Question

In a secondary school, there are 400 male students and 600 female students. 50% of the male students and 55% of the female students are in senior secondary curriculum, the others are in junior secondary curriculum. The school had appointed 8% of senior male students and 10% of senior female students to be the student leaders. No junior secondary student can be student leader.

(a) If a student is selected at random in the school, find the probability that

(i) the student is a senior student.

(ii) the student is a female student leader.

(b) Suppose a student is selected and known to be NOT a student leader, what is the probability that

(i) the student is a male student.

(ii) the student is a senior female student.

Answer 1

a.

i.

P(senior) = P(senior | male)*P(male) + P(senior | female)*P(female)

= (50%)*(no. of male students / no. of students) + (55%)*(no. of female students / no. of students)

= 0.50 * (400 / (400+600)) + 0.55*(600/(400+600))

= 0.50*0.4 + 0.55*0.6

= 0.53

P(senior) = 0.53

ii.

P(female student leader) = P(leader | senior female)*P(senior | female)*P(female)

= 0.10 * 0.55 * 0.6

= 0.033

b.

P(NOT student leader) = 1 - P(student leader)

= 1 - ( P(leader | senior female)*P(senior | female)*P(female) + P(leader | senior male)*P(senior | male)*P(male) )

= 1 - ( 0.10*0.55*0.6 + 0.08*0.50*0.4 )

= 0.951

i.

P(male | not student leader) = P(male and not student leader) / P(not student leader)

= [ P(male) - P(leader | senior male)*P(senior | male)*P(male) ] / P(not student leader)

= [ 0.4 - 0.08*0.50*0.4 ] / 0.951

= 0.4038

ii.

P(senior female student | not student leader) = P(senior female student AND not student leader) / P(not student leader)

= [ P(senior female student) - P(student leader | senior female student)*P(senior female student) ] / 0.951

= [ 0.55*0.6 - 0.10*0.55*0.6 ] / 0.951

= 0.3123

(please UPVOTE)