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Find the probability that the third head occurs on toss 10.

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

TOPIC 01

2023 JM01

Choosing three heads anywhere in ten does not enforce the last-head time.

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

01 / Standard#Your turn

Find the probability that the third head occurs on toss 10.

Official paper · jm01-2023 · II.1(c) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 6

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Exactly two heads must occur among the first nine tosses.
Hint 2
Fix a head on the last toss.
Worked solution
  1. Count the choices for the earlier two heads.

    P=(92)(1/4)2(3/4)7(1/4)P=\binom92(1/4)^2(3/4)^7(1/4)
  2. Simplify the powers.

    P=36⋅37410=(3/4)9P=\frac{36\cdot3^7}{4^{10}}=(3/4)^9

(3/4)⁹.

Checks and common pitfalls: Choosing three heads anywhere in ten does not enforce the last-head time.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Count the choices for the earlier two heads.
    P=(92)(1/4)2(3/4)7(1/4)P=\binom92(1/4)^2(3/4)^7(1/4)
  • Simplify the powers.
    P=36⋅37410=(3/4)9P=\frac{36\cdot3^7}{4^{10}}=(3/4)^9

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Curriculum and source notes ↗