Solveeit Logo
Login

Question

Mathematics Question on binomial distribution

Let X be a binomially distributed random variable with mean 4 and variance 4/3. Then, 54 P(X≤ 2) is equal to

A

7327\frac{73}{27}

B

14627\frac{146}{27}

C

14681\frac{146}{81}

D

12681\frac{126}{81}

Answer

14627\frac{146}{27}

Explanation

Solution

Mean = 4 = μ = np
 Variance=σ2=np(1−P)=43\ Variance =σ^2=np(1−P)=\frac{4}{3}
4(1−P)=434(1−P)=\frac{4}{3}
P=23P=\frac{2}{3}
n×23=4n×\frac{2}{3}=4
n=6
P(X=k)= nCkPk(1−P)n−kP(X=k)= \ ^nC_kP^k(1−P)^{n−k}
P(X≤2)=P(X=0)+P(X=1)+P(X=2)P(X≤2)=P(X=0)+P(X=1)+P(X=2)
=6C0P0(1−p)6+6C1P1(1−P)5+6C2P2(1−P)4= ^6C_0P^0(1−p)^6+ ^6C_1P^1(1−P)^5+ ^6C_2P^2(1−P)^4
=6C0(13)6+6C1(23)(13)5+6C2(23)2(13)4= ^6C_0(\frac{1}{3})^6+ ^6C_1(\frac{2}{3})(\frac{1}{3})^5+ ^6C_2(\frac{2}{3})^2(\frac{1}{3})^4
=(13)6[1+12+60]=7336=(\frac{1}{3})^6[1+12+60]=\frac{73}{3^6}
54P(X≤2)54P(X ≤ 2 )
=7336×54=14627= \frac{73}{3^6}\times 54 = \frac{146}{27}