The game of Clue involves 6 suspects, 6 weapons, and 9 rooms. One of each is randomly chosen and the object of the game is to guess the chosen three.

(a) How many solutions are possible? In one version of the game, the selection is made and then each of the players is randomly given three of the remaining cards. 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. Also, let X denote the number of solutions that are possible after that player observes his or her three cards.

(b) Express X in terms of S, W, and R.

(c) Find E[X]

The number of possible solution.

Six people have been named as suspects.

There are six weapons in all.

There are 9 rooms in all.

One of each is chosen at random.

The object of the game is to guess the three people that have been picked.

Let

In the set of three cards, S denotes the number of suspects.

In the set of three cards, W denotes the number of suspects.

In the set of three cards, R denotes the number of suspects.

And

Six people have been named as suspects.

Six people have been named as weapons

Six people have been named as rooms.

Therefore

Total possible solution = $$\begin{gathered} =S\times W\times R \\ =6\times 6\times 9 \\ =324 \end{gathered}$$

To express X in terms of$S,WandR$

Six people have been named as suspects.

There are six weapons in all.

There are 9 rooms in all.

One of each is chosen at random.

The object of the game is to guess the three people that have been picked.

In the set of three cards, S denotes the number of suspects.

In the set of three cards, W denotes the number of suspects.

In the set of three cards, R denotes the number of suspects.

Each player is given three of the remaining cards at random while making their choice.

Then, after a player sees three cards, let X denote the number of possible solutions.

Thus,

$X=(6-S)·(6-W)·(9-R)$

The value of$E\left[X\right]$

Six people have been named as suspects.

There are six weapons in all.

There are 9 rooms in all.

One of each is chosen at random.

Objective of game is to guess the chosen three.

For values of suspects, weapons, and rooms,

$S,W,R\in \{0,1,2,3\}$

$\begin{array}{r}\{3,0,0\},\{0,3,0\},\{0,0,3\},\\ \{1,1,1\},\{2,1,0\},\{2,0,1\},\\ \{1,2,0\},\{0,2,1\},\{1,0,2\},\\ \{0,1,2\},\end{array}$

Three cards can be combined in ten different ways.

localid="1647810190080" $\begin{array}{r}E\left[X\right]=\frac{1}{10}\sum _S \sum _W \sum _R (6-S)\cdot (6-W)\cdot (9-R)\\ =\frac{1}{10}\sum _S (6-S)\sum _W (6-W)\sum _R (9-R)\\ =\frac{1}{10}\left[\begin{array}{c}6\sum _{W+R=3} (6-W)(9-R)+5\sum _{W+R=2} (6-W)(9-R)\\ +4\sum _{W+R=1} (6-W)(9-R)+3\sum _{W+R=0} (6-W)(9-R)\end{array}\right]\\ =198.8\end{array}$

Therefore thelocalid="1647810203419" $E\left[X\right]=198.8$