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 $\frac{1}{2}$.
Important Questions from Discrete Probability
- 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
- Suppose two fair dice are thrown independently at random. Let $X$ and $Y$ be the numbers on the upper face of the first die and that of the second die, respectively. Then which of the following statements are true?
- A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability that the last ball drawn is black equals
Consider the problem of testing $H_0 : \theta = 1$ vs $H_1 : \theta = \frac{1}{2}$ where $\theta$ is the mean of a Poisson random variable. Let $X$ and $Y$ be a random sample from Poisson ($\theta$) distribution. Consider the following test procedure:
Reject $H_0$ if either $X = 0$ or $(X = 1 \text{ and } X + Y \leq 2)$; otherwise accept $H_0$.
Which of the following are true?
In a football league, the goals scored by home teams over 380 matches have the following frequency distribution.
Number of goals 0 1 2 3 4 5 Frequency 92 121 91 50 19 7 The average goals scored by home teams is 1.49. We want to test $H_0$: Goal distribution is Poisson. Based on observations the value of the $\chi^2$-statistic for goodness of fit is 1.27. Given $\chi^2_{0.05, 6} = 1.64, \chi^2_{0.05, 5} = 1.15, \chi^2_{0.95, 6} = 12.59$ and $\chi^2_{0.95, 5} = 11.07$, which of the following are true?
