Question

A board game is played as follows: a number cube numbered 1 to 6 is rolled. If the number 1 comes up, you get nothing. If the number 2, 3, or 5 comes up, you win 40 points. If the number 4 or 6 comes up, you lose 60 points. Use the expected value to prove whether or not the game is fair. Show all work for full credit.

Answer 1

Answer:

The game is fair.

Step-by-step explanation:

We are given that a board game is played as follows; a number cube numbered 1 to 6 is rolled.

Total cases=6

If the number 1 comes up , then you get nothing

Number of favourable cases that you get nothing =1

The probability that you get nothing =$\frac{1}{6}=0.167$

Using of probability formula : P(E)=$\frac{Number\;of\;favourable\;cases}{Total\;number\;of\;cases}$

If number 2,3 or 5 comes up you win 40 points

Number of favourable cases that you wins=3

The probability that you wins =$\frac{3}{6}=0.5$

Therefore, total number of points that you get =$0.5\times 40=20$

If number 4 or 6 comes up then you lose 60 points

Number of favourable cases that you lose=2

The probability that you lose =$\frac{2}{6}=0.3$

Therefore, total number of points that you get =$0.3\times 60=18$

Therefore, the probability of winning is greater than the probability of lose or you get nothing and the point that you get is greater than the points you lose.

Hence , the game is fair.

Answer 2

This is a fair game because you have a chance to win the same amount of money as you do to lose it.