Question

In class 12A there are 40 students, 12 have blue eyes, 4 have green and 24 have brown in class 12B there are 50 students, 21 have blue eyes, 2 have green and 27 have brown one person is chosen at random from each class to represent the sixth form what is the probability of:

P(different colours)
P(both blue)
P(one brown)

Answer 1

Answer: P(different colours) = 1 - P(same colours) = 1 - (3/10 * 42/100) - (1/10 * 4/100) - (6/10 * 54/100) = 1 - (3/25) - (1/250) - (27/25) = 1- 61/250 = 189/250

P(both blue) = (3/10) * (42/100) = 3/25

P(one brown) = (6/10) + (54/100) - (6/10) * (54/100) = (6/10) + (54/100) - (27/250) = (24/25) - (27/250) = 973/1000

Explanation:

To find the probability of different colours, we need to find the probability that the student chosen from class 12A has a different eye color than the student chosen from class 12B.

The probability of the student from class 12A having blue eyes is 12/40 = 3/10.

The probability of the student from class 12A having green eyes is 4/40 = 1/10.

The probability of the student from class 12A having brown eyes is 24/40 = 6/10.

The probability of the student from class 12B having blue eyes is 21/50 = 42/100.

The probability of the student from class 12B having green eyes is 2/50 = 4/100.

The probability of the student from class 12B having brown eyes is 27/50 = 54/100.

To find the probability of different colours, we need to find the probability that the student chosen from class 12A has a different eye color than the student chosen from class 12B.

P(different colours) = 1 - P(same colours) = 1 - (3/10 * 42/100) - (1/10 * 4/100) - (6/10 * 54/100) = 1 - (3/25) - (1/250) - (27/25) = 1- 61/250 = 189/250

To find the probability of both blue eyes, we need to find the probability that the student chosen from class 12A has blue eyes and the student chosen from class 12B also has blue eyes.

P(both blue) = (3/10) * (42/100) = 3/25

To find the probability of one brown eye, we need to find the probability that either the student chosen from class 12A or the student chosen from class 12B have brown eyes.

P(one brown) = (6/10) + (54/100) - (6/10) * (54/100) = (6/10) + (54/100) - (27/250) = (24/25) - (27/250) = 973/1000

It's important to note that these probability calculations assume that the two students are chosen independently and at random, one from each class. And all the students are different from each other.