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Complete the distribution of team A’s score for three independent shots with success probabilities 3/4, 2/3 and 1/2.

Read the idea, work independently, then explain what changed.

TOPIC 01

2026 JM01

The three success probabilities differ, so one binomial distribution does not apply to team A.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Complete the distribution of team A’s score for three independent shots with success probabilities 3/4, 2/3 and 1/2.

Official paper · jm01-2026 · II.1(a) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 7

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Each score is a count of successes.
Hint 2
Multiply three linear probability-generating factors.
Worked solution
  1. For each shooter, use miss probability plus hit probability times t.

    G(t)=(14+34t)(13+23t)(12+12t)G(t)=\left(\frac14+\frac34t\right)\left(\frac13+\frac23t\right)\left(\frac12+\frac12t\right)
  2. Expand; the coefficient of t to power k is the probability of score k.

    G(t)=1+6t+11t2+6t324G(t)=\frac{1+6t+11t^2+6t^3}{24}

Scores 0,1,2,3 have probabilities 1/24, 1/4, 11/24, 1/4.

Checks and common pitfalls: The three success probabilities differ, so one binomial distribution does not apply to team A.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • For each shooter, use miss probability plus hit probability times t.
    G(t)=(14+34t)(13+23t)(12+12t)G(t)=\left(\frac14+\frac34t\right)\left(\frac13+\frac23t\right)\left(\frac12+\frac12t\right)
  • Expand; the coefficient of t to power k is the probability of score k.
    G(t)=1+6t+11t2+6t324G(t)=\frac{1+6t+11t^2+6t^3}{24}

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Curriculum and source notes ↗