Question

Consider two events X and Y with probabilities, P(X) = 7/15, P( X Y)=1/3, and P(X/Y) = 2/3. Calculate P(Y), P(Y/X), and P(Y/X bar). State with reasons whether the events X and Y are dependent or mutually

exclusive.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Given :

P(X) = 7/15

P( X∩Y) = 1/3

P(X/Y) = 2/3

To find :

P(Y) , P(Y/X) and P(Y/X')

State with reasons whether the events X and Y are dependent or mutually exclusive.​

Solution :

a) P(Y)

P(X/Y) = P(X∩Y) / P(Y)

P(Y) = P(X∩Y) / P(X/Y)

P(Y) = (1/3) / (2/3)

P(Y) = 1/2

b) P(Y/X)

P(Y/X) = P(Y∩X) / P(X)

P(Y/X) = (1/3) / (7/15)

P(Y/X) = 5/7

c) P(Y/X') = P(Y∩X') / P(X')

We know, P(Y) = P(Y∩X) + P(Y∩X'), since X and X'  are mutually exclusive and exhaustive.

P(Y∩X') = P(Y) - P(Y∩X)

P(Y∩X') = 1/2 - 1/3

P(Y∩X') = 1/6

P(Y/X') = (1/6) / (1 - 7/15)

P(Y/X') = 5/16

X and Y are not mutually exclusive as P(Y∩X) ≠ 0

∵ P(X).P(Y) = 7/15 × 1/2 = 7/30 ≠ P(X∩Y) , they are not independent