maths question probability

1E,2E, 3E : 1st exam, 2nd exam and 3rd exam respectively
1I, 2I, 3I : 1st interview, 2nd interview, and 3rd interview respectively
P : pass
F : fail


=====
a)
P (entersthe university by sitting the entrance exam only once)
= P(1E-P) ´ P(1E-P)
= 60% ´ (1/3)
= 0.2


=====
b)
P(entersthe university by sitting the entrance exam only twice)
= P(F in the 1st attempt) ´ P(P in the 2nd attempt)
= (1 - 0.2) ´ [60%´ (1/3)]
= 0.08 + 0.08
= 0.16


=====
c)
P(entersthe university by sitting the entrance exam only thrice)
= P(F in the 1st and 2nd attempts) ´ P(P in the 3rd attempt)
= [1 - (0.2 + 0.16)] ´ [60% ´ (1/3)]
= 0.128

P(enters the university with one attempt | at most 3 attempts)
= 0.2 / (0.2 + 0.16 + 0.128)
= 0.2/0.488
= 25/61

2011-10-09 10:16:11 補充：
我也想過這問題。但我只當作是出問題者的語言表達不清。若果真的如 doraemonpaul 所說的語言陷阱，只是出題者的問題，不是我理解的問題了。

土扁，c部你中招了！

c部是問你只遇過1次interview就入到大學的概率，而不是問你在所有入到大學的學生之中抽到只遇過1次interview就入到大學的學生的概率。

由於只遇過1次interview就入到大學的學生只可能是只考1次entrance exam就入到大學的學生，因此c部的結果其實即是等於a部的結果。

緊記出現If這類的字眼並不是使用條件概率之金科玉律，需要視乎情況而定。

let say there are totally 100 students:
a) 100 x 60% x 1/3 = 20/100(students) = 1/5
b) 100 x 60% x 2/3 = 40/100(students) = 2/5
c) 100 x 60% x 1/3 x1/3 x 1/3 = 20/9
Q c I don't know the Q is exactly right or not