Question

Given that xis a binomial random variable, compute P(x)for each of the following cases:

a. n= 7, x= 3, p= .5

b. n= 4, x= 3, p= .8

c. n= 15, x= 1, p= .1

Short answer

a. The value of P(x) is 0.2734.

b. The value of P(x) is 0.4096.

c. The value of P(x) is 0.3431.

Step-by-step solution

Given information

X is a binomial random variable.

Computing the probability distribution p(x) when n = 7, x = 3, p = .5

a.

For a binomial random variable, the probability distribution

$P\left(x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x};x=0,1,2,...,n$

Here, n = 7 ,x = 3,p = 0.5

$$\begin{gathered} p\left(x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x} \\ =\left(\begin{array}{c}7\\ 3\end{array}\right){\left(0.5\right)}^3{\left(1-0.5\right)}^{7-3} \\ =\left(\begin{array}{c}7\\ 3\end{array}\right){\left(0.5\right)}^3{\left(0.5\right)}^4 \\ =0.2734 \end{gathered}$$

Hence, the probability distributionp(x)is 0.2734.

Computing the probability distribution P(x) when n = 4, x = 3, p = 8

b.

Here n = 4,x = 3, p = 0.8

$$\begin{gathered} p\left(x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x} \\ =\left(\begin{array}{c}4\\ 3\end{array}\right){\left(0.8\right)}^3{\left(1-0.8\right)}^{4-3} \\ =\left(\begin{array}{c}4\\ 3\end{array}\right){\left(0.8\right)}^3\left(0.2\right) \\ =4\times 0.512\times 0.2 \\ =0.4096 \end{gathered}$$

Hence, the probability distributionp(x)is 0.4096.

Computing the probability distribution p(x) when n =15, x =1, p = .1

c.

Here,n = 15,x = 1,p=0.1

$$\begin{gathered} p\left(x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x} \\ =\left(\begin{array}{c}15\\ 1\end{array}\right){\left(0.1\right)}^1{\left(1-0.1\right)}^{15-1} \\ =15\times 0.1\times {\left(0.9\right)}^{14} \\ =0.3431 \end{gathered}$$

Hence, the probability distributionp(x)is 0.3431.