標題:
Probability
發問:
John plays a shooting game, the probability that he hits the target is 0.63. It is given that among the first 4 hits, John hits the target 3 times. Find the probability that the 5th successful hit will be made by the 10th shooting. [Ans. : 0.03719] 更新: To : 翻雷滾天 風卷殘雲. John hit the target 3 times only, where is the 4th successful hit? Also, what happen to the 10th attempt (you said 5th to 9th)? Please explain, thanks.
最佳解答:
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Since the successive events are independent, for John's to make his 5th successful hit by the 10th shooting, he should make exactly 1 successful hit in the coming 5 attempts (5th to 9th), with prob: 5C1 x 0.63 x (1 - 0.63)^4 = 0.05904 Then make a successful hit at the 10th attempt, with total prob: 0.05904 x 0.63 = 0.03719 2010-11-30 08:43:28 補充： Since the question is asking for 5th success in 10th trial, so his 4th success should happen during his 5th to 9th attempt. Thus for making 1 success in his 5th to 9th attempt (total 5 attempts), the probability is 5C1 x 0.63 x 0.37^4 2010-11-30 08:43:33 補充： After that, he should succeed in his 10th attempt, i.e. prob = 0.63 Therefore overall probability = 5C1 x 0.63 x 0.37^4 x 0.63
其他解答:
Eleven是玫瑰崗作弊抄襲仔Benny Tse謝家昇|||||Totally correct 2010-11-29 17:15:39 補充： His answers means the 4th successful trial can be made in any 5th, 6th, 7th, 8th or 9th trial for he has put 5C1 ( i.e. 5 ) before, which means you have 5 possibilities for making this trial. But since the 10th trial must be successful, there is no such coefficient multiplied before.
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