3. what is the net amount that a contestant should expect to win per game if the game were to be played many
times?

0.55

Missing Statement:

At a carnival, one game costs $1 to play. The contestant gets one shot in an attempt to bust a balloon. Each balloon

contains a slip of paper with one of the following messages:

If the contestant does not bust a balloon, then the $1 cost is forfeited. The table below displays the probability

distribution of the discrete random variable, or net winnings for this game

Net Winnings −1         0        1          4            9

Probability      0.25     ?      0.3     0.08     0.02

Step-by-step explanation:

First we find of the missing probability for net winnings of 0, in which case:

When you receive the message, “Sorry, you do not win, but you get your dollar back,” then your net winnings is 0$.

The  probability of winning 0$ is = 1 - 0.25 - 0.3 - 0.08 -0.02 = 0.35

So now we have the complete details:

Net Winnings −1         0        1          4            9

Probability     0.25    0.35    0.3     0.08     0.02

Now we need to find ,the net amount that a contestant should expect to win is the expected value of the probability distribution.

Expected value of probability distribution = −1(0.25 ) +0 (0.35 ) + 1(0.3 ) +4 (0.08 ) + 9(0.02 )

Expected value of probability distribution = 0.55

The net amount that a contestant should expect to win per game if the game were to be played many  times is 0.55