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:

Sorry, you do not win, but you get your dollar back. (The contestant has not lost the $1 cost.) Congratulations, you win $2. (The contestant has won $1.) Congratulations, you win $5. (The contestant has won $4.) Congratulations, you win $10. (The contestant has won $9.)

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