AP EAMCET · Maths · Probability
7 coins are tossed simultaneously and the number of heads turned up is denoted by the random variable X . If \(\mu\) is the mean and \(\sigma^2\) is the variance of \(X\), then \(\frac{\mu \sigma^2}{P(X=3)}=\)
- A \(\frac{56}{5}\)
- B \(\frac{84}{5}\)
- C \(\frac{112}{5}\)
- D \(\frac{224}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{112}{5}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } n=7, p=\frac{1}{2}, q=\frac{1}{2}, \mu=n p=\frac{7}{2} \\ & \sigma^2=n p q=\frac{7}{4} \\ & p(X=3)={ }^7 \mathrm{C}_3 \times\left(\frac{1}{2}\right)^4 \times\left(\frac{1}{2}\right)^3 \\ & \frac{\mu \sigma^2}{p(X=3)}=\frac{7 \times 7 \times 2^7}{2…
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