# GATE | GATE-CS-2015 (Set 3) | Question 47

Suppose Xi for i = 1, 2, 3 are independent and identically distributed random variables whose probability mass functions are Pr[Xi = 0] = Pr[Xi = 1] = 1/2 for i = 1, 2, 3. Define another random variable Y = X1 X2 ⊕ X3, where ⊕ denotes XOR.

Then Pr[Y = 0 ⎪ X3 = 0] = ____________.
(A) 0.75
(B) 0.50
(C) 0.85
(D) 0.25

Explanation:

P (A|B) = P (A∩B) / P (B)

P (Y=0 | X3=0) = P(Y=0 ∩X3=0) / P(X3=0)

P(X3=0) = 1⁄2

Y = X1X2 ⊕ X3

The number of possibilities where Y = 0 can be obtained by constructing a table

From the above table, P(Y=0 ∩X3=0) = 3/8
And P (X3=0) = 1⁄2

P (Y=0 | X3=0) = P(Y=0 ∩X3=0) / P(X3=0) = (3/8) / (1/2) = 3⁄4 = 0.75

This solution is contributed by Anil Saikrishna Devarasetty .

Another Solution :

It is given X3 = 0.

Y can only be 0 when X1 X2 is 0. X1 X2 become 0 for X1 = 1, X2 = 0, X1 = X2 = 0 and X1 = 0, X = 1

So the probability is = 0.5*0.5*3 = 0.75

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