Answer to Question #129321 in Statistics and Probability for Umer

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

According to the Bayes' theorem

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

where $P(A|B)$ - cake failed to raise that was cooked by Khurram;

$P(B)$ - picked cake failed to raise;

$P(A)$ - probability that Khurram was cooking it;

$P(B|A)$ - probability that Khurram will fail any cake.

$P(B)$ can be found by the law of total probability:

$P(B) = 0.03*0.55+0.02*0.25+0.05*0.2=0.0315.$

$P(A) = 0.25$

$P(B|A) = 0.02$

Hence if the picked cake fails to rise, the probability it was baked by Khurram equals

$P(A|B) = \frac{0.02*0.25}{0.0315}=\frac{0.005}{0.0315}=0.159=15.9\%.$