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

times?

1 Answer

6 votes

Answer:

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

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

User Mark Erickson
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