8 × 0.75 = 0.06.
  2. Only Y: 0.9 × 0.2 × 0.75 = 0.135.
  3. Only Z: 0.9 × 0.8 × 0.25 = 0.18.
  4. Exactly one: 0.06 + 0.135 + 0.18 = 0.375.

Answer: B. 0.375.

Not quite. This is P(all three break down).

Solution

  1. Only X: 0.1 × 0.8 × 0.75 = 0.06.
  2. Only Y: 0.9 × 0.2 × 0.75 = 0.135.
  3. Only Z: 0.9 × 0.8 × 0.25 = 0.18.
  4. Exactly one: 0.06 + 0.135 + 0.18 = 0.375.

Answer: B. 0.375.

5.Lena hits a target with probability 13 on each shot, independently. She fires 4 shots. Find the probability that she hits the target at least once.

Correct.

Solution

  1. P(miss) = 23.
  2. P(no hits) = (23)⁴ = 1681.
  3. P(at least one hit) = 1 − 1681 = 6581.

Not quite. This adds 1/3 four times. A probability cannot be more than 1.

Solution

  1. P(miss) = 23.
  2. P(no hits) = (23)⁴ = 1681.
  3. P(at least one hit) = 1 − 1681 = 6581.

Answer: A. 6581.

Not quite. This is P(four hits).

Solution

  1. P(miss) = 23.
  2. P(no hits) = (23)⁴ = 1681.
  3. P(at least one hit) = 1 − 1681 = 6581.

Answer: A. 6581.

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

Solution

  1. P(miss) = 23.
  2. P(no hits) = (23)⁴ = 1681.
  3. P(at least one hit) = 1 − 1681 = 6581.

Answer: A. 6581.

6.A quiz has 3 questions. Each has 4 options, and only one option is right. Zara guesses every answer. Find the probability that she gets at least two right.

Not quite. This is exactly two right. Three right also counts.

Solution

  1. P(right) = 14 and P(wrong) = 34.
  2. Exactly two: 3 paths, each 14 × 14 × 34 = 364, so 964.
  3. Exactly three: (14)³ = 164.
  4. At least two: 964 + 164 = 1064 = 532.

Answer: D. 532.

Not quite. This is three right only.

Solution

  1. P(right) = 14 and P(wrong) = 34.
  2. Exactly two: 3 paths, each 14 × 14 × 34 = 364, so 964.
  3. Exactly three: (14)³ = 164.
  4. At least two: 964 + 164 = 1064 = 532.

Answer: D. 532.

Not quite. This is P(at least ONE right).

Solution

  1. P(right) = 14 and P(wrong) = 34.
  2. Exactly two: 3 paths, each 14 × 14 × 34 = 364, so 964.
  3. Exactly three: (14)³ = 164.
  4. At least two: 964 + 164 = 1064 = 532.

Answer: D. 532.

Correct.

Solution

  1. P(right) = 14 and P(wrong) = 34.
  2. Exactly two: 3 paths, each 14 × 14 × 34 = 364, so 964.
  3. Exactly three: (14)³ = 164.
  4. At least two: 964 + 164 = 1064 = 532.

7.A biased coin is thrown three times. The probability of getting no heads is 2764. Find the probability that the coin lands heads on one throw.

Not quite. This is P(tails). The question asks for heads.

Solution

  1. No heads means three tails: P(T)³ = 2764.
  2. Cube root: P(T) = 34, because 3³ = 27 and 4³ = 64.
  3. P(H) = 1 − 34 = 14.

Answer: B. 14.

Correct.

Solution

  1. No heads means three tails: P(T)³ = 2764.
  2. Cube root: P(T) = 34, because 3³ = 27 and 4³ = 64.
  3. P(H) = 1 − 34 = 14.

Not quite. This is P(at least one head) in three throws.

Solution

  1. No heads means three tails: P(T)³ = 2764.
  2. Cube root: P(T) = 34, because 3³ = 27 and 4³ = 64.
  3. P(H) = 1 − 34 = 14.

Answer: B. 14.

Not quite. This divides by 3. To undo a cube, take the cube root.

Solution

  1. No heads means three tails: P(T)³ = 2764.
  2. Cube root: P(T) = 34, because 3³ = 27 and 4³ = 64.
  3. P(H) = 1 − 34 = 14.

Answer: B. 14.

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

8.A fair dice is rolled n times. What is the smallest value of n for which the probability of at least one six is more than 12?

Not quite. With 3 rolls, P(at least one six) = 1 − 125216 = 91216, which is less than 12.

Solution

  1. P(at least one six) = 1 − (56)ⁿ. You need (56)ⁿ less than 12.
  2. n = 3: (56)³ = 125216. 125 is more than half of 216.
  3. n = 4: (56)⁴ = 6251296. 625 is less than half of 1,296 (648) ✓.

Answer: B. 4.

Correct.

Solution

  1. P(at least one six) = 1 − (56)ⁿ. You need (56)ⁿ less than 12.
  2. n = 3: (56)³ = 125216. 125 is more than half of 216.
  3. n = 4: (56)⁴ = 6251296. 625 is less than half of 1,296 (648) ✓.

