Question

Mr. Harrow has eight boys and six girls in his Honors Pre-calculus class. If he randomly chooses two students, one at a time, what is the probability that they are both girls?
A)
11/14

B)
15/91

C)
15/98

D)
74/91

Answer 1

$|\Omega|=C(14,2)=\dfrac{14!}{2!12!}=\dfrac{13\cdot14}{2}=91\\|A|=C(6,2)=\dfrac{6!}{2!4!}=\dfrac{5\cdot6}{2}=15\\\\P(A)=\dfrac{15}{91}\Rightarrow \text{B}$

Answer 2

Answer: 15/91 which is choice B

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There are two methods to find this answer.

Method 1) We have 6 girls and 8+6 = 14 students. The probability of picking a girl is 6/14 = 3/7. After the first girl is chosen, we have 5 girls left out of 14-1 = 13 students overall. The probability of picking another girl (assuming the first selection was a girl) is 5/13. Multiply these probabilities: (3/7)*(5/13) = (3*5)/(7*13) = 15/91

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Method 2) We can use the nCr combination formula. Order does not matter.

We have nCr = 6 C 2 = 15 ways to pick 2 girls. See the attached image below for the steps (figure 1)

Out of nCr = 14 C 2 = 91 ways to pick 2 students. See the attached image below for the steps (figure 2)

So that's another way to get the answer 15/91.