Chapter 11.4, Problem 24PPS

To determine

What is the probability that there will be at least $12$ successes?

Answer to Problem 24PPS

  $\boxed{0.3743}$

Explanation of Solution

Given information:

A binomial distribution has a $60\%$ rate of success. There are $18$ trials. What is the probability that there will be at least $12$ successes?

Calculation:

Generally, the probability of $k$ successes in $n$ trials of a binomial experiment is,

  $P(k){=}^n{C}_k{(p)}^k{(q)}^{n-k}$

Here $p$ represents the probability of success and $q$ represents the probability of failure.

Examine that a binomial distribution has $60\%$ rate of success and the total number of trials is $18$ .

So, the probability of success is,

  $\begin{array}{l}p=60\%\\ =0.6\end{array}$

The probability of failure is,

  $\begin{array}{l}q=1-p\\ =1-0.6\\ =0.4\end{array}$

The number of successful trials could be $12,13,14,15,16,17,or18.$

So, the probability that there will be at least $12$ sucesses is,

  $\text{P}\left(\text{At least 12}\right)$ = $P(12)+P(13)+P(14)+P(15)+P(16)+P(17)+P(18)$

Substitute the above values in $k$ in the above formula,

  $\text{P}\left(\text{At least 12}\right)$ = ${}^{18}{C}_{12}{(0.6)}^{12}{(0.4)}^{18-12}{+}^{18}{C}_{13}{(0.6)}^{13}{(0.4)}^{18-13}$

  $\begin{array}{l}{+}^{18}{C}_{14}{(}^{0.6}{(}^{0.4}{+}^{18}{C}_{15}{(}^{0.6}{(}^{0.4}\\ {+}^{18}{C}_{16}{(}^{0.6}{(}^{0.4}{+}^{18}{C}_{17}{(}^{0.6}{(}^{0.4}\\ {+}^{18}{C}_{18}{(}^{0.6}{(}^{0.4}\end{array}$

Again write each expression as,

  $\text{P}\left(\text{At least 12}\right)$ = $18,564{(0.6)}^{12}{(0.4)}^6+8,568{(0.6)}^{13}{(0.4)}^5+3,060{(0.6)}^{14}{(0.4)}^4$

  $\begin{array}{l}+816{(}^{0.6}{(}^{0.4}+153{(}^{0.6}{(}^{0.4}\\ +18{(}^{0.6}{(}^{0.4}+(1){(}^{0.6}{(}^{0.4}\\ =0.3743\end{array}$

Hence, the probability that there will be at least $12$ successes is $\boxed{0.3743}$ .