JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding the Mean and Variance of a Binomial Distribution

Q: Why do the binomial coefficients cancel directly?

By symmetry of combinations, binomnr = binomnn-r, so binom73 = binom77-3 = binom74 = 35.

Let X be a random variable having binomial distribution B(7, p). If P(X=3)=5 P(X=4), then the sum of the mean and the variance of X is:

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What feels right?

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A105/16B7/16C77/36D49/16

Step-by-step solutionView

Correct answer

Using the relation $P (X = 3) = 5 P (X = 4)$ , we find $p = \frac{1}{6}$ , giving a sum of mean and variance equal to $\frac{77}{36}$ .

Option analysis

Why each option works or fails

A · 105/16

Inverting the ratio of probabilities or confusing $p$ with $1 - p$ to get $p = \frac{3}{4}$ , which yields $E [X] + Var (X) = \frac{21}{4} + \frac{21}{16} = \frac{105}{16}$ . Equate $(\binom{7}{3}) p^{3} (1 - p)^{4} = 5 (\binom{7}{4}) p^{4} (1 - p)^{3}$ carefully and simplify to solve for $p$ correctly: $1 - p = 5 p$ , so $p = \frac{1}{6}$ .

B · 7/16

Calculating only the variance or confusing the parameter values, or finding a single component instead of the sum requested. Ensure you calculate the sum of both the mean $n p$ and the variance $n p (1 - p)$ rather than just one term or an incomplete combination.

C · 77/36

This is the correct option. Since $(\binom{7}{3}) = (\binom{7}{4}) = 35$ , the given relation simplifies to $(1 - p) = 5 p$ , yielding $p = \frac{1}{6}$ . Then the mean is $7 \cdot \frac{1}{6} = \frac{7}{6}$ and the variance is $7 \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{35}{36}$ , giving a sum of $\frac{42 + 35}{36} = \frac{77}{36}$ .

D · 49/16

Assuming $p = \frac{1}{4}$ or incorrectly setting up the coefficient ratio when solving the probability equation. Remember that $(\binom{7}{3}) = (\binom{7}{4})$ , so the binomial coefficients cancel directly without introducing an extra factor.

Reviewed route

Solution

StepWorking

01Given

A binomial distribution $X \sim B (n, p)$ with $n = 7$ and $P (X = 3) = 5 P (X = 4)$ .

02Goal

Find the sum of the mean and the variance of $X$ , given by $Mean + Var = n p + n p q$ .

03Approach

Use the probability mass function $P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r}$ . Note that $(\binom{7}{3}) = (\binom{7}{4})$ to simplify the equation $P (X = 3) = 5 P (X = 4)$ directly into a linear equation for $p$ . Then compute $Mean + Var = n p (1 + q)$ .

04Execute

Substitute $r = 3$ and $r = 4$ into the relation: $(\binom{7}{3}) p^{3} (1 - p)^{4} = 5 (\binom{7}{4}) p^{4} (1 - p)^{3}$ Since $(\binom{7}{3}) = (\binom{7}{4}) = 35$ , we divide both sides by $35 p^{3} (1 - p)^{3}$ (noting $0 < p < 1$ ): $1 - p = 5 p ⟹ 6 p = 1 ⟹ p = \frac{1}{6}, q = 1 - p = \frac{5}{6}$

05Execute

Compute the mean and variance: $Mean = n p = 7 \times \frac{1}{6} = \frac{7}{6}$ $Variance = n p q = 7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36}$ Sum them up: $Mean + Variance = \frac{7}{6} + \frac{35}{36} = \frac{42 + 35}{36} = \frac{77}{36}$

✓Verify

Check the condition: $P (X = 3) / P (X = 4) = (q / p) = (5 / 6) / (1 / 6) = 5$ , matching the given $P (X = 3) = 5 P (X = 4)$ . The sum $\frac{42 + 35}{36} = \frac{77}{36}$ is in simplest form and corresponds to Option C.

Hints that build this answer step by step

Using the binomial probability formula $P (X = k) = (\binom{n}{k}) p^{k} (1 - p)^{n - k}$ , what equation does $P (X = 3) = 5 P (X = 4)$ give?
$(\binom{7}{3}) p^{3} (1 - p)^{4} = 5 (\binom{7}{4}) p^{4} (1 - p)^{3}$
What is the value of $p$ obtained from solving $(\binom{7}{3}) p^{3} (1 - p)^{4} = 5 (\binom{7}{4}) p^{4} (1 - p)^{3}$ ?
$p = \frac{1}{6}$
For $n = 7$ and $p = \frac{1}{6}$ , what is the sum of the mean and the variance of $X$ ?
$\frac{77}{36}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why do the binomial coefficients cancel directly?

