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John and Anna each shoot three times, with independent hit probabilities 1/3 and 2/3. Find the probability of four total hits.

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

TOPIC 01

2025 JM01

The six shots do not share one success probability.

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

01 / Standard#Your turn

John and Anna each shoot three times, with independent hit probabilities 1/3 and 2/3. Find the probability of four total hits.

Official paper · jm01-2025 · I.12 · PDF 3

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option A1627\frac{16}{27}
  2. Option B6481\frac{64}{81}
  3. Option C2581\frac{25}{81}
  4. Option D58243\frac{58}{243}
  5. Option E34243\frac{34}{243}

Working and explanation

BUILD THE REASONING

Hint 1
List the possible hit-count pairs.
Hint 2
Use (1,3), (2,2), and (3,1).
Worked solution
  1. Calculate the required binomial probabilities for each person.

    PJ(1,2,3)=(49,29,127),PA(1,2,3)=(29,49,827)P_J(1,2,3)=\left(\frac49,\frac29,\frac1{27}\right),\quad P_A(1,2,3)=\left(\frac29,\frac49,\frac8{27}\right)
  2. Multiply within each independent pair, then add.

    P=49827+2949+12729=58243P=\frac49\frac8{27}+\frac29\frac49+\frac1{27}\frac29=\frac{58}{243}

D: 58/243.

Checks and common pitfalls: The six shots do not share one success probability.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Calculate the required binomial probabilities for each person.
    PJ(1,2,3)=(49,29,127),PA(1,2,3)=(29,49,827)P_J(1,2,3)=\left(\frac49,\frac29,\frac1{27}\right),\quad P_A(1,2,3)=\left(\frac29,\frac49,\frac8{27}\right)
  • Multiply within each independent pair, then add.
    P=49827+2949+12729=58243P=\frac49\frac8{27}+\frac29\frac49+\frac1{27}\frac29=\frac{58}{243}

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Curriculum and source notes ↗