Question:
In a shelf there are three mathematics and two physics books. A student takes a book randomly. If he randomly takes, successively for three times by replacing the book already taken every time, then the mean of the number of mathematics books which is treated as random variable is
In a shelf there are three mathematics and two physics books. A student takes a book randomly. If he randomly takes, successively for three times by replacing the book already taken every time, then the mean of the number of mathematics books which is treated as random variable is
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Recognizing the type of probability distribution is the key to solving such problems quickly. Look for the characteristics of a binomial distribution: fixed number of independent trials, only two outcomes per trial, and constant probability of success. The mean is simply $np$.
Updated On: Mar 30, 2026
- 3/2
- 129/125
- 9/5
- 174/125
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The Correct Option is C
Solution and Explanation
Step 1: Identify the type of probability distribution.
The experiment has a fixed number of trials $n = 3$.
Each trial has two outcomes: picking a mathematics book (success) or picking a physics book (failure).
The trials are independent because the book is replaced after each selection.
The probability of success is constant across trials.
Hence, this is a binomial distribution.
Step 2: Define the parameters of the binomial distribution.
Number of trials: $n = 3$.
Let a success be picking a mathematics book.
Total books = 3 (Math) + 2 (Physics) = 5.
Probability of success in a single trial: \[ p = P(\text{Math book}) = \frac{3}{5}. \]
Random variable $X$ = number of mathematics books selected in 3 trials.
Possible values: $X \in \{0,1,2,3\}$.
Step 3: State the formula for the mean of a binomial distribution.
For a binomial distribution, the mean is: \[ \mu = E(X) = n p. \]
Step 4: Calculate the mean.
Substitute the values of $n$ and $p$: \[ \mu = 3 \times \frac{3}{5} = \frac{9}{5}. \]
The experiment has a fixed number of trials $n = 3$.
Each trial has two outcomes: picking a mathematics book (success) or picking a physics book (failure).
The trials are independent because the book is replaced after each selection.
The probability of success is constant across trials.
Hence, this is a binomial distribution.
Step 2: Define the parameters of the binomial distribution.
Number of trials: $n = 3$.
Let a success be picking a mathematics book.
Total books = 3 (Math) + 2 (Physics) = 5.
Probability of success in a single trial: \[ p = P(\text{Math book}) = \frac{3}{5}. \]
Random variable $X$ = number of mathematics books selected in 3 trials.
Possible values: $X \in \{0,1,2,3\}$.
Step 3: State the formula for the mean of a binomial distribution.
For a binomial distribution, the mean is: \[ \mu = E(X) = n p. \]
Step 4: Calculate the mean.
Substitute the values of $n$ and $p$: \[ \mu = 3 \times \frac{3}{5} = \frac{9}{5}. \]
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