let O (n) be the total probability of nth head shows up at the odd number of toss E (n) for even number of toss we have E (n) = (1-p) * O (n) + p * O (n-1) E (n) + O (n) = 1 thus we get a recurrence relation for O (n) O (n) = -p/(2-p) * O (n-1) + 1/(2-p) Solving it we have O (n) = (-p/(2-p))^(n-1) * ( O (1) - 1/2) +1/2 with O( 1 ) = 1/(2-p) |
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