Problem 105 - A Complete Resource Book in Mathematics for JEE Main 2017

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 answer is (A) \(\frac{55}{1024}\).

Step 1: Identify Distribution Parameters

First, note that we have two binomial distributions: \(X\sim B\left(5, \frac{1}{2}\right)\) and \(Y\sim B\left(7, \frac{1}{2}\right)\). This means that \(X\) is a binomial random variable with 5 trials and probability of success \(\frac{1}{2}\), and \(Y\) is a binomial random variable with 7 trials and probability of success \(\frac{1}{2}\).

Step 2: Recognize the Sum of Two Binomials

Since \(X\) and \(Y\) are independent random variables, the sum \(X + Y\) is also binomially distributed with \(n = 5 + 7 = 12\) and \(p = \frac{1}{2}\). Therefore, \(X+Y\sim B\left(12,\frac{1}{2}\right)\).

Step 3: Use the Binomial Probability Formula

For a binomial distribution, the probability mass function is given by:\[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\] Here, we want to calculate \(P(X+Y=3)\), which translates to \(P(Z=3)\) where \(Z\sim B(12, \frac{1}{2})\). Thus, the formula becomes:\[P(Z=3) = \binom{12}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{9}\] Simplifying yields:\[P(Z=3) = \binom{12}{3} \left(\frac{1}{2}\right)^{12}\]

Step 4: Calculate \(\binom{12}{3}\)

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

Step 5: Substitute and Solve

Substitute the value from Step 4 back into the probability calculation:\[P(Z=3) = 220 \times \left(\frac{1}{2}\right)^{12} = 220 \times \frac{1}{4096}\]Now, simplify:\[P(Z=3) = \frac{220}{4096} = \frac{55}{1024}\]

Step 6: Answer the Question

We find that \(P(X+Y=3) = \frac{55}{1024}\). Therefore, the answer is choice (A).

Key Concepts

Independent Random Variables

In probability theory, when we say two random variables are independent, it means the outcome of one does not affect the outcome of the other. This can be a bit like tossing two separate coins: the result of the first toss doesn’t change the likelihood of heads or tails in the second toss.
For the variables \(X\) and \(Y\) in this exercise, being independent means that the trials in \(X\sim B(5, \frac{1}{2})\) do not influence the trials in \(Y\sim B(7, \frac{1}{2})\).

• Each variable comes from a binomial distribution with its own set of trials.
• The outcome of trials in \(X\) is unaffected by the trials in \(Y\), and vice versa.

This property is crucial for our calculations because it allows us to combine these two distributions. We can sum \(X\) and \(Y\) to produce a new variable, \(X+Y\), also a binomial distribution because of their independence.
As a result, we model their sum as a single binomial distribution with the sum of their trials and the same success probability.

Probability Mass Function

The probability mass function, or PMF, provides the probability of each possible value of a discrete random variable. For a binomial distribution, the PMF determines the likelihood of observing exactly \(k\) successes in \(n\) trials with a fixed probability of success \(p\) for each trial.
The PMF formula for a binomially distributed random variable \(X\sim B(n, p)\) is:
\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]

• \(\binom{n}{k}\) is the binomial coefficient, calculating the number of ways to choose \(k\) successes out of \(n\) trials.
• \(p^k\) represents the probability of \(k\) successes.
• \( (1-p)^{n-k}\) calculates the probability of \(n-k\) failures.

In the problem, we need the probability of \(X+Y=3\), factoring in the sum of independent variables, transforming our problem to finding \(P(Z=3)\) for \(Z\sim B(12, \frac{1}{2})\).
By plugging into the PMF, we see that it allows for calculating precise probabilities for these types of random events.

Binomial Coefficient

The binomial coefficient is a key component in calculating probabilities in a binomial distribution. It is represented as \(\binom{n}{k}\) and tells us in how many different ways \(k\) successes can occur in \(n\) trials, regardless of order.
It's calculated using the following formula:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

• \(!\) denotes a factorial, which is the product of an integer and all the integers below it.
• It's crucial for determining the number of favorable sequences of success and failure in binomial settings.

In our exercise, the binomial coefficient \(\binom{12}{3}\) calculated as \(\frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\), accounts for the various possible outcomes that result in exactly 3 successes across 12 trials.
By understanding the use of the binomial coefficient, students can see how it factors into larger calculations, making the mathematics behind binomial distributions work smoothly.