F.5 math 1 question

👨🏻‍🏫 學術台數學

用戶174067

2007-12-08 7:15 am

我唔明點解e條o既Probability全部都等於1over3.
請解釋
Amy has three keys, only one of them can open the door. If she tries the keys at random and discards those that cannot open the door, find the probablilty that she can open the door in the first trial, second trial and last trial.

Thz~

更新1:

不過佢discards those that cannot open the door~咁唔係第2次變左1over2, 第3次變左1咩?

更新2:

書Answer係1over3

回答 (4)

[匿名]

2007-12-08 7:23 am

✔ 最佳答案

the probablilty that she can open the door in the first trial
P(A) = 1/3
3條抽一條要中當然係1/3

the probablilty that she can open the door in the second trial
P(B) = 2/3 X 1/2 = 1/3
第一次要唔中,所以係要抽3條中的其中2條 (P = 2/3),第2次得番2條,中唔中都係1/2,乘埋佢

the probablilty that she can open the door in the last trial.
P(C) = 2/3 x 1/2 x 1/1 = 1/3
同上,不過第3次得番一條,所以係乘1/1

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sammy_0330

2007-12-08 7:30 am

First Trial
Probability that Amy can open the door in the trial
= 1 - P (can't open on 1st trial)
= 1-2/3 = 1/3
Second Trial
Probability
= P (can't open on 1st trial) x P (can open on 2nd trial)
= 2/3 x 1/2
= 1/3
Third Trial
Probability
= P (can't open on 1st trial) x P (can't open on 2nd trial) x P (can open on 3rd trial)
= 2/3 x 1/2 x 1
= 1/3

參考: myself

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iloveshow

2007-12-08 7:23 am

probability open on first trial = 1/3

probability open on second trial
= P(fail at first time)*P(succeed on 2nd time)
= (2/3)*(1/2)
= 1/3

probability open on third trial
=P(fail at first time)*P( fail on 2nd time)
or
= 1 - sum of the above probability
= 1/3

參考: myself

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chunman

2007-12-08 7:22 am

有冇replacement先

如果冇，第一次係1/3
最尾一次係 1

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