By symmetry of combinations, $(\binom{n}{r}) = (\binom{n}{n - r})$ , so $(\binom{7}{3}) = (\binom{7}{7 - 3}) = (\binom{7}{4}) = 35$ .

Answer

Using the relation $P (X = 3) = 5 P (X = 4)$ , we find $p = \frac{1}{6}$ , giving a sum of mean and variance equal to $\frac{77}{36}$ .

Why each option works or fails

A: 105/16 - Inverting the ratio of probabilities or confusing $p$ with $1 - p$ to get $p = \frac{3}{4}$ , which yields $E [X] + Var (X) = \frac{21}{4} + \frac{21}{16} = \frac{105}{16}$ . Equate $(\binom{7}{3}) p^{3} (1 - p)^{4} = 5 (\binom{7}{4}) p^{4} (1 - p)^{3}$ carefully and simplify to solve for $p$ correctly: $1 - p = 5 p$ , so $p = \frac{1}{6}$ .
B: 7/16 - Calculating only the variance or confusing the parameter values, or finding a single component instead of the sum requested. Ensure you calculate the sum of both the mean $n p$ and the variance $n p (1 - p)$ rather than just one term or an incomplete combination.
C · correct: 77/36 - This is the correct option. Since $(\binom{7}{3}) = (\binom{7}{4}) = 35$ , the given relation simplifies to $(1 - p) = 5 p$ , yielding $p = \frac{1}{6}$ . Then the mean is $7 \cdot \frac{1}{6} = \frac{7}{6}$ and the variance is $7 \cdot \frac{1}{6} \cdot \frac{5}{6} = \frac{35}{36}$ , giving a sum of $\frac{42 + 35}{36} = \frac{77}{36}$ .
D: 49/16 - Assuming $p = \frac{1}{4}$ or incorrectly setting up the coefficient ratio when solving the probability equation. Remember that $(\binom{7}{3}) = (\binom{7}{4})$ , so the binomial coefficients cancel directly without introducing an extra factor.

Step-by-step solution

Given: A binomial distribution $X \sim B (n, p)$ with $n = 7$ and $P (X = 3) = 5 P (X = 4)$ .
Goal: Find the sum of the mean and the variance of $X$ , given by $n p$ .
Approach: Use the probability mass function $(\binom{7}{3}) = (\binom{7}{4}) = 35$ . Note that $(\binom{7}{3}) = (\binom{7}{4})$ to simplify the equation $P (X = 3) = 5 P (X = 4)$ directly into a linear equation for $p$ . Then compute $Mean + Var = n p (1 + q)$ .
Execute: Substitute $r = 3$ and $r = 4$ into the relation: $(\binom{7}{3}) p^{3} (1 - p)^{4} = 5 (\binom{7}{4}) p^{4} (1 - p)^{3}$ Since $(\binom{7}{3}) = (\binom{7}{4}) = 35$ , we divide both sides by $35 p^{3} (1 - p)^{3}$ (noting $0 < p < 1$ ): $1 - p = 5 p ⟹ 6 p = 1 ⟹ p = \frac{1}{6}, q = 1 - p = \frac{5}{6}$
Execute: Compute the mean and variance: $Mean = n p = 7 \times \frac{1}{6} = \frac{7}{6}$ $Variance = n p q = 7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36}$ Sum them up: $Mean + Variance = \frac{7}{6} + \frac{35}{36} = \frac{42 + 35}{36} = \frac{77}{36}$
Verify: Check the condition: $P (X = 3) / P (X = 4) = (q / p) = (5 / 6) / (1 / 6) = 5$ , matching the given $P (X = 3) = 5 P (X = 4)$ . The sum $\frac{42 + 35}{36} = \frac{77}{36}$ is in simplest form and corresponds to Option C.

Finding the Mean and Variance of a Binomial Distribution

What feels right?

Using the relation $P (X = 3) = 5 P (X = 4)$ , we find $p = \frac{1}{6}$ , giving a sum of mean and variance equal to $\frac{77}{36}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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