Chapter 11.4, Problem 37PPS

To determine

To find the probability of $s$ successes.

Answer to Problem 37PPS

  $\boxed{0.324}$ Or $\boxed{32.4\%}$

Explanation of Solution

Given information:

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

  $n=10,p=0.2,s>2$

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:

  $p=0.2,n=10$

In the above expression, the probability of failure is,

  $\begin{array}{l}q=1-p\\ =1-0.2\\ =0.8\end{array}$

Now, we will consider the probability that there are at least $2$ successes out of $8$ trials. The probability that there are at least $2$ successes out of $8$ trials will be $1$ minus the probability that there are $0,1$ or $2$ successes.

Thus, the probability that there are more than $2$ successes out of $8$ trials is,

  $p\left(k>2\right)=1-\left(p\left(0\right)+p\left(1\right)+p\left(2\right)\right)$

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

  $p\left(k>2\right)=1-\left(C{}_0{\left(0.2\right)}^0{\left(0.8\right)}^{10-0}+C{}_1{\left(0.2\right)}^1{\left(0.8\right)}^{10-1}+C{}_2{\left(0.2\right)}^2{\left(0.8\right)}^{10-2}\right)$

  $=1-\left(1{\left(0.8\right)}^{10}+10\left(0.2\right){\left(0.8\right)}^9+45{\left(0.2\right)}^2{\left(0.8\right)}^8\right)$

  $=1-\left(0.107+0.268+0.301\right)$

  $=0.324$

Hence, the value of $p\left(k>2\right)$ for $p=0.2,n=10$ is $\boxed{0.324}$ or $\boxed{32.4\%}$ .