110 students were picked at random and asked whether they had homework in English, History and Maths during a particular week:

Question

English has 25 students

history has 10 students

maths has 11 studens

H + M has 9 students

H + E has 20 students

M + E has 17 students

(This means they had 2 homework not added)

13 students had homework from every single class

What is the probability of:

a)   A student having no homework?

b)   A student having 2 pieces of homework?

c)   A student having English homework?

Answer 1

(a) We know that there are 110 students sampled, and among them, 25 + 10 + 11 + 9 + 20 + 17 + 13 = 105 had one or more pieces of homework. So 110 - 105 = 5 students had no homework. So P(no homework) = 5 / 110 = 1 / 22, or approximately 0.045.

(b) I would offer two answers here, one assuming that the question is asking the probability of a student having exactly two pieces of homework, and the second assuming that the question is asking the probability of having two or three pieces of homework (since students with 3 pieces of homework have two pieces of homework plus another piece of homework).

P(exactly two pieces of homework) = (9 + 20 + 17) / 110 = 46 / 110 = 23 / 55, or approximately 0.418.

P(two or more pieces of homework) = (9 + 20 + 17 + 13) / 110 = 59 / 110, or approximately 0.536.

(c) P(English homework) = (25 + 20 + 17 + 13) / 110 = 75 / 110 = 15 / 22, or approximately 0.682.