Six heads in nine tosses given five in the first six

Question

A fair coin is tossed repeatedly; $H_n$ counts heads among the first $n$ tosses. Given that $H_6=5$ , what is the probability that $H_9=6$ ?

Short answer

$3/8$ . The condition fixes the first six tosses' head count, so we need exactly one head among the next three independent tosses: $\binom31/2^3=\tfrac38$ .

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Lessons behind this question

Conditional Probability, Total Probability, and Bayes' Rule
Worked Examples — Probability Trees, Monty Hall, and Bayesian Updating
Populations, Samples, and Sampling Distributions

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