Question

If mean and variance of Binomial distribution are 4 and 2.4 respectively, then probability of success 'p' is equal to

Answer 1

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

Given that

• Mean of Binomial Distribution = 4

• Variance of Binomial Distribution = 2.4

Let assume that

• n be the number of independent trials

• p be the probability of success

• q be the probability of failure

We know that

In Binomial Distribution,

$\red{\bf :\longmapsto\:Mean = np}$

$\rm :\implies\:np = 4 - - - (1)$

And

$\purple{\bf:\longmapsto\:Variance = npq}$

$\rm :\implies\:npq = 2.4$

$\rm :\longmapsto\:4q = 2.4 \: \: \: \: \: \: \{ \: using \: (1) \: \}$

$\rm :\longmapsto\:q = \dfrac{2.4}{6}$

$\rm :\longmapsto\:q = \dfrac{24}{60}$

$\bf\implies \:q = \dfrac{2}{5} - - - (2)$

We know that,

$\bf :\longmapsto\:p + q = 1$

$\rm :\longmapsto\:p + \dfrac{2}{5} = 1 \: \: \: \: \: \{ \: using \: (2) \: \}$

$\rm :\longmapsto\:p = 1 - \dfrac{2}{5}$

$\rm :\longmapsto\:p = \dfrac{5 - 2}{5}$

$\bf\implies \:p = \dfrac{3}{5}$

Additional Information :-

The Binomial Distribution for random variable 'r' is given by

$\green{\boxed{ \bf \: P(r) = \: ^nC_r\: {p}^{r} {q}^{n - r} \: \: where \: 0 \leqslant r \leqslant n}}$