Problem 68 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is (a) \(8 / 3\) (b) \(3 / 8\) (c) \(4 / 5\) (d) \(5 / 4\)

Step-by-Step Solution

Answer

The variance of the distribution of success is \(\frac{5}{4}\) (option d).

Step 1: Determine the Probability of Success

When a standard 6-sided dice is rolled, the possible outcomes are 1, 2, 3, 4, 5, and 6. Among these, the odd numbers are 1, 3, and 5. Thus, there are 3 odd numbers out of 6 possible outcomes. Therefore, the probability of getting an odd number (success) on a single toss is \( P(\text{success}) = \frac{3}{6} = \frac{1}{2} \).

Step 2: Identify the Binomial Experiment

Tossing the dice 5 times can be framed as a binomial experiment with each toss being an independent trial. Here, each toss has two possible outcomes - success (getting an odd number) or failure (getting an even number). Thus, we can denote the number of successes in 5 trials as a binomial random variable \( X \) with parameters \( n = 5 \) (number of trials) and \( p = \frac{1}{2} \) (probability of success).

Step 3: Calculate the Variance of the Binomial Distribution

The formula for the variance of a binomial distribution is given by \( \text{Var}(X) = n \cdot p \cdot (1 - p) \). Substituting in the values from our problem, \( n = 5 \) and \( p = \frac{1}{2} \), we get:\[ \text{Var}(X) = 5 \cdot \frac{1}{2} \cdot \left(1 - \frac{1}{2}\right) = 5 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{5}{4}. \]

Step 4: Compare and Choose the Correct Answer

Review the provided options:(a) \(\frac{8}{3}\),(b) \(\frac{3}{8}\),(c) \(\frac{4}{5}\),(d) \(\frac{5}{4}\).From our calculation, the variance is \( \frac{5}{4} \). Therefore, the correct answer is option (d): \( \frac{5}{4} \).

Key Concepts

Probability of Success

The probability of success is a fundamental concept in statistics, especially in binomial distribution. It represents the likelihood of a desired outcome occurring in a single trial.

In the context of a binomial experiment, such as tossing a dice, we define 'success' based on the criteria we set. For instance, when we toss a standard 6-sided die and consider rolling an odd number as a success, what is the probability that we will achieve success in one toss?

Given a standard die, the possible outcomes are: 1, 2, 3, 4, 5, and 6. Among these, 1, 3, and 5 are odd numbers, which are considered a success based on our criteria.

• Number of successful outcomes (odd numbers): 3
• Total possible outcomes: 6

Thus, the probability of success, which is the chance of rolling an odd number, is calculated as:
\[ P( ext{success}) = \frac{3}{6} = \frac{1}{2} \].

This means there's a 50% chance of rolling an odd number on a single toss of the die.

Variance of Distribution

Variance in a distribution provides insight into how much the outcomes differ from the expected value. In a binomial distribution, it describes how much the success count will fluctuate in repeated experiments.

The variance for a binomial distribution with parameters \( n \) (number of trials) and \( p \) (probability of success) is given by the formula:
\[ \text{Var}(X) = n \cdot p \cdot (1 - p) \].

For our dice tossing problem:

• Number of trials \( n = 5 \)
• Probability of success \( p = \frac{1}{2} \)

Substituting these values into the variance formula, we get:
\[ \text{Var}(X) = 5 \cdot \frac{1}{2} \cdot \left(1 - \frac{1}{2}\right) = 5 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{5}{4} \].

This means that in 5 tosses of the die, the number of odd numbers we can expect to get will have a variance of \( \frac{5}{4} \). This value indicates the variability in the number of successes across different trials.

Binomial Experiment

A binomial experiment consists of a series of independent trials, where each trial results in either a 'success' or 'failure.'

To identify an experiment as binomial, it must meet certain criteria:

• The number of trials \( n \) is fixed.
• Each trial is independent of the others.
• Each trial has exactly two possible outcomes: success or failure.
• The probability of success \( p \) remains constant for each trial.

In the case of our dice problem, we are asked to toss the dice 5 times, which serves as a perfect illustration of a binomial experiment.

Here:

• The number of trials, \( n = 5 \), is fixed.
• Each toss of the die is independent of the others— the outcome of one does not influence another.
• Each toss results in either rolling an odd number (success) or not (failure).
• We calculated that the probability of success, \( p = \frac{1}{2} \), remains the same for each toss.

Thus, tossing the dice multiple times fits perfectly within the framework of a binomial experiment, making it easier to apply binomial probability concepts to find solutions.