The mean and variance of a binomial distribution are α and (α/3) respectively. If P(X=1)=(4/243), then P(X=4 or 5) is equal to :

Question

The mean and variance of a binomial distribution are α and (α/3) respectively. If P(X=1)=(4/243), then P(X=4 or 5) is equal to : (A) (5/9) (B) (64/81) (C) (16/27) (D) (145/243)

Tardigrade Admin · Accepted Answer

n p =α .... (1) npq =α / 3.....(2) From (1) (2) q =1 / 3 p =2 / 3 n C 1 q n -1 p 1=(4/243) ( n /3 n )=(2/243) n =6 P (4 text or 5) = 6 C 4((2/3))4((1/3))2+ 6 C 5((2/3))5 ⋅((1/3))0 =(16/27)

Q. The mean and variance of a binomial distribution are $α$ and $\frac{α}{3}$ respectively. If $P (X = 1) = \frac{4}{243}$ , then $P (X = 4$ or 5 $)$ is equal to :

$\frac{5}{9}$

$\frac{64}{81}$

$\frac{16}{27}$

$\frac{145}{243}$

$n p = α ....$ (1)
$n pq = α /3.....$ (2)
From (1) & (2)
$q = 1/3& p = 2/3$
$^{n} C_{1} q^{n - 1} p^{1} = \frac{4}{243}$
$\frac{n}{3 ^{n}} = \frac{2}{243}$
$n = 6$
$P (4 or 5)$
$=^{6} C_{4} (\frac{2}{3})^{4} (\frac{1}{3})^{2} +^{6} C_{5} (\frac{2}{3})^{5} \cdot (\frac{1}{3})^{0}$
$= \frac{16}{27}$

Q. The mean and variance of a binomial distribution are α and 3α​ respectively. If P(X=1)=2434​, then P(X=4 or 5) is equal to :

95​

8164​

2716​

243145​

np=α.... (1) npq=α/3.....(2) From (1) & (2) q=1/3&p=2/3 nC1​qn−1p1=2434​ 3nn​=2432​ n=6 P(4 or 5) =6C4​(32​)4(31​)2+6C5​(32​)5⋅(31​)0 =2716​

Q. The mean and variance of a binomial distribution are $α$ and $\frac{α}{3}$ respectively. If $P (X = 1) = \frac{4}{243}$ , then $P (X = 4$ or 5 $)$ is equal to :

$\frac{5}{9}$

$\frac{64}{81}$

$\frac{16}{27}$

$\frac{145}{243}$