SOLUTION: Please help me solve this question: A game involves rolling a die once. If a number less than 3 appears,you lose $60. If a 4 or 6 appears,you win $120. If a 5 appears,nothing h

Question 765574: Please help me solve this question:
A game involves rolling a die once. If a number less than 3 appears,you lose $60. If a 4 or 6 appears,you win $120. If a 5 appears,nothing happens.
A.) show the probability distribution for the money.
B.) calculate the expected monetary value of the game.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your probabilities are incomplete.
there is no condition for a value of 3 being chosen.
i will assume that if a 3 or a 5 appears, nothing happens.
that makes the probabilities come out correctly.
p(<3) = 2/6
p(4 or 6) = 2/6
p(3 or 5) = 2/6
your expected value is:
2/6 * -60 + 2/6 * 120 + 0 which is equal to -120/6 + 240/6 = 120/6 = 20.
on the average, if you play the game an infinite number of times, you should collect 20 dollars each time.
example:
you play the game 1200 times.
2/6 * 1200 = 2400/6 = 400 times you will lose 60 dollars.
400 times you will win 120 dollars
400 times you will get nothing.
your total winning is 400 * 120 = 48,000.
your total losing is 400 * 60 = 24,000
your net gain is 24,000
divide that by 1200 and you have average a net gain of 20 per game.

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