Question

Ryerson’s badminton team has 4 male members and 7 female members; Ryerson’s tennis team has 4 male members and
3 female members. These two groups have different members. The university decides to make the two teams have equal
number of members by randomly moving two persons from the badminton group to the tennis group. It then randomly
selects a person from the tennis group. What is the probability to get a female?

Answer 1

Badminton team= 4M + 7F= total 11

Tennis team= 4M + 3F= total 7

There are 3 conditions for randomly moving 2 people from the badminton team to the tennis team: (i) both are males, (ii) both are famales, (iii) one is a male and other is a female.

Probability of (i): 4C2/11C2 = 4!/ 2! 2! * 2! 9!/ 11!

= 4!/ 2! * 9!/11!

= 12/ 11*10

= 12/110

Probability of (ii): 7C2/11C2= 7!/2! 5! * 2! 9!/ 11!

= 7!/5! * 9!/11!

= 42/110

Probability of(iii): 4/11* 7/11= 28/121

Furhter the total no. of people in tennis team

after condition (i) : 6M+ 3F= total 9

after condition (ii) : 4M+ 5F= total 9

after condition (iii): 5M+ 4F= total 9

Again probability of selecting a female member from tennis team,

after condition (i) : 3/9= 1/3

after condition (ii): 5/9

after condition (iii): 4/9

Thus , the required probability of selecting a female member from tennis team after randomly moving two members from the badminton team to the tennis team is given by

= Probability of condition (i)* probability of selecting a female after condition (i) + probability of condition (ii)* probability of selecting a female after condition (ii)+ probability of condition (iii)* probability of selecting a female after condition (iii)

= 12/110* 1/3 + 42/110* 5/9 + 28/121* 4/9

= 4/110+ 210/990+ 112/1089

= 2/55+ 7/33 + 112/1089

= 0.3513314968

which is the required probability.