Question

A game of chance involves rolling an unevenly balanced 4-sided die. The probability that a roll comes up 1 is 0.22, the probability that a roll comes up 1 or 2 is 0.42, and the probability that a roll comes up 2 or 3 is 0.54 . If you win the amount that appears on the die, what is your expected winnings? (Note that the die has 4 sides.)

Answer 1

Answer:

Expected Winnings = 2.6

Step-by-step explanation:

Since the probability of rolling a 1 is 0.22 and the probability of rolling either a 1 or a 2 is 0.42, the probability of rolling only a 2 can be determined as:

$P_{1,2} = P_{1}+P_{2}\\P_{2} = 0.42 - 0.22 = 0.20$

The same logic can be applied to find the probability of rolling a 3

$P_{2,3} = P_{2}+P_{3}\\P_{3} = 0.54 - 0.20 = 0.34$

The sum of all probabilities must equal 1.00, so the probability of rolling a 4 is:

$P_{4} =1- P_{1}+P_{2}+P_{3} = 1-0.22+0.20+0.34\\P_{4}=0.24$

The expected winnings (EW) is found by adding the product of each value by its likelihood:

$EW=1*P_{1}+2*P_{2}+ 3*P_{3}+ 4*P_{4} \\EW=1*0.22+2*0.20+ 3*0.34+ 4*0.24\\EW=2.6$

Expected Winnings = 2.6