Answer to Question #129283 in Statistics and Probability for Zain Akbar

Question

Three cooks Mohsin, Khurram, and Shahid bake a special kind of cake. In the restaurant where they work, Mohsin, bakes 55 percent of these cakes, Khurram bakes 25 percent and Shahid bakes 20 percent and with respective probabilities of 0.03, 0.02 and 0.05 they fail to rise. A cake is selected and fails to rise what is probability that it was backed by cook Khurram

Answer 1

Let $M$ be an event 'Mohsin  bakes the cake', $K$ be an event 'Khurram  bakes the cake', and $S$ be an event 'Shahid bakes the cake'. Given

$$P(M)=0.55, P(K)=0.25,P(S)=0.20$$

Let $A$ be an event 'cake fails to rise'. Given

$$P(A|M)=0.03,P(A|K)=0.02.P(A|S)=0.05$$

By Bayes' Theorem

$P(K|A)=$

$=\dfrac{P(K)P(A|K)}{P(M)P(A|M+P(K)P(A|K)+P(S)P(A|S)}$

$P(K|A)=\dfrac{0.25(0.02)}{0.55(0.03)+0.25(0.02)+0.20(0.05)}=$

$=\dfrac{10}{63}\approx0.1587$

If a cake is selected and fails to rise then the probability that it was backed by cook Khurram is $\dfrac{10}{63}\approx0.1587.$