# Assume that a procedure yields a binomial distribution with a trial repeated

Assume that a procedure yields a binomial distribution with a trial repeated $$\displaystyle{n}={5}$$ times. Use some form of technology to find the cumulative probability distribution given the probability $$\displaystyle{p}={0.16}$$ of success on a single trial.
(Report answers accurate to 4 decimal places.)
$$\begin{array}{|c|c|}\hline k & P(X=k) \\ \hline 0 & \\ \hline 1 & \\ \hline 2 & \\ \hline 3 & \\ \hline 4 & \\ \hline 5 & \\ \hline \end{array}$$

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Step 1
The binomial probability distribution is,
$$P(X=k)=\left(\begin{array}{c}n\\ x\end{array}\right)(p)^{k}(1-p)^{n-k}$$
In the formula, n denotes the number of trails, p denotes probability of success, and k denotes the number of success.
Step 2
The probability of $$\displaystyle{k}={0}$$ is,
$$P(X=0)=\left(\begin{array}{c}5\\ 0\end{array}\right)(0.16)^{0}(1-0.16)^{5-0}$$
$$\displaystyle={0.4182}$$
The probability of $$\displaystyle{k}={1}$$ is,
$$P(X=1)=\left(\begin{array}{c}5\\ 1\end{array}\right)(0.16)^{1}(1-0.16)^{5-1}$$
$$\displaystyle={0.3983}$$
The probability of $$\displaystyle{k}={2}$$ is,
$$P(X=2)=\left(\begin{array}{c}5\\ 2\end{array}\right)(0.16)^{2}(1-0.16)^{5-2}$$
$$\displaystyle={0.1517}$$
The probability of $$\displaystyle{k}={3}$$ is,
$$P(X=3)=\left(\begin{array}{c}5\\ 3\end{array}\right)(0.16)^{3}(1-0.16)^{5-3}$$
$$\displaystyle={0.0289}$$
The probability of $$\displaystyle{k}={4}$$ is,
$$P(X=4)=\left(\begin{array}{c}5\\ 4\end{array}\right)(0.16)^{4}(1-0.16)^{5-4}$$
$$\displaystyle={0.0028}$$
The probability of $$\displaystyle{k}={5}$$ is,
$$P(X=5)=\left(\begin{array}{c}5\\ 5\end{array}\right)(0.16)^{5}(1-0.16)^{5-5}$$
$$\displaystyle={0.0001}$$
Thus, the probability distribution given the probability $$\displaystyle{p}={0.16}$$ of success on a single trial is,
$$\begin{array}{|c|c|} \hline k & P(X=k) \\ \hline 0 & 0.4182\\ \hline 1&0.3983\\ \hline2&0.1517\\ \hline3&0.0289\\ \hline4&0.0028\\ \hline5&0.0001 \\ \hline \end{array}$$