Problema Solution
For a certain game of chance, a player loses $4 with a probability of 0.35, breaks even with probability 0.10, gains $1 with probability 0.20, gains $2 with probability 0.30, and gains $6 with probability 0.05.
If played game of chance one time. What is the probability the player will win some money? If the player plays the game many times, what is the player’s expected value of the probability distribution?
Answer provided by our tutors
P(win)=P(gain $1)+P(gain $2)+P(gain $6)=0.20+0.30+0.05=0.55
So the probability the player will win some money is 0.55
E(x)=(-4)*0.35+0*0.10+1*0.20+2*0.30+6*0.05 = -0.30
The expected value of the probability distribution is -$0.30