Question

A game of chance involves rolling a 15-sided die once. If a number from 1 to 3 comes up, you win 2 dollars. If the number 4 or 5 comes up, you win 8 dollars. If any other number comes up, you lose. If it costs 5 dollars to play, what is your expected net winnings?

Answer #1

solution:

total possible outcome in rolling a 15 sided dice = 15

**for winning 2 dollor**

number of ways of rolling 1 to 3 (1 or 2 or 3) = 3 ways

probability of rolling 1 to 3 = 3/15

**for winning 8 dollor**

number of ways of rolling 4 or 5 = 2 ways

so probability of rollilng 4 or 5 = 2/15

**for none winning**

number of ways of rolling 6 to 15 = 10

probability of rollling 6 to 10 = 10/15

so payment for a play = $5

probability distribution of winning is

win(X) | prob.(X) |

$2 | 3/15 |

$8 | 2/15 |

$0 | 10/15 |

so expected winning =

so expected winning = **- $3.53**

**it is an expected loss of $ 3.53**

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