## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.3 - Limits and Continuity - Exercise Set - Page 1162: 62

#### Answer

The probability of winning the prize is $0.0000004720$.

#### Work Step by Step

Consider the provided information. We first calculate the total possible combinations of choosing the five numbers. \begin{align} & _{50}{{C}_{5}}=\frac{50!}{\left( 50-5 \right)!5!} \\ & =\frac{50!}{45!5!} \\ & =\frac{\left( 50 \right)\left( 49 \right)\left( 48 \right)\left( 47 \right)\left( 46 \right)45!}{45!5!} \\ & =\frac{\left( 50 \right)\left( 49 \right)\left( 48 \right)\left( 47 \right)\left( 46 \right)}{5!} \end{align} Further solve the above expression. \begin{align} & _{50}{{C}_{5}}=\frac{\left( 50 \right)\left( 49 \right)\left( 48 \right)\left( 47 \right)\left( 46 \right)}{\left( 5 \right)\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right)} \\ & =\left( 10 \right)\left( 49 \right)\left( 4 \right)\left( 47 \right)\left( 23 \right) \\ & =2118760 \end{align} There can be just one outcome that can match the five numbers drawn randomly, Thus, $~\text{The probability of winning the prize}=\frac{\text{Number of outcomes in favour of win}}{\text{Total outcomes}}$ $~\text{The probability of winning the prize}=\frac{1}{2118760}\approx 0.0000004720$ Hence, the probability of winning the prize is $0.0000004720$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.