Ms-08

2. A restaurant is experiencing discontentment among its customers. It analyses that there are three factors responsible viz. food quality, service quality and interior décor. By conducting an analysis, it assesses the probabilities of discontentment with the three factors as 0.40, 0.35 and 0.25 respectively. By conducting a survey among the customers, it also evaluated the probabilities of a customer going away discontented on account of these factors as 0.6, 0.8 and 0.5, respectively. With this information, the restaurant wants to know that, if a customer is discontented, what are the probabilities that it is so due to food, service or interior décor?

Solution : Let

“A” be the case of discontentment due to food.

“B” be the case of discontentment due to service.

& “C” be the case of discontentment due to interior décor.

Now, it’s given that

P(A) = 0.40

P(B) = 0.35

P( C ) = 0.25

Let the Probability of a CUSTOMER going away discontented on account be representated as “E”.

Then, it’s given that,

P(E/A) = 0.6 ( customer going away discontented on due to food)

P(E/B) = 0.8 ( “ “ “ “ “ “ “ due to service)

P(E/C) = 0.5 (“ “ “ “ “ “ “ “ due to interior décor)

Now, Probability of a customer discontented ,

due to FOOD is given by P(A/E)

due to SERVICE P(B/E)

due to INTERIOR DÉCOR P(C/E)

According to Baye’s theorem,

a) P(A/E) = {P(E/A) × P(A)} ÷ {P(E/A) × P(A) + P(E/B) × P(B) + P(E/C) × P(C )}

= {0.6 × 0.40} ÷ {0.6 × 0.40 + 0.8 × 0.35 + 0.5 × 0.25}

= 0.372 (approx.)

b) P(B/E) = {P(E/B) × P(B)} ÷ {P(E/A) × P(A) + P(E/B) × P(B) + P(E/C) × P(C )}

= {0.8×0.35}÷ {0.6 × 0.40 + 0.8 × 0.35 + 0.5 × 0.25}

= 0.434 (approx.)

c) P(C/E) = {P(E/C) × P(C)} ÷ {P(E/A) × P(A) + P(E/B) × P(B) + P(E/C) × P(C )}

= {0.5×0.25}÷{0.6 × 0.40 + 0.8 × 0.35 + 0.5 × 0.25}

= 0.194 (approx.)