Problem 105 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(X\) and \(Y\) are independent binomial variates \(B\left(5, \frac{1}{2}\right)\) and \(B\left(7, \frac{1}{2}\right)\) then the value of \(P(X+Y=3)\) is (A) \(\frac{55}{1024}\) (B) \(\frac{44}{1024}\) (C) \(\frac{33}{1024}\) (D) none of these

Step-by-Step Solution

Answer

The probability \( P(X+Y=3) \) is \( \frac{55}{1024} \), which is option (A).

Step 1: Define the Binomial Distribution Formula

Recall that the probability of a binomial random variable \( X \) taking a particular value \( x \) is given by the formula: \[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \]where \( n \) is the number of trials and \( p \) is the probability of success in each trial.

Step 2: Apply the Formula to X and Y

For \( X \sim B(5, \frac{1}{2}) \), the probability is: \[ P(X = x) = \binom{5}{x} \left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^{5-x} = \binom{5}{x} \left(\frac{1}{2}\right)^5 \].Similarly, for \( Y \sim B(7, \frac{1}{2}) \), \[ P(Y = y) = \binom{7}{y} \left(\frac{1}{2}\right)^y \left(\frac{1}{2}\right)^{7-y} = \binom{7}{y} \left(\frac{1}{2}\right)^7 \].

Step 3: Convolution of Two Independent Binomial Variables

When \( X \) and \( Y \) are independent, the sum \( X + Y \) follows a binomial distribution with parameters \( n = 12 \) and \( p = \frac{1}{2} \). So, \( X + Y \sim B(12, \frac{1}{2}) \).

Step 4: Calculate P(X + Y = 3)

Using the binomial formula for \( X + Y \sim B(12, \frac{1}{2}) \):\[ P(X + Y = 3) = \binom{12}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{9} = \binom{12}{3} \left(\frac{1}{2}\right)^{12} \].This simplifies to \[ P(X + Y = 3) = \binom{12}{3} \times \left(\frac{1}{4096}\right) \].

Step 5: Calculate Binomial Coefficient

Calculate the binomial coefficient \( \binom{12}{3} \): \[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \].

Step 6: Compute the Final Probability

Substitute \( \binom{12}{3} \) into the probability:\[ P(X + Y = 3) = 220 \times \frac{1}{4096} = \frac{220}{4096} = \frac{55}{1024} \].

Step 7: Match the Answer to Choices

The probability \( \frac{55}{1024} \) matches option (A).

Key Concepts

Independent Variables

In probability and statistics, independent variables are random variables that do not influence each other's outcomes. This means that the occurrence of one does not affect the probability of the occurrence of the other.
For example, if two events, A and B, are independent, then the probability of both A and B occurring is the product of their individual probabilities:

• The formula for independent variables is: \( P(A \cap B) = P(A) \times P(B) \)
  This concept is crucial in understanding the binomial distributions for independent variables.

In the case of independent binomial variables, often denoted as \( X \) and \( Y \), they can be manipulated separately when determining their probability distributions. This key feature is utilized when calculating probabilities like \( P(X + Y = k) \), where \( X \) and \( Y \) are summed but do not affect one another directly.

Probability Calculation

Calculating the probability of a specific outcome in a binomial distribution involves using a specific formula. This formula is applicable to scenarios where an event occurs a fixed number of times, each with the same probability of success.

• The probability of exactly \( k \) successes in \( n \) independent Bernoulli trials (experiments with two possible outcomes) is calculated as:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]
  Where \( \binom{n}{k} \) is the binomial coefficient representing the number of ways to choose \( k \) successes from \( n \) trials, \( p \) is the probability of success on a single trial, and \( 1-p \) is the probability of failure.

When you apply this formula, it enables you to compute the probability of achieving exactly a certain number of successes in a series of independent and identical trials.
This method was used to determine the probability for both \( X \) and \( Y \) and later on, to calculate \( P(X + Y = 3) \) by finding probabilities for individual distributions first and then summing them up.

Convolution of Distributions

The convolution of distributions gives us a way to combine two independent random variables. For two independent binomial distributions, say \( X \sim B(n_1, p) \) and \( Y \sim B(n_2, p) \), their sum \( X + Y \) itself is a binomial distribution.

• The combined distribution is given by:
  \( X + Y \sim B(n_1 + n_2, p) \)
  This reflects that the total number of trials and the probability of success remain consistent for the sum.

In the solution provided, the distribution of \( X + Y \) was simplified to \( B(12, \frac{1}{2}) \) because the sum of the trials from the individual distributions were added together.
This new distribution allowed us to calculate \( P(X + Y = 3) \) using the same binomial formula, thus simplifying the calculation by leveraging the merged distribution properties.