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
Answer: (A)
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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Last Updated :
14 Feb, 2018
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