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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.

A Raffle At A Local Fair Has Tickets For Purchase For10 Each There Are 100 Total Tickets In The RaffleThe Tickets Are Shuffled Then 3 Are Selected To Wina 100 P class=

Sagot :

SOLUTION:

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:

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

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:

[tex]E(X)\text{ = -3.96 \lparen correct to 2 decimal places\rparen}[/tex]

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