JEE PYQ: Permutation And Combination Question 44
Question 44 - 2019 (11 Jan Shift 2)
A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If $X$ be the number of white balls drawn, then $\left(\frac{\text{mean of } X}{\text{standard deviation of } X}\right)$ is equal to:
(1) $4$
(2) $4\sqrt{3}$
(3) $3\sqrt{2}$
(4) $\frac{4\sqrt{3}}{3}$
Show Answer
Answer: (2)
Solution
$P(\text{white}) = \frac{30}{40} = \frac{3}{4}$, $n = 16$. Mean $= np = 12$. SD $= \sqrt{npq} = \sqrt{16 \times \frac{3}{4} \times \frac{1}{4}} = \sqrt{3}$. Ratio $= \frac{12}{\sqrt{3}} = 4\sqrt{3}$.
BETA
