[中五數學]Probability一問

this's an conditional prob. question
P(second correct | >1 correct)
=P(second correct and >1 correct)/P(>1 correct)
=P(second correct and >1 correct)/[1-P(all wrong)]
=(1/3)/[1-(1-1/4)(1-1/3)]
=2/3

2008-03-24 14:23:45 補充：
> 即at least

Let A = David answers the first question correctly,
and B = David answers the second question correctly.

Given P(A) = 1/4 and P(B) = 1/3.

P(B | given at least one correct), i.e. P(B given at least one correct)
= P(at least one correct ∩ B)/P(at least one correct)
= P(AB or ~AB)/[1-P(both not correct)]
= P(B) / [1-(1-1/4)(1-1/3)]
= (1/3) / (1-3/4*2/3)
= (1/3) / (1-1/2)
= (1/3) / (1/2)
= 2/3

p(answer second question correctly)=3/12

there are two cases for his answering the second question
correctly. the first one is first question wrong and the second
question correct. the second one is both questions are correct.
for the first case, the probability is 3/4 x 1/3=1/4
for the second case, the probability is 1/4 x 1/3=1/12
therefore, the required probability is 1/4 + 1/12 =1/3