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probability again!

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by winniethepooh » Wed Jul 13, 2011 11:52 am
A, B And C each try independently to solve a problem. If their individual probability for success are 1/4, 1/2, and 5/8, respectively , what is the probability that A and B, but not C, will solve the problem

1. 11/8
2. 7/8
3. 9/64
4. 5/64
5. 3/64
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by Frankenstein » Wed Jul 13, 2011 12:00 pm
Hi,
p(A) = 1/4
p(B) = 1/2
p(C) = 5/8. So, p(not C) = 1 - 5/8 = 3/8
As A,B,C are independent events, p(AnBnNotC) = p(A)p(B)p(not C) = 3/64

Hence, E
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by winniethepooh » Wed Jul 13, 2011 12:21 pm
why don't we just take the total probability and divide the combined of A and B b y it?
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by amit2k9 » Wed Jul 13, 2011 10:10 pm
probability = 1/4*1/2*3/8 = 3/64

what you are essentially saying is

(1/4*1/2*3/8)/ 1/4*1/2 = 3/8 essentially wrong. I hope I have interpreted you correctly.
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