Q48 / 90 · JEE Main 2022 (26 Jul Shift 2) · Mathematics

Let $X$ be a binomially distributed random variable with mean $4$ and variance $\frac{4}{3}$ . Then $54 P (X \leq 2)$ is equal to

A
$\frac{73}{27}$
Correct answer
B
$\frac{146}{27}$
C
$\frac{146}{81}$
D
$\frac{126}{81}$

Solution

Mean $n p = 4$
Variance $n p q = \frac{4}{3}$
i.e. $n = 6, p = \frac{2}{3}, q = \frac{1}{3}$
Now $54 P (X \leq 2) = 54 [P (X = 2) + P (X = 1) + P (X = 0)]$
$= 54 ({}^{6}C_{2} {(\frac{2}{3})}^{2} {(\frac{1}{3})}^{4} + {}^{6}C_{1} {(\frac{2}{3})}^{1} {(\frac{1}{3})}^{5} + {}^{6}C_{0} {(\frac{2}{3})}^{0} {(\frac{1}{3})}^{6})$
$= 54 (15 \times \frac{4}{3^{6}} + 6 \times \frac{2}{3^{6}} + \frac{1}{3^{6}}) = 2 \times 3^{3} (\frac{73}{3^{6}}) = \frac{146}{27}$

Hence, the correct answer is option: 'B'.

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Let X be a binomially distributed random variable with mean 4 and variance 43. Then 54 PX≤2 is equal to

Solution

Relevant Questions

Let $X$ be a binomially distributed random variable with mean $4$ and variance $\frac{4}{3}$ . Then $54 P (X \leq 2)$ is equal to