We are given two independent random variables:
We need to find the probability $P(X < Y)$.
Let $Y$ represent the number of successes in $m=11$ independent Bernoulli trials, each with probability $p=\frac{1}{2}$. We can decompose $Y$ into two parts:
So, $Y = Z + W$. The variable $X$ is independent of $Y$, and thus independent of $Z$ and $W$. $X$ and $Z$ are independent and identically distributed.
We can find $P(X < Y)$ by conditioning on the value of $W$:
$ P(X < Y) = P(X < Z + W) $
$ = P(X < Z + W | W=0) P(W=0) + P(X < Z + W | W=1) P(W=1) $
$ = P(X < Z) \times \frac{1}{2} + P(X < Z + 1) \times \frac{1}{2} $
$ = \frac{1}{2} [ P(X < Z) + P(X \le Z) ] $
Since $X$ and $Z$ are independent and identically distributed with $p=\frac{1}{2}$, their distribution is symmetric. Therefore:
Also, $P(X \le Z) = P(X < Z) + P(X = Z)$.
Substitute these back into the equation for $P(X < Y)$:
$ P(X < Y) = \frac{1}{2} [ P(X < Z) + P(X \le Z) ] $
$ = \frac{1}{2} [ P(X < Z) + (P(X < Z) + P(X = Z)) ] $
$ = \frac{1}{2} [ 2 P(X < Z) + P(X = Z) ] $
Now substitute the expression for $P(X < Z)$:
$ P(X < Y) = \frac{1}{2} \left[ 2 \left( \frac{1 - P(X = Z)}{2} \right) + P(X = Z) \right] $
$ = \frac{1}{2} [ (1 - P(X = Z)) + P(X = Z) ] $
$ = \frac{1}{2} [ 1 ] $
$ = \frac{1}{2} $
The probability $P(X < Y)$ is equal to $\frac{1}{2}$.