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

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

Answer 2

The game is fair.

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 =
`(1)/(6)=0.167`

Using of probability formula : P(E)=
`(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 =
`(3)/(6)=0.5`

Therefore, total number of points that you get =
`0.5* 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 =
`(2)/(6)=0.3`

Therefore, total number of points that you get =
`0.3* 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.