Chapter 11.4, Problem 38PPS

To determine

To find the probability of $s$ successes.

Answer to Problem 38PPS

  $\boxed{0.766}$ Or $\boxed{76.6\%}$

Explanation of Solution

Given information:

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

  $n=6,p=0.6,s\le 4$

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:

  $p=0.6,n=6$

In the above expression, the probability of failure is,

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

Now, we will consider the probability that there are less than or equal to $4$ successes out of $6$ trials. The probability that there are less than or equal to $4$ successes out of $6$ trials will be $1$ minus the probability that there are $5$ or $6$ successes.

Thus, the probability that there are less than or equal to $4$ successes out of $6$ trials is,

  $p\left(k\le 4\right)=1-\left(p\left(5\right)+p\left(6\right)\right)$

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

  $p\left(k\le 4\right)=1-\left(C{}_5{\left(0.6\right)}^5{\left(0.4\right)}^{6-5}+C{}_6{\left(0.6\right)}^6{\left(0.4\right)}^{6-6}\right)$

  $=1-\left(6{\left(0.6\right)}^5\left(0.4\right)+1{\left(0.6\right)}^6\right)$

  $=1-\left(0.187+0.047\right)$

  $=0.766$

Hence, the value of $p\left(k\le 4\right)$ for $p=0.6,n=6$ is $\boxed{0.766}$ or $\boxed{76.6\%}$ .