F.5 math 1 question

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

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

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

有冇replacement先

如果冇，第一次係1/3
最尾一次係 1