Question

Out of a class of 5 girls and 7 boys, 6 are to be chosen randomly to visit the state house. What is the probability thisgroup of 6 will consist of 4 girls and 2 boys?7/665/442/2311/2

Answer 1

We need to combine 4 out of 5 girls, and 2 out of 7 boys. The number of ways each of these combinations can be made is:

`C(5,4)=(5!)/(4!\cdot(5-4)!)=(5!)/(4!\cdot1!)=5`
`C(7,2)=(7!)/(2!\cdot(7-2)!)=(7!)/(2\cdot5!)=(7\cdot6)/(2)=7\cdot3=21`

Then, the number of ways both combinations will happen is the product:

`C(5,4)\cdot C(7,2)=5\cdot21=105`

Now, the total number of ways we could choose any 6 students out of 12 (5 girls + 7 boys) is:

`C(12,6)=(12!)/(6!\cdot(12-6)!)=(12!)/(6!\cdot6!)=(12\cdot11\cdot10\cdot9\cdot8\cdot7)/(6\cdot5\cdot4\cdot3\cdot2)=2\cdot11\cdot2\cdot3\cdot7=924`

Finally, the required probability is found dividing the number of events of interest (105) by the total number of events (924):

P = 105/924 = 15/132 = 5/44