Question

An engineering school reports that 56% of its students were male (M), 32% of its students were between the ages of 18 and 20 (A), and that 26% were both male and between the ages of 18 and 20.

What is the probability of choosing a random student who is a female or between the ages of 18 and 20? Assume P(F) = P(not M).

Answer 1

Answer:

7/10

Step-by-step explanation:

Using set notation;

Let n(U) be the total number of students in the school = 100%

Let n(M) be the percentage of male students in the school = 56%

Let n(A) be the percentage of students between the ages of 18 and 20 (A) in the school = 32%

Let n(M∩A) be the percentage of both male and between the ages of 18 and 20 = 26%

The n(MUA)' be the number of female students in the school

Using the formula to get n(MUA)

n(MUA) = n(M)+n(A)- n(M∩A)

n(MUA) = 56+32-26

n(MUA) = 62%

Also, n(U) = n(MUA)+n(MUA)'

100 = 62+n(MUA)'

n(MUA)' = 100-62

n(MUA)' = 38%

This means that there are 38% of students in the school.

The probability of choosing a random student who is a female or between the ages of 18 and 20 is expressed as;

P(F or A) = P(F)+P(A) (mutually exclusive event i.e both cannot occur at the same time)

P(F or A)  = 38/100 + 32/100

P(F or A) = (38+32)/100

P(F or A) = 70/100

P(F or A)  = 7/10

Hence the probability of choosing a random student who is a female or between the ages of 18 and 20 is 7/10.