Probability 5 Question 2
2. Let a random variable $X$ have a binomial distribution with mean 8 and variance 4 . If $P(X \leq 2)=\frac{k}{2^{16}}$, then $k$ is equal to
(2019 Main, 12 April I)
(a) 17
(b) 121
(c) 1
(d) 137
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Answer:
Correct Answer: 2. (d)
Solution:
- Let for the given random variable ’ $X$ ’ the binomial probability distribution have $n$-number of independent trials and probability of success and failure are $p$ and $q$ respectively. According to the question, Mean $=n p=8$ and variance $=n p q=4$
$$ \begin{aligned} & \therefore \quad q=\frac{1}{2} \Rightarrow p=1-q=\frac{1}{2} \\ & \text { Now, } \quad n \times \frac{1}{2}=8 \Rightarrow n=16 \\ & P(X=r)=^{16} C _r \frac{1}{2}^{16} \\ & \therefore \quad P(X \leq 2)=P(X=0)+P(X=1)+P(X=2) \\ & { }^{16} C _0 \frac{1}{2}^{16}+{ }^{16} C _1 \frac{1}{2}^{16}+{ }^{16} C _2 \frac{1}{2}^{16} \\ & =\frac{1+16+120}{2^{16}}=\frac{137}{2^{16}}=\frac{k}{2^{16}} \\ & \Rightarrow \quad k=137 \end{aligned} $$
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