P(x_1 = 1) = a P(x_2 = 1) = b P(x_3 = 1) = c P(A) = P(A^B) + P(A^B) P(x_1 = 0, x
ID: 3175842 • Letter: P
Question
P(x_1 = 1) = a P(x_2 = 1) = b P(x_3 = 1) = c P(A) = P(A^B) + P(A^B) P(x_1 = 0, x_1 = 0) = 1- a - b + c + ab P(x_1 = 0, x_2 = 1) = b - c - ab P(x_1 = 0, x_2 = 0) = a - c - ab P(x_1 = 0, x_2 = 1) = c + ab P(x_1 = 1 = E(x_1)=1/4 = [a] P(x_2 = 1 = E(x_2)=1/3 = [b] P(x_3 = 1 = E(x_3)=1/2 = [c] E(x_1 -x_2)=1/10 E(x_1 -x_3)=1/15 E(x_2 -x_3)=1/20 E(x_1 -x_2 x_3)=1/100 Cos (x_1, x_3) = -7/120 = [e] Cos (x_2, x_3) = -7/60 = [f] Cos (x, x_2, x_3) = 29/3000 = [g] P(A) = P(A^B^C) + P(A^B^C) + P(A^B^C) + P(A^B^C) + Find P(x_1 =1, x_2 =1, x_3 = 1)? P(x_1 =1, x_2 =1, x_3 = 1)? P(x_1 =1, x_2 =0, x_3 = 1)? P(x_1 =1, x_2 =1, x_3 = 0)? P(x_1 =0, x_2 =1, x_3 = 1)?Explanation / Answer
We have
P(X1 = 0 , X2 = 1) = b - c - ab
= 1 / 3 - 1 / 2 - 1 / 4 * 1 / 3
= - 0.25
P(X1 = 0 , X2 = 1) = - 0.25 < 0
But 0 <= P(X1 = 0 , X2 = 1) <= 1
So P(X1 = 0 , X2 = 1) = b - c - ab is invalid.
Please check and resend the problem.
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