In a company there are 7 executives: 4 women and 3 men. Three are selected to attend a management seminar. Find these probabilities.
a. All 3 selected will be women
b. All 3 selected will be men
c. 2 men and 1 woman will be selected
d. 1 man and 2 women will be selected
2 Answers

a) Probability that all 3 selected will be women = 4C3*3C0/7C3 = 4*1/35 = 4/35 = 0.1143
b) Probability that all 3 selected will be men = 4C0*3C3/7C3 = 1*1/35 = 1/35 = 0.0286
c) Probability that 2 men and 1 women will be selected = 3C2*4C1/7C3 = 3*4/35 = 12/35 = 0.3429
d) Probability that 1 man and 2 women will be selected = 3C1*4C2/7C3 = 3*6/35 = 18/35 = 0.5143

First calculate the total no. of ways in which the selection can be done.
That would be 7C3 = 35
a.)
No. of ways the 3 women could be selected = 4C3 = 4
As a probability, = 4/35 = 0.11428 = 0.114 (3 sf)
b.)
No. of ways the 3 men could be selected = 3C3 = 1
As a probability, = 1/35 = 0.028571 = 0.0286 (3 sf)
c.)
No. of ways the 2 men could be selected = 3C2 = 3
No. of ways the woman could be selected = 4C1 = 4
No. of ways 2 men and 1 woman will be selected = 3*4 = 12
As a probability, = 12/35 = 0.34285 = 0.343 (3 sf)
d.)
No. of ways the man could be selected = 3C1 = 3
No. of ways the 2 women could be selected = 4C2 = 6
No. of ways the man and 2 women will be selected = 3*6 = 18
As a probability, = 18/35 = 0.51428 = 0.514 (3 sf)