At Sally's Hair Salon there are three hair stylists. 36% of the hair cuts are done by Chris, 26% are done by Karine, and the rest are done by Amy. Chris finds that when he does hair cuts, 5% of the customers are not satisfied. Karine finds that when she does hair cuts, 6% of the customers are not satisfied. Amy finds that when she does hair cuts, 8% of the customers are not satisfied. Suppose that a customer leaving the salon is selected at random. If the customer is not satisfied, what is the probability that their hair was done by Amy

We are given that Sally's Hair Salon there are three hair stylists. 36% of the hair cuts are done by Chris, 26% are done by Karine, and the rest are done by Amy.

Let the Probability of hair cutting done by Chris = P(C) = 0.36

Probability of hair cutting done by Karine = P(K) = 0.26

Probability of hair cutting done by Amy = P(A) = 0.368

Also, let NS = event that customer is not satisfied with his cutting

So, Probability that customers are not satisfied given that their hair cutting is done by Chris = P(NS/C) = 0.05

Probability that customers are not satisfied given that their hair cutting is done by Karine = P(NS/K) = 0.06

Probability that customers are not satisfied given that their hair cutting is done by Amy = P(NS/A) = 0.08

Now, a customer leaving the salon is selected at random. If the customer is not satisfied, the probability that their hair was done by Amy is given by = P(A/NS)

For finding the above probability we will use the concept of Bayes' Theorem;

SO, P(A/NS) = [tex]\frac{\text{P(A) \times P(NS/A)}}{\text{P(C) \times P(NS/C) + P(K) \times P(NS/K) +P(A) \times P(NS/A) }}[/tex][tex]\frac{P(A) \times P(NS/A) }{P(C) \times P(NS/C) +P(K) \times P(NS/K) +P(A) \times P(NS/A) }[/tex]

= [tex]\frac{0.38 \times 0.08}{0.36 \times 0.05+0.26 \times 0.06+0.38 \times 0.08}[/tex]

= [tex]\frac{0.0304}{0.064}[/tex]

= 0.475

Hence, the probability that their hair was done by Amy is 0.475.