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baily106
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Probability Question

by baily106 Sat Jul 26, 2014 12:09 pm

Hello,

I am referring to question #14 on page 298 of The Official Guide Book. The question asks "Let A, B, C and D be events for which P(A or B) = 0.6, P(A) = 0.2, P (C or D) = 0.6, and P© = 0.5. The events A and B are mutually exclusive, and the events C and D are independent. a) Find P(B) which I believe is 0.4, but how do you get P(D) = 0.2?

Thanks.
kwame.appau
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Re: Probability Question

by kwame.appau Sat Jul 26, 2014 7:58 pm

Since p(c) and p(D) are independent then it means that
p(C or D)= p(C) + p(D)-p(C) x p(D)
0.6 = 0.5+ p(D)- 0.5 x p(D)
0.1 = P(D)(1-0.5)
p(D)= 0.1/0.5
p(D)= 0.2
I hope Tommy might add little explanation to that
tommywallach
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Re: Probability Question

by tommywallach Tue Jul 29, 2014 4:14 pm

If P(A or B) = .6

Then the P(A) + P(B) - P(both) = .6

Make sure you understand why. We can't just add probabilities, otherwise the odds that it will rain tomorrow (say 1/3) or that it will be cold tomorrow (say 1/3) or that I will be friendly tomorrow (say 1/3) would be 100%.

So, to continue:

.2 + P(B) - P(A and B) = .6

HOWEVER, the question says A and B can't happen at the same time, so we can ignore that crossover:

.2 + P(B) = .6
P(B) = .4

For, C and D, it's the same as above.

.5 + P(D) - P (C and D) = .6
.5 + P(D) - .5 * P(D) = .6
P(D) - .5 * P(D) = .1
.5 * P(D) = .1
P(D) = .1/.5 = .2

Hope that helps! (And great explanation, Kwame!)

-t
baily106
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Re: Probability Question

by baily106 Fri Aug 01, 2014 6:19 am

That helps; thanks.
tommywallach
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Re: Probability Question

by tommywallach Fri Aug 01, 2014 10:35 pm

Sweet!