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

Answer is $0.75$
 
As $X_3 = 0$ is given, to have Y = 0, $X_1X_2$ should be $0$, meaning $\left(X_1, X_2 \right)$ should be one of $\left\{(0,0) (0,1) (1,0)\right\}$

So, required probability $=  3 \times \dfrac{1}{2} \times \dfrac{1}{2} = 0.75 \because $ we can choose any of the $3$ possibilities in $3$ ways and then probability of each set of two combination is $  \dfrac{1}{2} \times \dfrac{1}{2}$.

We can also do like follows:

There are totally $4$ possibilities -  $\left\{(0,0) (0,1) (1,0), (1,1)\right\}$, out of which $3$ are favourable cases.

So, required probability $ = \dfrac{3}{4} = 0.75$.

Comments

@Arjun Sir in P[Y=0|X=0], how to know whether "|" means "OR" (union) or "/" (bayes theorem). I have always seen 
"/"

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@srestha

 

P(A intersect B) = P(A) * P(B) only valid for independent event and not for dependent. Yes this formula of Bayes theorem not just for independent it works on dependent also.

Like in this case we can take –

Probability= P(A intersect B)/P(B)

= ((0.5*0.5*0.5)+(0.5*0.5*0.5)+ (0.5*0.5*0.5))/ 0.5

Here 0.5 *0.5*0.5 is probabiliyy of choosing 0 or 1.

Which i choosen for{ 000,100,010}

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@Arjun sir is my way of approach correct?

x1,x2,x3 can take values 000,010,100 to make Y=0   [ Y =x1 AND x2  EXOR x3 ]
so P[Y=0 AND x3=0]=3/8
P[x3=0]=4/8;
according to formula
P(A intersect B) / P(B)  
 = P[Y=0 AND x3=0] / P[x3=0]
=(3/8) / (4/8) = 0.75

Answer 2

We have to find $P\left ( \frac{Y=0}{X_{3}=0} \right ) = ? $ 

$X_{1}$	$X_{2}$	$X_{3}$	$Y$
0	0	0	0
0	0	1	1
0	1	0	0
0	1	1	1
1	0	0	0
1	0	1	1
1	1	0	1
1	1	1	0

$P\left ( Y=0 \cap X_{3}=0\right ) = \frac{3}{8}$ and $P\left ( X_{3} = 0 \right ) = \frac{4}{8}$

So $P\left ( \frac{Y=0}{X_{3}=0} \right ) = \frac{P\left ( Y=0 \cap X_{3}=0\right )}{P\left ( X_{3} = 0 \right )} = \frac{3}{4}$

Comments

but

$Y = X_1X_2 \oplus X_3$

have u considered that?

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great👍

Answer 3

Answer : 3 /4 

Comments

Good One

Answer 4

There are 3 different random variables = X1,X2, and X3 and each have a 1/2 probability of being either 0 or 1.

Y = X1 . X2 ⊕ X3     [. = AND, ⊕ = XOR]

Since each value (0,1) is equally possible for each variable (X1,X2,X3), we can create a truth table.

X1	X2	X3	X1.X2	Y = X1 . X2 ⊕ X3
0	0	0	0	0
0	0	1	0	1
0	1	0	0	0
0	1	1	0	1
1	0	0	0	0
1	0	1	0	1
1	1	0	1	1
1	1	1	1	0

Now,

$P(Y=0 | X_{}3=0) = \frac{P(Y=0 \bigcap X_{}3=0)}{P(X_{}3=0)} = \frac{\frac{3}{8}}{\frac{4}{8}} = \frac{3}{4}$