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)=\frac{5!}{4!\cdot(5-4)!}=\frac{5!}{4!\cdot1!}=5$$C(7,2)=\frac{7!}{2!\cdot(7-2)!}=\frac{7!}{2\cdot5!}=\frac{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:

$C(12,6)=\frac{12!}{6!\cdot(12-6)!}=\frac{12!}{6!\cdot6!}=\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7}{6\cdot5\cdot4\cdot3\cdot2}=2\cdot11\cdot2\cdot3\cdot7=924$

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