$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
$$P(X=2) = \binom{5}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^3 = 10 \cdot \frac{1}{9} \cdot \frac{8}{27} = \frac{80}{243}$$
Back to JEE Main 2024 Probability & Statistics
JEE Main 2024 — Mathematics Probability & Statistics
medium
mcq
2024
Official previous-year question
Verified 30 May 2026.
Question
In a binomial distribution with $n = 5$ and $p = \frac{1}{3}$, the probability of exactly $2$ successes is:
Options
- A
$\binom{5}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^3$
- B
$\binom{5}{2}\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2$
- C
$\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^3$
- D
$5\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^3$
Solution
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