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Question

Suppose $X \sim \text{Binomial}(10, \frac{1}{2})$, $Y \sim \text{Binomial}(11, \frac{1}{2})$, where $X$ and $Y$ are independent. Then, $P(X < Y)$ is

The correct answer is
equal to $\frac{1}{2}$

Binomial Probability $P(X < Y)$ Calculation

We are given two independent random variables:

  • $X \sim \text{Binomial}(n=10, p=\frac{1}{2})$
  • $Y \sim \text{Binomial}(m=11, p=\frac{1}{2})$

We need to find the probability $P(X < Y)$.

Decomposing the Binomial Variable 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:

  • Let $Z$ be the number of successes in the first $n=10$ trials. Then $Z \sim \text{Binomial}(10, \frac{1}{2})$.
  • Let $W$ be the outcome of the 11th trial, where $W=1$ if success (with probability $\frac{1}{2}$) and $W=0$ if failure (with probability $\frac{1}{2}$).

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.

Using Conditional Probability

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) ] $

Applying Symmetry

Since $X$ and $Z$ are independent and identically distributed with $p=\frac{1}{2}$, their distribution is symmetric. Therefore:

  • $P(X < Z) = P(Z < X)$
  • The total probability is $P(X < Z) + P(Z < X) + P(X = Z) = 1$.
  • Substituting $P(Z < X)$ with $P(X < Z)$, we get $2 P(X < Z) + P(X = Z) = 1$.
  • This implies $P(X < Z) = \frac{1 - P(X = Z)}{2}$.

Also, $P(X \le Z) = P(X < Z) + P(X = Z)$.

Combining Results

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} $

Conclusion

The probability $P(X < Y)$ is equal to $\frac{1}{2}$.

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Important Questions from Discrete Probability

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  2. Let $X$ and $Y$ be independent Poisson random variables with means $4$ and $2$, respectively. Then, which of the following statements are true?
  3. Consider the M/M/1 queue in which customers arrive according to a Poisson process with rate $3$ and successive service times are independent exponential random variables having mean $\frac{1}{9}$. Let $P_n$ be the long run probability that there are exactly $n$ customers in the system. Then, which of the following statements are true?
  4. Let $X$ be a Binomial$(n, p)$ random variable, where $n \in \{5,6\}$ and $p\in \{\frac{1}{4}, \frac{3}{4}\}$. If $X = 3$ is observed, then the maximum likelihood estimate of $(n, p)$ is
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