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Official PYQ · 2015

GATE CS 2015 Set 3 Q37 — Probability: Random Variables and Distributions

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Suppose XiX_i for i=1,2,3i=1, 2, 3 are independent and identically distributed random variables whose probability mass functions are Pr[Xi=0]=Pr[Xi=1]=12 for i=1,2,3Pr[X_i = 0] = Pr[X_i = 1] = \frac{1} {2} \text{ for } i = 1, 2, 3. Define another random variable Y=X1X2⊕X3Y = X_1X_2 \oplus X_3, where ⊕\oplus denotes XOR. Then Pr[Y=0∣X3=0]=Pr[Y=0 \mid X_3 = 0] =______.
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Answer: 0.75

We are given that X1,X2,X3X_1, X_2, X_3 are independent and identically distributed (i.i.d.) random variables with:

Pr[Xi=0]=Pr[Xi=1]=12for i=1,2,3Pr[X_i = 0] = Pr[X_i = 1] = \frac{1}{2} \quad \text{for } i = 1, 2, 3

We need to find the conditional probability Pr[Y=0∣X3=0]Pr[Y=0 \mid X_3 = 0], where Y=X1X2⊕X3Y = X_1X_2 \oplus X_3.

Given the condition X3=0X_3 = 0, we can substitute X3=0X_3 = 0 into the expression for YY:

Y=X1X2⊕0=X1X2Y = X_1X_2 \oplus 0 = X_1X_2

Since X3X_3 is independent of X1X_1 and X2X_2, the conditional probability simplifies to:

Pr[Y=0∣X3=0]=Pr[X1X2=0]Pr[Y=0 \mid X_3 = 0] = Pr[X_1X_2 = 0]

The product X1X2=0X_1X_2 = 0 if and only if at least one of X1X_1 or X2X_2 is 00. Thus:

Pr[X1X2=0]=1−Pr[X1X2=1]Pr[X_1X_2 = 0] = 1 - Pr[X_1X_2 = 1]

Since X1X_1 and X2X_2 are independent:

Pr[X1X2=1]=Pr[X1=1]⋅Pr[X2=1]=12×12=0.25Pr[X_1X_2 = 1] = Pr[X_1 = 1] \cdot Pr[X_2 = 1] = \frac{1}{2} \times \frac{1}{2} = 0.25

Therefore:

Pr[Y=0∣X3=0]=1−0.25=0.75Pr[Y=0 \mid X_3 = 0] = 1 - 0.25 = 0.75

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