A jewelry shop put all the expensive jewls in a safe when it is closed. The two managers of the shop keep the two keys of the safe and they have to arrive on time to open the safe every morning. The probability that manager X and manager Y will arrive on time is 0.93 and 0.88 respectively. It could be assumed that whether one manager is late or not is independent of the other being late or not. On any particular day:
(a) what is the probability that manager Y will be late but not manager X?
(b) what is the probability that the safe could be opened on time if it could be opened by either one of the keys?
(c) what is the probability that the safe could not be opened on time if it could only be opened using both of the keys?
Please answer these questions in detail, thanks.
Statistic in business
2012-10-21 8:54 am
回答 (2)
2012-10-21 9:22 am
✔ 最佳答案
(a)P(X on time) = 0.93
P(X late) = 1 - 0.93 = 0.07
P(Y on time) = 0.88
P(Y late) = 1 - 0.88 = 0.12
The required probability
= P([Y late] and [X on time])
= P(Y late) x P(X on time)
= 0.12 x 0.93
= 0.1116
(b)
The required probability
= P([X on time] or [Y on time])
= P(X on time) + P(Y on time) - P([X on time] and [Y on time])
= 0.93 + 0.88 - 0.93 x 0.88
= 0.9916
(c)
The required probability
= P([X on time] and [Y on time])
= P(X on time) x P(Y on time)
= 0.93 x 0.88
= 0.8184
參考: 土扁
2012-11-02 3:53 am
hi 我都係做嚟份assignment
我覺得(c) part 答錯
我覺得(c) part 答錯
收錄日期: 2021-04-13 19:04:16
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