Question 231683: For mutually exclusive events X1, X2, and X3, let P(X1) = .22, P(X2) = .35 and P(X3) = .43. Also, P(Y | Y1) = .40 P(Y|X2) = .30 and P(Y|X3) = .60. Find P(X3|Y)
A) . 20 B) . 57 C) . 38 D) . 23
This theorem is so complicated for me that I dont know where to begin.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! For mutually exclusive events X1, X2, and X3,
let P(X1) = .22, P(X2) = .35 and P(X3) = .43.
Also, P(Y | X1) = .40 P(Y|X2) = .30 and P(Y|X3) = .60.
Find P(X3|Y)
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P(X3 | Y) = P(X3 and Y)/P(Y)
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You are told P(Y | X3) = 0.6
So P(Y and X3)/P(X3) = 0.6
and P(Y and X3)/0.43 = 0.6
So, P(Y and X3) = 0.43*0.6 = 0.258
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Using that result you have:
P(X3 | Y) = 0.258/P(Y)
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But P(Y) = P(Y|X1) + P(Y|X2) + P(Y|X3) = 0.4 + 0.3 + 0.6 = 1.3
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Final Answer: P(X3 |Y) = 0.258/1.3 = 0.198
Round you be 0.20
which is answer "A"
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Cheers,
Stan H.
A) . 20 B) . 57 C) . 38 D) . 23
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