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]
