Question

A raffle at a local fair has tickets for purchase for$10 each. There are 100 total tickets in the raffle.The tickets are shuffled, then 3 are selected to wina $100 prize. Use the table below to help youcalculate the expected value of this raffle.

Answer 1

Step 1:

In this question, we are given the following:

Step 2:

The details of the solution are as follows:

The expected value of this raffle is calculated thus:

`E(X)\text{ = x P\lparen x\rparen}`
`\begin{gathered} ((3)/(100)\text{ x 90 \rparen + \lparen }(2)/(99)\text{ x 90 \rparen + \lparen}(1)/(98)\text{ x 90 \rparen + } \\ (\text{ 1 - \lparen}(3)/(100)\text{ + }(2)/(99)+\text{ }\frac{1}{98\text{ }})\text{ x -10 \rparen} \end{gathered}`
`((270)/(100))\text{ + \lparen}(180)/(99))\text{ + \lparen}(90)/(98))\text{ + \lparen0.939593898 x -10 \rparen}`
`2.7\text{ + 1.818181818 + 0.918367346 - 9.39593898}`
`E(X)\text{ = -3.959389816 }`
`E(X)\text{ }\approx-3.96\text{ \lparen correct to 2 decimal places\rparen}`

Note:

Can the expected value of a random variable be negative?

It is also called the expected value.

The expected value of a discrete random variable is equal to the mean of the random variable.

Probabilities can never be negative, but the expected value of the random variable can be negative.

CONCLUSION:

The final answer is:

`E(X)\text{ = -3.96 \lparen correct to 2 decimal places\rparen}`