Question

An advertising company has 8 men and 5 women. Suppose the company has to select a team of 4 members to work on the new hybrid car, Hyper Geo Metro 2013. If the members of the team are selected at random, what is the probability that 2 men and 2 women will be selected? (enter answer in decimals; bear in mind the 1% error tolerance).

What is the probability that men will constitute a majority in the team?

Answer 1

An advertising company has 8 men and 5 women.
Suppose the company has to select a team of 4 members to work on the new hybrid car,
Hyper Geo Metro 2013.
If the members of the team are selected at random
the probability that 2 men and 2 women will be selected.
general formula for probability
probability = number of favourable events / total number of events
favourable events are 8c2 *5c2
we use formula ncr = n!/(n-r)!*r!
8c2 value is 8!/(8-2)!*2! = 8*7*6!/(6!)*2!
8c2 = 56/2 = 28
5c2 = 5!/(5-2)!*2!
5c2 = 5*4*3!/(3!)*2!
5c2=20/2 =10
total number of events = 13c4 = 13!/(13-4)!*4! = 13*12*11*10*9!/(9!)*4!
total number of events = 13*12*11*10/(4*3*2*1) = 715
the probability that 2 men and 2 women will be selected = 8c2 *5c2/13c4
a.
the probability that 2 men and 2 women will be selected = 28*10/715 = 0.3916
b.
the probability that men will constitute a majority in the team = 8c3 *5c1/13c4
because men majority so that the only possible case
8c3 value = 8!/(8-3)!*3! = 8*7*6*5!/(5!)*3! = 8*7*6/(6)= 56
5c1 = 5
the probability that men will constitute a majority in the team = 8c3 *5c1/13c4 = 56*5/715 = 280/715
the probability that men will constitute a majority in the team =0.3916