Problem 56 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is: \(\quad\) Online April 10, 2016] (a) \(\frac{496}{729}\) (b) \(\frac{192}{729}\) (c) \(\frac{240}{729}\) (d) \(\frac{256}{729}\)

Step-by-Step Solution

Answer

The correct answer is (d) \( \frac{256}{729} \).

Step 1: Understand the Problem

We are given an experiment that succeeds twice as often as it fails. We need to find the probability of at least 5 successes in 6 trials. The problem supplies multiple choice answers.

Step 2: Define Probability of Success and Failure

Let the probability of success be \( p \) and the probability of failure be \( q \). Since the experiment succeeds twice as often as it fails, \( p = 2q \). Also, \( p + q = 1 \). Solving the equations gives \( p = \frac{2}{3} \) and \( q = \frac{1}{3} \).

Step 3: Define the Probability Distribution

This is a binomial distribution problem where \( n = 6 \) trials and the probability of success is \( p = \frac{2}{3} \). The probability of getting exactly \( k \) successes is given by the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]

Step 4: Calculate Probability of Exactly 5 Successes

Calculate the probability of getting exactly 5 successes using:\[ P(X = 5) = \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^1 \]This simplifies to:\[ P(X = 5) = 6 \times \left(\frac{32}{243}\right) \times \frac{1}{3} = \frac{192}{729} \]

Step 5: Calculate Probability of Exactly 6 Successes

Calculate the probability of getting exactly 6 successes using:\[ P(X = 6) = \binom{6}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^0 \]This simplifies to:\[ P(X = 6) = 1 \times \frac{64}{729} = \frac{64}{729} \]

Step 6: Calculate Probability of At Least 5 Successes

The probability of at least 5 successes in 6 trials is the sum of the probabilities of exactly 5 and exactly 6 successes:\[ P(X \geq 5) = P(X = 5) + P(X = 6) = \frac{192}{729} + \frac{64}{729} = \frac{256}{729} \]

Step 7: Verify with Options

Compare our calculated result \( \frac{256}{729} \) with the given options: (a) \( \frac{496}{729} \), (b) \( \frac{192}{729} \), (c) \( \frac{240}{729} \), (d) \( \frac{256}{729} \). The solution matches option (d).

Key Concepts

Binomial Distribution

The Binomial Distribution is a statistical method used to model the number of successes in a fixed number of trials. Each trial is independent, with only two possible outcomes: success or failure.
In this scenario, the number of trials is six, and each trial is trying to determine if the experiment succeeds or fails. Binomial distribution is perfect for this type of problem because it handles discrete events that have binary outcomes.
The distribution is defined by two parameters:

• The total number of trials, denoted by \( n \).
• The probability of success in each trial, denoted by \( p \).

The probability of getting exactly \( k \) successes in \( n \) trials is given by the formula:\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \]where \( \binom{n}{k} \) is a binomial coefficient that calculates the number of ways to choose \( k \) successes from \( n \) trials.

Success and Failure Probability

Success and Failure Probability are two fundamental elements in binomial experiments.
Each trial has a certain probability of resulting in a success or failure. These probabilities are fixed across all trials.
In the given exercise, it is mentioned that the experiment succeeds twice as often as it fails.

• This means that if the probability of failure is \( q \), then the probability of success \( p \) would be \( 2q \).

Additionally, since the sum of success and failure probabilities must equal one, we can solve these equations to find that \( p = \frac{2}{3} \) and \( q = \frac{1}{3} \).
This relationship highlights the dependence of success and failure probabilities on each other.

Combinatorics

Combinatorics plays a critical role in calculating probabilities in the binomial distribution. It deals with counting, arranging, and finding patterns.
In the binomial distribution formula, the term \( \binom{n}{k} \) represents the binomial coefficient.

• This coefficient counts how many different combinations of \( k \) successes can occur in \( n \) trials.

For instance, if you have six trials and you want to find the probability of exactly five successes, \( \binom{6}{5} \) provides the number of distinct arrangements in which you can achieve this.
The formula for the binomial coefficient is given by:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where \( n! \) is n factorial, representing the product of all positive integers up to \( n \).

Probability of Success

The Probability of Success is crucial in determining the outcomes of a binomial experiment.
It defines the likelihood that a single trial results in success. This probability remains constant for each trial within an experiment.
In the example given, the probability of success, \( p \), is \( \frac{2}{3} \).
This means each trial has a 66.67% chance of being successful, reflecting the condition that success occurs twice as often as failure.

• The higher \( p \) is, the more likely you are to get a greater number of successes.

Using the probability of success in combination with the binomial formula enables the calculation of the probability of attaining any specified number of successes across multiple trials.

Probability of Failure

The Probability of Failure complements the probability of success in a binomial distribution.
It represents the likelihood that a single trial does not succeed. For this experiment, the probability of failure, \( q \), is \( \frac{1}{3} \).
This means there is a 33.33% chance of failure in each trial.
It is essential to remember that the sum of the probabilities of success and failure must equal one:

• \( p + q = 1 \)

This relationship ensures that all possible outcomes are accounted for.
In our example, given that success is twice as likely as failure, \( q \) is inherently half of \( p \).
Understanding this balance helps in setting up the equations for each problem scenario.