Not quite. Six rolls is the average wait for a six. It is not where the probability first passes a half.

Solution

  1. P(at least one six) = 1 − (56)ⁿ. You need (56)ⁿ less than 12.
  2. n = 3: (56)³ = 125216. 125 is more than half of 216.
  3. n = 4: (56)⁴ = 6251296. 625 is less than half of 1,296 (648) ✓.

Answer: B. 4.

Not quite. This uses (16)ⁿ, the chance of ALL sixes, instead of (56)ⁿ.

Solution

  1. P(at least one six) = 1 − (56)ⁿ. You need (56)ⁿ less than 12.
  2. n = 3: (56)³ = 125216. 125 is more than half of 216.
  3. n = 4: (56)⁴ = 6251296. 625 is less than half of 1,296 (648) ✓.

Answer: B. 4.

9.A team plays three matches. P(wins the first) = 0.6. After a win, P(wins the next) = 0.7. After a loss, P(wins the next) = 0.4. Find the probability that the team wins exactly two of the three matches.

Not quite. This is only the path win, win, lose. There are three paths.

Solution

  1. WWL: 0.6 × 0.7 × 0.3 = 0.126.
  2. WLW: 0.6 × 0.3 × 0.4 = 0.072.
  3. LWW: 0.4 × 0.4 × 0.7 = 0.112.
  4. Exactly two: 0.126 + 0.072 + 0.112 = 0.31.

Answer: C. 0.31.

Not quite. This multiplies one path by 3, but the three paths are NOT equal here.

Solution

  1. WWL: 0.6 × 0.7 × 0.3 = 0.126.
  2. WLW: 0.6 × 0.3 × 0.4 = 0.072.
  3. LWW: 0.4 × 0.4 × 0.7 = 0.112.
  4. Exactly two: 0.126 + 0.072 + 0.112 = 0.31.

Answer: C. 0.31.

Correct.

Solution

  1. WWL: 0.6 × 0.7 × 0.3 = 0.126.
  2. WLW: 0.6 × 0.3 × 0.4 = 0.072.
  3. LWW: 0.4 × 0.4 × 0.7 = 0.112.
  4. Exactly two: 0.126 + 0.072 + 0.112 = 0.31.

Not quite. This uses 0.6 for every match. The probabilities change after each result.

Solution

  1. WWL: 0.6 × 0.7 × 0.3 = 0.126.
  2. WLW: 0.6 × 0.3 × 0.4 = 0.072.
  3. LWW: 0.4 × 0.4 × 0.7 = 0.112.
  4. Exactly two: 0.126 + 0.072 + 0.112 = 0.31.

Answer: C. 0.31.

10.An archer hits the target with probability 0.9 on each shot, independently. She takes three shots. Which working shows that P(at most one hit) = 0.028?

Not quite. This is P(at least one miss). "At most one hit" means 0 or 1 hits.

Solution

  1. At most one hit means 0 hits or exactly 1 hit.
  2. No hits: 0.1³ = 0.001.
  3. Exactly one hit: 3 paths of 0.9 × 0.1 × 0.1, so 3 × 0.009 = 0.027.
  4. 0.001 + 0.027 = 0.028 ✓.

Answer: D. 0.1³ + 3 × 0.9 × 0.1² = 0.028.

Not quite. This uses one path for exactly one hit. There are three.

Solution

  1. At most one hit means 0 hits or exactly 1 hit.
  2. No hits: 0.1³ = 0.001.
  3. Exactly one hit: 3 paths of 0.9 × 0.1 × 0.1, so 3 × 0.009 = 0.027.
  4. 0.001 + 0.027 = 0.028 ✓.

Answer: D. 0.1³ + 3 × 0.9 × 0.1² = 0.028.

Not quite. This is exactly one hit. It leaves out no hits at all.

Solution

  1. At most one hit means 0 hits or exactly 1 hit.
  2. No hits: 0.1³ = 0.001.
  3. Exactly one hit: 3 paths of 0.9 × 0.1 × 0.1, so 3 × 0.009 = 0.027.
  4. 0.001 + 0.027 = 0.028 ✓.

Answer: D. 0.1³ + 3 × 0.9 × 0.1² = 0.028.

Correct.

Solution

  1. At most one hit means 0 hits or exactly 1 hit.
  2. No hits: 0.1³ = 0.001.
  3. Exactly one hit: 3 paths of 0.9 × 0.1 × 0.1, so 3 × 0.009 = 0.027.
  4. 0.001 + 0.027 = 0.028 ✓.

Apply

Three sets of lights

On her way to school, Nadia passes three sets of traffic lights. They are green with probabilities 12, 35 and 34, independently.

1. Draw a tree diagram for the three sets of lights.

2. Find the probability that all three are green.

3. Find the probability that at least one is not green.

4. Find the probability that exactly two are green.

G = green, N = not green. First, second, then third light.
  1. All green: 12 × 35 × 34 = 940.
  2. At least one not green: 1 − 940 = 3140.
  3. GGN: 12 × 35 × 14 = 340. GNG: 12 × 25 × 34 = 640. NGG: 12 × 35 × 34 = 940.
  4. Exactly two: 1840 = 920.

Next: the Lesson 2 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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