JEE Main 2022 · 26 Jul · Shift 2 · Mathematics Q18

MathematicsSingle correct+4 / -1Question 18 of 90

Let $X$ be a binomially distributed random variable with mean 4 and variance $\dfrac43$ . Then, $54\,P(X\le2)$ is equal to

\dfrac{73}{27}

\dfrac{146}{27}

\dfrac{146}{81}

\dfrac{126}{81}

Answer and worked solution

Official answer: B

Parameters. $np=4$ , $npq=\frac43$ : $q=\frac13$ , $p=\frac23$ , $n=6$ .

$P(X\le2)$ $=\frac{1+6\cdot2+15\cdot4}{729}=\frac{73}{729}$ , so $54P=\frac{146}{27}$ .

(B)

Topics:Binomial distribution

JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 18