F.7 probability [10pts]

2011-10-02 9:18 pm

A box contains ten light bulbs, two of them are defective.
Two bulbs are selected at random, one by one without replacement.
Consider the following three events:
A: a defective bulb is drawn on the first drawing
B: a defective bulb is drawn on the second drawing
C: at least one non-defective bulb is drawn

1) P(A or B), P(A or C), P(B or C)
2) P( Aand B), P( A and C), P(B and C)

Ans
1) 14/45, 1,1
2) 1/45, 8/45, 8/45

thanks

更新1:

for 1) not 14/45 is 17/45

回答 (1)

myisland8132

2011-10-02 9:50 pm

✔ 最佳答案

Your answer is not true

2011-10-02 13:50:32 補充：
(1) P(A or B)

= P(there is at least one defective bulb)

= 1 - C(8,2)/C(10,2)

= 1 - 28/45

= 17/45

As P(A or C) and P(B or C) have the same meaning of P(A or B)

We conclude that P(A or C) = P(B or C) = 17/45

(2) P(A and B)

= P(both defective bulbs are drawn)

= C(2,2)/C(10,2)

= 1/45

P(A and C)

= P(A)

= (2/10)(8/9)

= 8/45

P(B and C)

= P(B)

= (8/10)(2/9)

= 8/45

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