pedzenekO

2021-08-15

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. (1) 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. (2) Express X in terms of S, W, and R. (3) Find E[X]

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(1)
The given information:
-The number of suspects is 6
-The number of rooms is 9

-The number if weapons is 6
Each is given one at random. Identifying the three choices is the goal of the game.
Let's determine the total number of potential answers.
The total number of solutions = The number of suspects * The number of weapons* The number of rooms=6*6*9=324
(2)
Three of the remaining cards are randomly distributed to each player after the selection is completed.
-Let S represents the number of suspects in the set of three cards
-Let W represents the number of weapons in the set of three cards
-Let R represents the number of rooms in the set of three cards
-Let x represents the number of solutions that are possible after a player observes given three cards.
X=(6-S)*(6-W)*(9-R)
(3)
The value of $S,W,R⇒\left\{0,1,2,3\right\}$
And we have, S+W+R=3
{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}, these are combinations of 3 cards.
So, there are 10 possible combinations of 3 cards.
E[X]=1/10sum_Ssum_Wsum_R(6-S)*(6-W)*(9-R)=1/10sum_S(6-S)sum_W(6-W)sum_R(9-R)=1/10[6sum_(W+R=3) (6-W)(9-R)+5sum_(W+R=2) (6-W)(9-R)+4sum_(W+R=1) (6-W)(9-R)+3sum_(W+R=0) (6-W)(9-R)]=190.4

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