Chapter 3, Problem 180SE

To determine

Find the expected winnings from the lottery.

Check whether it is worth 33 cents to enter this lottery or not.

Answer to Problem 180SE

The expected winning from the lottery is $0.031.

Explanation of Solution

Calculation:

Mean:

For a discrete random variable Y and probability function $p\left(y\right)$, the expected value is,

$E\left(Y\right)={\displaystyle \sum _{y}yp\left(y\right)}$

The number of combinations that contains two alphabets and four numbers is,

$\begin{array}{c}\text{Combinations}=26\times 26\times 10\times 10\times 10\times 10\\ =6,760,000\end{array}$

The probability of first price is,

$P\left(\text{First}\right)=\frac{1}{6,760,000}$

The probability of second price is,

$P\left(\text{Second}\right)=\frac{2}{6,760,000}$

The probability of third price is,

$P\left(\text{Third}\right)=\frac{10}{6,760,000}$

The expected winning from the lottery is,

$\begin{array}{c}E\left(\text{Winning}\right)=\left(100,000\times \frac{1}{6,760,000}\right)+\left(50,000\times \frac{2}{6,760,000}\right)+\left(1,000\times \frac{10}{6,760,000}\right)\\ =0.015+0.015+0.001\\ =0.031\end{array}$

             $=\$10$

Hence, the expected winning from the lottery is $0.031.

It is clear that expected winning from the lottery is $0.031 which means 3 cents $\left(=0.031\times 100\text{ Cents}\right)$ this indicates that the price of the lottery is high.

Hence, the expected winning from the lottery is very less than the price of lottery.