Chapter 11.4, Problem 35PPS

To determine

To find the probability that he wins at least one prize.

Answer to Problem 35PPS

  $\boxed{0.603}$ Or $\boxed{60.3\%}$

Explanation of Solution

Given information:

A beverage company is having a sweepstakes. The probabilities of winning selected prizes are shown at the right. If Ernesto purchases $8$ beverages, what is the probability that he wins at least one prize?

  $\begin{array}{l} Oddsofwinning\\ \begin{array}{ll}beverage\hfill & 1in10\hfill \\ CD\hfill & 1in200\hfill \\ hat\hfill & 1in250\hfill \\ MP3Player\hfill & 1in20,000\hfill \\ Car\hfill & 1in25,000,000\hfill \end{array}\end{array}$

Calculation:

Here, we will consider a beverage company that has announced a sweepstakes.

The probability of winning a beverage as prize is $1$ in $10$ , that is,

  $\frac{1}{10}=0.1$

The probability of winning a CD as prize is $1$ in $200$ , that is,

  $\frac{1}{200}=0.005$

The probability of winning a hat as prize is $1$ in $250$ , that is,

  $\frac{1}{250}=0.004$

The probability of winning a MP3 Player as prize is $1$ in $20,000$ , that is,

  $\frac{1}{20,000}=0.00005$

The probability of winning a car Player as prize is $1$ in $25,000,000$ , that is,

  $\frac{1}{25,000,000}=0.00000004$

Thus, the probability of winning any of the above prizes will be the sum of all the above probabilities, that is,

  $0.1+0.005+0.004+0.00005+0.00000004=0.10905004$

  $\approx 0.1091$

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

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

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

In the above situation, the probability of success is,

  $p=0.1091$

The probability of failure is,

  $\begin{array}{l}q=1-p\\ q=1-0.1091\\ q=0.8909\end{array}$

Now, we will consider the probability that there will be at least one prize in $8$ beverages a person $E$ buys. The probability that there will be at least one prize in $8$ beverages will be $1$ minus the probability that there will be no prize in $8$ beverages.

Thus, the probability that there will be at least one prize in $8$ beverages is,

  $p\left(Atleast1prize\right)=1-p\left(NoPrize\right)$

  $=1-p\left(0\right)$

Now, substitute $p=0.1091,$

  $q=0.8909,$

  $n=8,$

  $k=0$ in the above expression,

  $p\left(Atleast1prize\right)=1-C{}_0{\left(0.1091\right)}^0{\left(0.8909\right)}^{8-0}$

  $=1-1{\left(0.8909\right)}^8$

  $=1-0.397$

  $=0.603$

Hence, the probability that there will be at least one prize in $8$ beverages is $\boxed{0.603}$ or $\boxed{60.3\%}$ .