Question

The mean and variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is​

Answer 1

Answer:

The mean and variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is. n= 8. = 28/256.

Answer 2

$\large\underline{\sf{Solution-}}$

We know that,

In a binomial distribution,

$\boxed{ \bf{ \: Mean = np}}$

and

$\boxed{ \bf{ \: Variance = npq}}$

where,

• n is number of independent trials

• p is probability of success

• q is probability of failure

So,

According to statement, it is given that

Mean of Binomial Distribution = 4

$\bf :\longmapsto\:np = 4 - - - (1)$

and

Variance of Binomial Distribution = 2

$\bf :\longmapsto\:npq = 2$

$\rm :\longmapsto\:4q = 2$

$\bf\implies \:q = \dfrac{1}{2}$

We know that,

$\red{\rm :\longmapsto\:p + q = 1}$

$\rm :\longmapsto\:p + \dfrac{1}{2} = 1$

$\rm :\longmapsto\:p = 1 - \dfrac{1}{2}$

$\rm :\longmapsto\:p = \dfrac{2 - 1}{2}$

$\bf\implies \:p = \dfrac{1}{2}$

On substituting the value of p in equation (1), we get

$\rm :\longmapsto\:\dfrac{1}{2} \times n = 4$

$\bf\implies \:n = 8$

So, we have the following parameters with us

$\rm :\longmapsto\:n = 8$

$\rm :\longmapsto\:p = \dfrac{1}{2}$

$\rm :\longmapsto\:q = \dfrac{1}{2}$

Now, we know that

Probability of getting r success out of n independent trials with probability of success p and failure q is

$\boxed{ \bf{ \: P(r) \: = \: ^nC_r \: {p}^{r} \: {q}^{n - r} }}$

So,

Probability of getting 2 successes is given by

$\rm :\longmapsto\:P(2) \: = \: ^8C_2 \times {\bigg[\dfrac{1}{2} \bigg]}^{2} {\bigg[\dfrac{1}{2} \bigg]}^{8 - 2}$

$\rm \: = \:\dfrac{8 \times 7}{2 \times 1} \times {\bigg[\dfrac{1}{2} \bigg]}^{2} \times {\bigg[\dfrac{1}{2} \bigg]}^{6}$

$\rm \: = \:28 \times {\bigg[\dfrac{1}{2} \bigg]}^{6 + 2}$

$\rm \: = \:28 \times {\bigg[\dfrac{1}{2} \bigg]}^{8}$

$\rm \: = \:\dfrac{28}{256}$

$\rm \: = \:\dfrac{7}{64}$

$\bf\implies \:\boxed{ \bf{ \: P(2) = \:\dfrac{7}{64} \: \: }}$