For a certain game of chance a player loses 3 with a probabi

For a certain game of chance a player loses $3 with a probability of 0.40, breaks even with probability 0.15 gains $1 with probability 0.30, gains $4 with probability 0.05, and gains $5 with probability 0.10.

A player plays this game of chance one time. What is the probability that the player will win some money?

If the player plays the game many times, what is the player

Solution

i) For a single game, for the player to win some money he has to earn, either $1 or $4 or $5,

hence the required probability = 0.3 + 0.05 + 0.10 = 0.45

ii) Player\'s expected value is the summation of the products of the (cash outcome*the respective probability),

hence the expected value = (-3*0.4) + (0*0.15) + (1*0.30) + (4*0.05) + (5*0.10) = -0.2