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Aloiza [94]
2 years ago
13

Clue is a board game in which you must deduce three details surrounding a murder. In the original game of Clue, the guilty perso

n can be chosen from 66 people, and there are 66 different possible weapons and 99 possible rooms. At one point in the game, you have narrowed the possibilities down to 55 people, 55 weapons, and 66 rooms. What is the probability of making a random guess of the guilty person, murder weapon, and location from your narrowed-down choices, and the guess being correct
Mathematics
1 answer:
solong [7]2 years ago
7 0

Answer:

The probability of the guess being correct is 1/150.

Step-by-step explanation:

The information provided for the game Clue is:

  • Clue is a board game in which you must deduce three details surrounding a murder.
  • The guilty person can be chosen from 6 people.
  • There are 6 different possible weapons.
  • There are 9 possible rooms.

Now it is provided that at some point the possibilities have been narrowed down to 5 people, 5 weapons, and 6 rooms.

The number of ways to select a guilty person from the 5 selected people is,

n (P) = 5

The number of ways to select a weapon from the 5 selected weapon is,

n (W) = 5

The number of ways to select a room from the 6 selected rooms is,

n (R) = 6

Then the total number of random guesses is,

N = n (P) × n (W) × n (R)

   = 5 × 5 × 6

   = 150

And only one of these 150 guesses is correct.

So, the probability of the guess being correct is 1/150.

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Answer:

a) 0.05

b) 0.9826

c) 0.000039308

Step-by-step explanation:

a) P_X(0) = \frac{e^{-3}3^0}{0!} = e^{-3} =  0.05

b) For two minutes, the mean is doubled, hence it is 6. In order to calculate the probability of al least two calls arriving, we calculate first the probability of the complementary event: At most 1 call will arrive. For that probability, we need to sum the probabilities of 0 and 1.

P_X(0) = e^{-6}

P_X(1) = \frac{e^{-6}*6^1}{1!} = 6*e^{-6}

Hence,

P(X \geq 2) = 1-P(X < 2) = 1- 7*e^{-6} = 0.9826

c) For five minutes the mean is 15. We need to sum the probabilities of 0, 1 and 2.

P_X(0) = e^{-15}

P_X(1) = e^{15}*15

P_X(2) = \frac{e^{-15}*15^2}{2!} = 112.5*e^{-15}

As a result,

P(X \leq 2) = e^{-15}(1+15+112.5) = 0.000039308

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