Chapter 11.4, Problem 39PPS

To determine

To find the probability of $s$ successes.

Answer to Problem 39PPS

  $\boxed{0.99}$ Or $\boxed{99\%}$

Explanation of Solution

Given information:

Each binomial distribution has $n$ trials and $p$ probabilities of success. Determine the probability of $s$ successes.

  $n=9,p=0.25,s\le 5$

Calculation:

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

  $p\left(k\right)=C{}_k{\left(p\right)}^k{\left(q\right)}^{n-k}$

Thus, $p$ represents the probability of success and $q$ represents the probability of failure.

Now, consider a binomial probability with the following attributes:

  $p=0.25,n=9$

In the above expression, the probability of failure is,

  $\begin{array}{l}q=1-p\\ =1-0.25\\ =0.75\end{array}$

Now, we will consider the probability that there are less than or equal to $5$ successes out of $9$ trials. The probability that there are less than or equal to $5$ successes out of $9$ trials will be $1$ minus the probability that there are $6,7,8$ or $9$ successes.

Thus, the probability that there are at least $5$ successes out of $9$ trials is,

  $p\left(k\le 5\right)=1-\left(p\left(6\right)+p\left(7\right)+p\left(8\right)+p\left(9\right)\right)$

Now, substitute the above $k$ values in the above expression,

  $p\left(k\le 5\right)=1-\left(C{}_6{\left(0.25\right)}^6{\left(0.75\right)}^{9-6}+C{}_7{\left(0.25\right)}^7{\left(0.75\right)}^{9-7}+C{}_8{\left(0.25\right)}^8{\left(0.75\right)}^{9-8}+C{}_9{\left(0.25\right)}^9{\left(0.75\right)}^{9-9}\right)$

  $=1-\left(0.009+0.001+0.000+0.000\right)$

  $=1-\left(0.01\right)$

  $=0.99$

Hence, the value of $p\left(k\le 5\right)$ for $p=0.25,n=9$ is $\boxed{0.99}$ or $\boxed{99\%}$ .