Question

Question 5 The following table presents the probabilities associated with the outcomes for the sets of bivariate events X and Y. The marginal probabilities are in the margins of the table. Y1 Y2 X1 ? 0.09 ? 0.41 X2 0.10 ? 0.03 X3 ? 0.11 ? 0.41 0.25 Y3 Assume P(Y3|X1) = 36/123 and P(Yl)=0.36. a) Compute P(Yin X2). (5 points) b) Compute Px2). (5 points) c) Compute P1X2 42)(6 points) d) Are the sets of events X and Y independent? (6 points)

To rewrite in case not clear:

c) Compute P(X2 | Y2)

d) Are the sets of events X and Y independent?

Answer 1

We have given the sets o bivariate events Y 84 Xi X2 X3 EY Yi 0.2 0.10 0.06 0.36 Y₂ 0.09 0.05 0.25 0.12 Y3 Exi 0.03 0.24 0.99 0.41 10.18 0.41 1 P(Y3 (xi) = 5367 P(43, XI). 36 Plu) 20.36 P(x3 123 P(43, X;) = 36x0.91 - 19.76 = 0.12 123 123 3 3 ΣΣ' χι lal So By this P(X2) = |- Phi) - P(x3) -| -6.91-0.01 20.18 P(4/2, 82) = 0.18-0.03-0.10 20.05 P(A/2) 2 0:09+0.0570.l1 € 0.25 after So by completing table @ PLYNX) = Ply, 5X2) = 0.10 6 P(X2) = 1- P(xi) - P(x3) = 0.18 © P(x2/42) = P1X2 NY) P/X2, 4) olos Ply) Ply 0 (28s

Now (AN-2 (a) Sets of events X ay indep. ift P(X=xi, 4 = j) = P(x=xi). Ply=Zj) Flag P(X= 4 = 429,1 = 0.2 & PCX700)-P147 %) 20.4 180:36 Honce P(x=x1 = 4:41) & PIX=24) - P14 441) So y and y are not indep- first = 0.19