Probability | Part IA, 2002

A coin shows heads with probability pp on each toss. Let πn\pi_{n} be the probability that the number of heads after nn tosses is even. Show carefully that πn+1=(1p)πn+p(1πn)\pi_{n+1}=(1-p) \pi_{n}+p\left(1-\pi_{n}\right), n1n \geq 1, and hence find πn\pi_{n}. [The number 0 is even.]

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