Morgan is playing a board game that requires three standard dice to be thrown at one time. each die has six sides, with one of the numbers 1 through 6 on each side. she has one throw of the dice left, and she needs a 17 to win the game. what is the probability that morgan wins the game (order matters)?

The probability of having a 17 is the number of ways of having 17 divided by the number of different possible outcomes.

Ways of having 17:

Die 1     Die 2     Die 3    sum
6            6            5          6+6+5 = 17
6            5            6          6+5+6 = 17
5            6            6          5+6+6 = 17

Those are the only 3 possibilities of having 17 (you can only loose 1 point).

The total number of outcomes are 6*6*6 = 216

So the chances (probability) to win is 3 / 216 = 1/72 = 0.0139 

Ican't do any of the other stuff, but i can get you started. the answer, i believe, is $2.50.