JEE Main 2022 · 26 Jul · Shift 1 · Mathematics Q16

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MathematicsSingle correct+4 / -1Question 16 of 90

The mean and variance of a binomial distribution are $\alpha$ and $\dfrac\alpha3$ respectively. If $P(X=1)=\dfrac4{243}$ , then $P(X=4\text{ or }5)$ is equal to :

\dfrac59

\dfrac{64}{81}

\dfrac{16}{27}

\dfrac{145}{243}

Answer and worked solution

Official answer: C

q $=\frac13$ , $p=\frac23$ . $n\cdot\frac23\cdot3^{-(n-1)}=\frac4{243}$ gives $n=6$ .

P(4 or 5) $=\frac{15\cdot16+6\cdot32}{729}=\frac{432}{729}=\frac{16}{27}$ .

(C)

Topics:Binomial distribution

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JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 16