Name: Conditional Probability
Uploaded: 2019-07-19
Duration: 6 min 14 s
Description: What is the probability of occurrence of an event when another event associated with it has already occurred? Check out this video to learn about it.

Question

The sum and product of the mean and variance of a binomial distribution are respectively

10

and

24

. Find the distribution.

Accepted Answer

Given that,

Mean $+$ Variance $= 10$

Mean $\times$ Variance $= 24$

Let Mean $= α$ , Variance $= β$

So, Mean $+$ Variance $= 10 = α + β$

Mean $\times$ Variance $= 24 = α \times β$

Now, Any quadratic equation can be written as

$\Rightarrow x^{2} - (α + β) x + α \times β = 0$

$\Rightarrow x^{2} - 10 x + 24 = 0$

$\Rightarrow x^{2} - 6 x + 4 x + 24 = 0$

$∴ x = 4, 6$

Case(i): Mean $= 4$ , Variance $= 6$

$∴ n p = 4, n p q = 6$

$\Rightarrow \frac{n p}{n p q} = \frac{4}{6}$

$\Rightarrow q = \frac{3}{2} = 1.5$

It is not possible $(∴ p + q = 1)$

Case(ii): Mean $= 6$ , Variance $= 4$

$∴ n p = 6, n p q = 4$

$\Rightarrow \frac{n p}{n p q} = \frac{6}{4}$

$\Rightarrow q = \frac{2}{3}$

So, $p = 1 - q = 1 - \frac{2}{3} = \frac{2}{3}$

$∴ n p = 6,$

$\Rightarrow n (\frac{1}{3}) = 6$

$\Rightarrow n = 18$

$∴ n = 18 and P = \frac{1}{3}$ , is a binomial distribution.

The sum and product of the mean and variance of a binomial distribution are respectively $10$ and $24$ . Find the distribution.

Important Questions on Probability (Part-II)

Learn and Prepare for any exam you want !

The sum and product of the mean and variance of a binomial distribution are respectively 10 and 24. Find the distribution.

Important Questions on Probability (Part-II)

Learn and Prepare for any exam you want !

The sum and product of the mean and variance of a binomial distribution are respectively $10$ and $24$ . Find the distribution.