Question

If the sum of the mean and variance of a Binomial distribution of 5 trials is 9/5,find p(x>=1).

Answer 1

The mean and variance of the Binomial distribution are given by the following formulas: $\mu=np$ and $\sigma^2=np(1-p)$ .

5 trials means that $n=5$ .

Substitute this value into two formulas:

$\mu=5p$

$\sigma^2=5p(1-p)$

Since the sum of the mean and variance is $9/5$ :

$5p+5p(1-p)=9/5$

This is a quadratic equation that we need to solve in order to find $p$

$5p+5p-5p^2=9/5$

$-5p^2+10p=9/5$

Divide both sides by $-5$ :

$p^2-2p=-9/25$

Complete the square:

$p^2-2p+1=-9/25+1$

$(p-1)^2=16/25$

$p-1=-4/5$ or $p-1=4/5$

$p=1/5=0.2$ or $p=9/5=1.8$

Since the probability can't be greater than 1, the only solution is $p=0.2$ .

Now, that we know the value of $p$ and $n$ , we can find $P(x \ge 1)$ . Since it's the complement to $P(x=0)$ ,

$$P(x \ge 1)=1-P(x=0)$$

General formula for $P(x)$ for Binomial distribution is $P(x)=C(n,x)p^x(1-p)^{n-x}$ , thus:

$$P(x \ge 1)=1-P(x=0)=$$ $$=1-C(5,0)\cdot 0.2^0\cdot (1-0.2)^{5-0}=$$ $$=1-1\cdot 1\cdot 0.8^5=0.67232$$

Answer: $P(x \ge 1)=0.67232$ .