MHT CET · Maths · Probability
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
- A 121
- B 17
- C 137
- D 1
Answer & Solution
Correct Answer
(C) 137
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& n p=8, n p q=4 \\
& \Rightarrow q=\frac{1}{2}, p=\frac{1}{2} \text { and } n=16
\end{aligned}\)
Now \(P(x \leq 2)=n_{C_0} p^0 q^n+n_{C_1} p^1 q^{n-1}+n_{C_2} p^2 q^{n-2}\)
\(\begin{aligned}
& =\left\{{ }^{16} C_0+{ }^{16} C_1+{ }^{16} C_2\right\}\left(\frac{1}{2}\right)^{16} \\
& =\frac{137}{2^{16}} \\
& \Rightarrow k=137
\end{aligned}\)
& n p=8, n p q=4 \\
& \Rightarrow q=\frac{1}{2}, p=\frac{1}{2} \text { and } n=16
\end{aligned}\)
Now \(P(x \leq 2)=n_{C_0} p^0 q^n+n_{C_1} p^1 q^{n-1}+n_{C_2} p^2 q^{n-2}\)
\(\begin{aligned}
& =\left\{{ }^{16} C_0+{ }^{16} C_1+{ }^{16} C_2\right\}\left(\frac{1}{2}\right)^{16} \\
& =\frac{137}{2^{16}} \\
& \Rightarrow k=137
\end{aligned}\)
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