Yahoo 知識+ 將於 2021 年 5 月 4 日 (美國東岸時間) 停止服務。從 2021 年 4 月 20 日 (美國東岸時間) 起,Yahoo 知識+ 網站將轉為僅限瀏覽模式。其他 Yahoo 資產或服務,或你的 Yahoo 帳戶將不會有任何變更。你可以在此服務中心網頁進一步了解 Yahoo 知識+ 停止服務的事宜,以及了解如何下載你的資料。

匿名
匿名 發問於 科學及數學數學 · 1 十年前

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

評分
  • 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
還有問題嗎?立即提問即可得到解答。