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Question: If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that...

If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1, is

A

23\frac { 2 } { 3 }

B

45\frac { 4 } { 5 }

C

78\frac { 7 } { 8 }

D

1516\frac { 15 } { 16 }

Answer

1516\frac { 15 } { 16 }

Explanation

Solution

We have mean (X)=np=2( X ) = n p = 2

and variance ⇒ q=12q = \frac { 1 } { 2 } or p=12p = \frac { 1 } { 2 }

and n=4n = 4

Thus p(X≥1)=1−p(X=0)=1−4C0(12)4=1516p ( X \geq 1 ) = 1 - p ( X = 0 ) = 1 - { } ^ { 4 } C _ { 0 } \left( \frac { 1 } { 2 } \right) ^ { 4 } = \frac { 15 } { 16 }.