GATE CSE 2015 Set 3 | Question: 37

Question

Suppose $X_i$ for $i=1, 2, 3$ are independent and identically distributed random variables whose probability mass functions are $Pr[X_i = 0] = Pr[X_i = 1] = \frac{1} {2} \text{ for } i = 1, 2, 3$. Define another random variable $Y = X_1X_2 \oplus X_3$, where $\oplus$ denotes XOR. Then $Pr[Y=0 \mid X_3 = 0] =$______.

Comments

pls explain bayes method.

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@kavita

c thz http://quiz.geeksforgeeks.org/gate-gate-cs-2015-set-3-question-47/

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Notice -> $X_i$'s are independent But X and Y are not independent.

Answer 1

Given,Y = 0 and X3 = 0

Y = X1.X2 $\bigoplus$ X3

0 = 1.0 $\bigoplus$ 0

0 = 0.1 $\bigoplus$ 0

0 = 0.0 $\bigoplus$ 0

there is only three possibility out of 4

P(E) = Favorable Outcomes / Total Outcomes

P(E) = 3 / 4

P(E) = 0.75

 

Comments

$P(y=0| x3=0)=P(x1 x2=0)= 3/4$

Answer 2

Given, Probability Mass Function:

$ => Pr[X_1 = 0] = Pr[X_1 = 1] = 1/2$       $… (1)$

$ => Pr[X_2 = 0] = Pr[X_2 = 1] = 1/2$       $… (2)$

$ => Pr[X_3 = 0] = Pr[X_3 = 1] = 1/2$       $… (3)$

Also it is given that random variable $Y=X_1X_2\bigoplus X_3$

We need to find: $Pr[Y=0 | X_3=0]$ = ?

$=> Pr[Y=0 | X_3=0]$

$=> Pr[ (X_1X_2\bigoplus X_3) = 0 | X_3=0]$      $(\because Y=X_1X_2\bigoplus X_3)$

$=> Pr[ (X_1X_2\bigoplus 0) = 0| X_3=0]$          (Put $X_3=0$ as it is given that $X_3=0$ then we have to find $Y=0$)

$=> Pr[ (X_1X_2) = 0| X_3=0]$       $… (4)$

It is given in the question that $X_i$ for $i=1,2,3$ are independent and identically distributed random variables

$=> Pr[ (X_1X_2) = 0]$ can be written as:

$=> Pr( X_1 =0 , X_2 = 0) + Pr( X_1 =1 , X_2 = 0) + Pr( X_1 =0 , X_2 = 1)$

Because we can say that $X_1 X_2 = 0 $ if either $X_1 = 0 $ or $X_2 = 0 $ or both $X_1 = 0$ and $X_2 = 0$

$=> [Pr( X_1 =0) \times Pr(X_2 = 0)] + [Pr( X_1 =1) \times Pr(X_2 = 0)] + [Pr( X_1 =0) \times Pr(X_2 = 1)]$

$=> (½ \times ½) +  (½ \times ½) + (½ \times ½)$

$ => ¼ + ¼ + ¼ $

$=> ¾ $ (answer)