If $X$ is a binomial variate with mean 6 and variance 2 , then the value of $P(5 \leq X \leq 7)$ is

If $X$ is a binomial variate with mean 6 and variance 2 , then the value of $P(5 \leq X \leq 7)$ is
  1. $\frac{4762}{6561}$
  2. $\frac{4672}{6561}$
  3. $\frac{5264}{6561}$
  4. $\frac{5462}{6651}$

Solution

We have, $ \begin{aligned} & n p=6 \text { and } n p q=2 \\ & \Rightarrow \quad 6 \times q=2 \\ & \Rightarrow \quad q=\frac{1}{3} \\ & \Rightarrow \quad p=1-q=1-\frac{1}{3}=\frac{2}{3} \\ & \Rightarrow \quad n \times \frac{2}{3}=6 \\ & \Rightarrow \quad n=9 \\ & \therefore P(5 \leq x \leq 7)=P(x=5)+P(x=6)+P(x=7) \\ & ={ }^9 C_5\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right)^4+{ }^9 C_6\left(\frac{2}{3}\right)^6\left(\frac{1}{3}\right)^3+{ }^9 C_7\left(\frac{2}{3}\right)^7\left(\frac{1}{3}\right)^2 \\ & =\frac{1}{3^9}\left[\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times 32+\frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 64+\frac{9 \times 8}{2 \times 1} \times 128\right] \\ & =\frac{1}{3^9}[9 \times 8 \times 7 \times 2 \times 4+3 \times 8 \times 7 \times 32+9 \times 4 \times 128] \\ & =\frac{3 \times 8 \times 7 \times 2 \times 4+8 \times 7 \times 32+3 \times 4 \times 128}{3^8} \\ & =\frac{1344+1792+1536}{6561}=\frac{4672}{6561} . \\ & \end{aligned} $

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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