Part A:
Given that t<span>he game of clue involves 6 suspects, 6 weapons, and 9 rooms.
The number of ways that one of each is randomly chosen is given by:
Therefore, the number of solutions possible is 324.
Part B:
Given that a </span>players is randomly given three of the remaining cards, <span>let s, w, and r be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player.
The number of suspects, weapons, and rooms remaining respectively after the player observes his or her three cards are: 6 - s, 6 - w, and 9 - r.
Let x denote the number of solutions that are possible after that player observes his or her three cards, then:
Therefore, x in terms of s, w, and r is given by x = (6 - s)(6 - w)(9 - r).
Part C:
The expected value E(x) of a data set
with probabilities 
is given by 
There are </span>
possible combinations s, w and r. They are (3, 0, 0), (0, 3, 0), (0, 0, 3), (2, 1, 0), (0, 2, 1), (1, 0, 2), (2, 0, 1), (1, 2, 0), (0, 1, 2), (1, 1, 1)
Thus the expected value is given by
Given that t<span>he game of clue involves 6 suspects, 6 weapons, and 9 rooms.
The number of ways that one of each is randomly chosen is given by:
Therefore, the number of solutions possible is 324.
Part B:
Given that a </span>players is randomly given three of the remaining cards, <span>let s, w, and r be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player.
The number of suspects, weapons, and rooms remaining respectively after the player observes his or her three cards are: 6 - s, 6 - w, and 9 - r.
Let x denote the number of solutions that are possible after that player observes his or her three cards, then:
Therefore, x in terms of s, w, and r is given by x = (6 - s)(6 - w)(9 - r).
Part C:
The expected value E(x) of a data set
There are </span>
Thus the expected value is given by