Question

Restaurants A & B are the only places open for breakfast on Saturday mornings in Small Town, USA. 60% of the people in Small Town that go out for breakfast on Saturday mornings go to restaurant A. Restaurant A has a customer satisfaction rating of 98% (2% unsatisfied). Restaurant B has a customer satisfaction rating of 95%. What is the probability that a customer ate breakfast at Restaurant A if they had an unsatisfactory experience (unsatisfactory rating)? (Round to nearest percentage) A. 38 B. 62 C. 24 D. 40 Reset Selection

Answer 1

Solution:

Given:

A = Restaurant A and B = Restaurant B

S = Satisfied

UNS = Unsatisfied

P(A) = 60%

P(B) = 100%-60%= 40%

P(S | A) =98%

thus

P( UNS | A) =2%

P(S | B) =95%

P(UNS | B ) = 5%

We have to find:

P( A | UNS) = ..........?

Using Bayes rule of probability:

$P( A | UNS) = \frac{P(UNS | A) \times P(A)}{P(UNS | A) \times P(A)+P(UNS | B) \times P(B)}$

$P( A | UNS) = \frac{98\% \times 60\%}{98\% \times 60\%+95\% \times 40\%}$

$P( A | UNS) = \frac{2\% \times 60\%}{2\% \times 60\%+ 5\% \times 40\%}$

$P( A | UNS) = \frac{1.2\%}{1.2\%+2\%}$

$P( A | UNS) = \frac{1.2\%}{3.2\%}$

$\mathbf{{\color{DarkOrange} P( A | UNS) = 38\%}}$

Thus correct answer is: A. 38