Random Variables and Binomial Distribution | NIMCET

Quick answer: A random variable X assigns a number to each outcome. Its mean is $E(X) = \sum x_ip_i$ and its variance is $\text{Var}(X) = E(X^2) - [E(X)]^2$. In a binomial distribution with n independent trials and success probability p, $P(X = r) = {}^nC_r\,p^rq^{n - r}$ where $q = 1 - p$, with mean $np$ and variance $npq$.

Syllabus note: Several admission portals report that random variables and the binomial distribution were dropped from the revised NIMCET syllabus. Check the official information brochure for your exam year; our revised syllabus update has the details. The topic remains useful for older papers and other entrance exams.

This builds on conditional probability and independent events.

Random variables and distributions

A discrete random variable takes countable values $x_1, x_2, \ldots$ with probabilities $p_1, p_2, \ldots$, where every $p_i \ge 0$ and $\sum p_i = 1$.

Example: X = number of heads in 2 tosses.

Mean and variance

For the two-toss example: $E(X) = 1$, $E(X^2) = \frac{1}{2} + 1 = \frac{3}{2}$, $\text{Var}(X) = \frac{1}{2}$.

Binomial distribution

Conditions (Bernoulli trials):

• a fixed number n of trials,
• each trial has two outcomes, success or failure,
• the trials are independent,
• the success probability p is the same in every trial.

$P(X = r) = {}^nC_r\,p^r\,q^{n - r}, \quad r = 0, 1, \ldots, n$

The probabilities are the terms of the binomial expansion of $(q + p)^n$.

Worked examples

Example 1: exact number of successes

A fair coin is tossed 6 times. Find the probability of exactly 4 heads.

$^6C_4\left(\frac{1}{2}\right)^6 = \frac{15}{64}$

Example 2: at least one

A die is rolled 4 times. Find the probability of at least one six.

$1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296}$

Example 3: finding n and p

A binomial distribution has mean 4 and variance 3. Find n and p.

Step 1: $q = \frac{npq}{np} = \frac{3}{4}$, so $p = \frac{1}{4}$.

Step 2: $n = \frac{4}{1/4} = 16$.

Example 4: an impossible pair

Can a binomial distribution have mean 3 and variance 4?

No. That would need $q = \frac{4}{3} > 1$.

Example 5: expectation of a game

A player wins ₹10 if a die shows 6 and loses ₹2 otherwise. Find the expected gain.

$E = 10 \cdot \frac{1}{6} - 2 \cdot \frac{5}{6} = \frac{10 - 10}{6} = 0$

Summary table

Exam tips

• Check the four Bernoulli conditions before using the binomial formula.
• Divide variance by mean to get q directly.
• For "at least one", use $1 - q^n$.

See all topics in the NIMCET Probability and Statistics chapter.

Common mistakes

• Writing variance as $np$ or as $\sqrt{npq}$.
• Forgetting $^nC_r$ in the binomial formula.
• Using the binomial formula for draws without replacement, where trials are not independent.
• Writing $\text{Var}(aX + b) = a\,\text{Var}(X) + b$.

Frequently Asked Questions

What is the formula for the binomial distribution?

P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, where q = 1 − p.

What are the mean and variance of a binomial distribution?

The mean is np and the variance is npq.

How do you find the mean of a random variable?

Multiply each value by its probability and add: E(X) = Σ xᵢpᵢ.

Can the variance of a binomial distribution exceed its mean?

No. Since q < 1, npq is always less than np.

Is the binomial distribution in the NIMCET syllabus?

Several portals report it was removed in the revised syllabus. Confirm with the official information brochure for your exam year.