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# Probability

For an unfair coin, probability of getting head is 0.6. Assume the fact that getting head or tail is independent from throw to throw. What is the mean number of throws of getting a tail before getting a head?

### 1 個解答

• 最愛解答

P(Getting no tail) = 0.6

P(Getting 1 tail) = 0.4 x 0.6

P(Getting 2 tails) = 0.42 x 0.6

P(Getting 3 tails) = 0.43 x 0.6

......

Thus the mean no. of tails got before getting a head is given by:

1 x 0.4 x 0.6 + 2 x 0.42 x 0.6 + 3 x 0.43 x 0.6 + ... + n x 0.4n x 0.6 + ,,,

= 0.24 x (1 + 2 x 0.4 + 3 x 0.42 + 4 x 0.43 + ...)

For the expression 1 + 2x + 3x2 + 4x3 + ... with 0 < x < 1, we have:

1 + 2x + 3x2 + 4x3 + ... = d(x + x2 + x3 + ,,,)/dx

= d[x/(1 - x)]/dx (By sum of geometric series to infinitely many terms)

= 1/(1 - x)2

So sub x = 0.4, we have 1 + 2 x 0.4 + 3 x 0.42 + 4 x 0.42 + ... = 1/0.36

So the mean no. of throws of getting a tail is 0.24/0.36 = 2/3

資料來源： 原創答案