Problem 75 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(X\) and \(Y\) are the independent random variables for \(B\left(5, \frac{1}{2}\right)\) and \(B\left(7, \frac{1}{2}\right)\), then \(P(X+Y \geq 1)=\) (A) \(\frac{4095}{4096}\) (B) \(\frac{309}{4096}\) (C) \(\frac{4032}{4096}\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (A) \(\frac{4095}{4096}\).

Step 1: Understanding the Problem

We have two independent random variables, following binomial distributions: \(X \sim B\left(5, \frac{1}{2}\right)\) and \(Y \sim B\left(7, \frac{1}{2}\right)\). We need to find the probability \(P(X+Y \geq 1)\).

Step 2: Calculate Probability of Complement Event

The complement of \(P(X+Y \geq 1)\) is \(P(X+Y = 0)\). Since both \(X\) and \(Y\) are binomial distributions, the probability that both are zero is given by \(P(X = 0) \cdot P(Y = 0)\).

Step 3: Calculate \(P(X = 0)\) and \(P(Y = 0)\)

\(P(X = 0) = \left(\frac{1}{2}\right)^5 = \frac{1}{32}\) and \(P(Y = 0) = \left(\frac{1}{2}\right)^7 = \frac{1}{128}\).

Step 4: Compute \(P(X+Y = 0)\)

Since \(X\) and \(Y\) are independent, \(P(X+Y = 0) = P(X = 0) \cdot P(Y = 0) = \frac{1}{32} \cdot \frac{1}{128} = \frac{1}{4096}\).

Step 5: Calculate \(P(X+Y \geq 1)\)

Using the complement rule, \(P(X+Y \geq 1) = 1 - P(X+Y = 0) = 1 - \frac{1}{4096} = \frac{4095}{4096}\).

Step 6: Conclusion: Selecting the Correct Option

The probability \(P(X+Y \geq 1)\) is \(\frac{4095}{4096}\), which corresponds to option (A).

Key Concepts

Independent Random Variables

Random variables are key components in probability theory and statistics. A random variable is independent if the outcome of one does not affect the outcome of the other. In simpler terms, two events are independent if the occurrence of one does not change the probability of the other. In the context of our exercise, both random variables \(X\) and \(Y\) are independent.

• \(X\) follows a binomial distribution with parameters 5 and \(\frac{1}{2}\), denoted as \(B(5, \frac{1}{2})\).
• \(Y\) also follows a binomial distribution with parameters 7 and \(\frac{1}{2})\), denoted as \(B(7, \frac{1}{2})\).

The significance of independence here is crucial because it allows us to use certain mathematical properties, such as multiplying probabilities when calculating a joint probability for independent events. This simplifies calculations and makes solving problems much quicker.

Complement Rule

In probability, understanding the complement of an event can often make problems much simpler. The complement of an event \(A\), denoted as \(A'\), is composed of all the outcomes that are not in \(A\). The main principle is that the probability of an event plus the probability of its complement always sums to 1: \[P(A) + P(A') = 1\]In our problem, we are interested in finding \(P(X+Y \geq 1)\). What we do instead is first find the complement, \(P(X+Y = 0)\). By using the formula:\[P(X+Y \geq 1) = 1 - P(X+Y = 0)\]We effectively transform the problem into a simpler version, as \(P(X+Y = 0)\) is easier to calculate. This approach is particularly helpful when finding probabilities of at least one happening, especially when directly counting all possible outcomes is cumbersome.

Probability Calculation

The process of probability calculation involves determining the likelihood of an event happening. In this exercise, our main goal was to determine \(P(X+Y \geq 1)\). Let's break it down step by step:1. **Identify the Complement**:- We found the complement, \(P(X+Y = 0)\), since it is simpler and involves calculating zero cases in both distributions.
2. **Calculate Individual Probabilities**:- For \(P(X = 0)\), the probability is calculated as \(\left(\frac{1}{2}\right)^5 = \frac{1}{32}\).- For \(P(Y = 0)\), it becomes \(\left(\frac{1}{2}\right)^7 = \frac{1}{128}\).
3. **Use Independence Property**:- Multiply individual probabilities: \[ P(X+Y = 0) = \frac{1}{32} \times \frac{1}{128} = \frac{1}{4096} \]
4. **Apply the Complement Rule**:- Finally calculate:\[ P(X+Y \geq 1) = 1 - \frac{1}{4096} = \frac{4095}{4096} \]It's through these systematic steps that we're able to effectively solve problems involving binomial distributions and independent random variables.