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55 votes
55 votes
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 $Pr[Y=0 \mid X_3 = 0] =$______.

6 Answers

Best answer
65 votes
65 votes
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$.
edited by
62 votes
62 votes

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

8 votes
8 votes

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,

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

 

Answer:

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