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Category: Business and Industry
Date Submitted: 04/09/2013 01:43 PM
January 24, 2011
Pedro D. Arlando, Jr.
DBA 7310 Statistics for Business and Research
Professor Richard Gray
CASE 6.1 Answer to Case Study: Let’s Make a Deal
1. Before Monty shows what is behind curtain C, the probability that the car is behind curtain A is 1/3. The probability that the car is behind curtain B is also 1/3. The reason being is that the car can be behind any of the three curtains with equal probability.
2. After Monty shows what is behind curtain C, the probability that the car is behind curtain A does not change because he did not provide new information about curtain A, and there is an increase in probability that the car is behind curtain B because Monty did not show what is behind curtain B, raising the possibility that car is behind curtain B. After Monty revealed what is behind curtain C there is a probability that the car is behind curtain B. Theoretically this is given by P(Car =B)|(Monty choses C) and using Bayes’ Law Formula becomes:
P(Car in B|(Monty choses C) = P(Car in B) x P(Monty choses C|Car in B)
P(Monty choses C)
= _________________________P(Car in B)_x_P(Monty choses C|Car_in_B)__________________________________________
P(Car in A) x P(Monty choses C|Car in A) + P(Car in B) x P(Monty choses C|Car in B) + P(Car in C) x P(Monty choses C)| Car in C
With,
P(Car in A) = P(Car in B)=P(Car in C) = 1/3
P(Monty choses C|Car in A)=1/2, due to Monty could have chosen any of B and C if car was in A)
P(Monty choses |Car in B =1, due to Monty could have chosen only C if the car was in B, as same as he wouldn’t choose B if car was in B,
P(Monty choses C|car in C) = 0, because if the car is in C, he would never reveal it…what is behind of C.
Substituting the values,
P(Car in B|(Monty choses C) = _______1/3 x 1_________
(1/3x1/2)+(1/3x1)+(1/3x0)...