Answer to Question #325298 in Statistics and Probability for Chanel

Question

A game costs $9.00 to play. In the game, there is a bag that contains 14 balls; 4 are red, 8 are green, and 2 are yellow. The player of the game draws one ball at random from the bag, and the player wins $15.00 if they draw a red ball, nothing if they draw a green ball, and $19.00 if they draw a yellow ball. What is the expected value of this game for the player? Round your answer to the nearest cent

Answer 1

Probability to get red ball = 4/14=2/7

Probability to get green ball = 8/14=4/7

Probability to get yellow ball = 2/14=1/7

Expected value =(4/7)(-9)$=-5.14 $ if we get green ball.

Expected value =(2/7)15 $=4.29 $ if we get red ball. And -9$ for play. It is -4.71$.

Expected value =(1/7)19 $=2.71 $ if we get yellow ball. And -9$ for play. It is -6.29$

Expected value of game is -5.14-6.29-4.71=-16.14 $