Probability-3 (FRM)

Of all the employees at an office, 40% prefer coffee and 60% prefer tea. Of those who prefer coffee, 30% are females and of those who prefer tea, 40% are female. What is the probability that a randomly selected employee prefers coffee, given that the person selected is a female?

(a)1/3          (b) 2/3        (c) 1/9              (d) 2/9

Solution follows here:

Solution:

n(Coffee) = 40%      => n(Female, Coffee) = 30% of 40% = 12%

                                    => n(Male, Coffee)    = 70% of 40% = 28%

n(Tea) = 60%            => n(Female, Tea) = 40% of 60% = 24%

                                    => n(Male, Tea) = 60% of 60% = 36%

Probability of finding female employee preferring coffee = P(Female,Coffee) = 12/100

Number of Female employees = n(Female) = 12%+24% = 36%

∴Probability of finding female employee = P(Female) = 36/100

Probability that a randomly selected employee prefers coffee, given that the person selected is a female

= P(Female,Coffee) / P(Female)

= (12/100) / (36/100) = 1/3

Answer (a)