Suppose x is a binomial random variable with n = 3 and p = .3.

1. Calculate the value of $\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$,role="math" localid="1653657859012" $\mathbf{x}\mathbf{=0,1,2,3,}$using the formula for a binomial probability distribution.
2. Using your answers to part a, give the probability distribution for x in tabular form.

The value of q in a binomial probability distribution, in this context, can be found by subtracting the value of p from 1.Here, the value of n (number of trials) is given. As the value of p (number of successes) is 0.3, subtracting 0.3 from 1, we get 0.7, which is the value of q (number of failures).

The value of $\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$is calculated by the formula $\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{x}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{!}}{\mathbf{\left(}\mathbf{p}\mathbf{\right)}}^{\mathbf{x}}{\mathbf{\left(}\mathbf{q}\mathbf{\right)}}^{\mathbf{n}\mathbf{-}\mathbf{x}}$ when the value of x is 0:

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)=\frac{\mathrm{n}!}{\mathrm{x}!(\mathrm{n}-\mathrm{x})!}{\left(\mathrm{p}\right)}^{\mathrm{x}}{\left(\mathrm{q}\right)}^{\mathrm{n}-\mathrm{x}} \\ \mathrm{p}\left(0\right)=\frac{3!}{0!(3-0)!}{\left(0.3\right)}^0{\left(0.7\right)}^{3-0} \\ =\frac{3\times 2\times 1}{1\left(3\right)!}{\left(0.3\right)}^0{\left(0.7\right)}^3 \\ =0.343 \end{gathered}$$

The value of $\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$ is calculated with the formula $\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\text{=}}\frac{\mathbf{\text{n!}}}{\mathbf{\text{x!}}\mathbf{\left(}\mathbf{\text{n-x}}\mathbf{\right)}\mathbf{\text{!}}}{\mathbf{\left(}\mathbf{\text{p}}\mathbf{\right)}}^{\mathbf{\text{x}}}{\mathbf{\left(}\mathbf{\text{q}}\mathbf{\right)}}^{\mathbf{\text{n-x}}}$ when the value of x is 1:

$$\begin{gathered} \mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{x}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{!}}{\mathbf{\left(}\mathbf{p}\mathbf{\right)}}^{\mathbf{x}}{\mathbf{\left(}\mathbf{q}\mathbf{\right)}}^{\mathbf{n}\mathbf{-}\mathbf{x}} \\ \mathbf{p}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}\mathbf{!}}{\mathbf{1}\mathbf{!}\mathbf{(}\mathbf{3}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathbf{\left(}\mathbf{0.3}\mathbf{\right)}}^{\mathbf{1}}{\mathbf{\left(}\mathbf{0.7}\mathbf{\right)}}^{\mathbf{3}\mathbf{-}\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}}{\mathbf{1}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{!}}{\mathbf{\left(}\mathbf{0.3}\mathbf{\right)}}^{\mathbf{1}}{\mathbf{\left(}\mathbf{0.7}\mathbf{\right)}}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0.441} \end{gathered}$$

The value of $\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ is calculated with the formula $\mathbf{\text{p}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{x}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{!}}{\mathbf{\left(}\mathbf{p}\mathbf{\right)}}^{\mathbf{x}}{\mathbf{\left(}\mathbf{q}\mathbf{\right)}}^{\mathbf{n}\mathbf{-}\mathbf{x}}$ when the value of x is 2:

$$\begin{gathered} \mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{x}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{!}}{\mathbf{\left(}\mathbf{p}\mathbf{\right)}}^{\mathbf{x}}{\mathbf{\left(}\mathbf{q}\mathbf{\right)}}^{\mathbf{n}\mathbf{-}\mathbf{x}} \\ \mathbf{p}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\frac{\mathbf{3}\mathbf{!}}{\mathbf{2}\mathbf{!}\mathbf{(}\mathbf{3}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathbf{\left(}\mathbf{0.3}\mathbf{\right)}}^{\mathbf{2}}{\mathbf{\left(}\mathbf{0.7}\mathbf{\right)}}^{\mathbf{3}\mathbf{-}\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{1}}{\mathbf{2}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{!}}{\mathbf{\left(}\mathbf{0.3}\mathbf{\right)}}^{\mathbf{2}}{\mathbf{\left(}\mathbf{0.7}\mathbf{\right)}}^{\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0.189} \end{gathered}$$

The value of $\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is calculated with the formula $\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{x}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{!}}{\mathbf{\left(}\mathbf{p}\mathbf{\right)}}^{\mathbf{x}}{\mathbf{\left(}\mathbf{q}\mathbf{\right)}}^{\mathbf{n}\mathbf{-}\mathbf{x}}$ when the value of x is 3:

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)=\frac{\mathrm{n}!}{\mathrm{x}!(\mathrm{n}-\mathrm{x})!}{\left(\mathrm{p}\right)}^{\mathrm{x}}{\left(\mathrm{q}\right)}^{\mathrm{n}-\mathrm{x}} \\ \mathrm{p}\left(3\right)=\frac{3!}{3!(3-3)!}{\left(0.3\right)}^3{\left(0.7\right)}^{3-3} \\ =\frac{3\times 2\times 1}{6\left(0\right)!}{\left(0.3\right)}^3{\left(0.7\right)}^0 \\ =0.027 \end{gathered}$$

Therefore, when p(0) is 0.343, p(1) is 0.441, p(2) is 0.189, and p(3) is 0.027.

b.

The values calculated above have been shown below in the form of a table.

As shown above, when x is replaced by 0, the probability becomes 0.343.

When x is replaced by 1, the probability becomes 0.441.

When x is replaced by 2, the probability becomes0.189, and when x is replaced by 3, the probability becomes 0.027.