Question

1. mrs. stevens calculus class has 23 students in it, 12 girls and 11 boys. she uses a calculator to select two students at random to solve a problem on the board. (a. what is the probability that both students are boys? b. given that the second student chosen is a boy, what is the probability that the first student was also a boy? (5 points))

Explanation:

Step1: Calculate total 2-student selections

Total ways to choose 2 students: $\binom{23}{2} = \frac{23 \times 22}{2} = 253$

Step2: Calculate boy-only 2-student selections

Ways to choose 2 boys: $\binom{11}{2} = \frac{11 \times 10}{2} = 55$

Step3: Find probability for part a

Probability both are boys: $\frac{\text{Boy selections}}{\text{Total selections}} = \frac{55}{253} = \frac{5}{23}$

Step4: Define events for part b

Let $A$ = first student is boy, $B$ = second student is boy. Use conditional probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$

Step5: Calculate $P(B)$ (second is boy)

$P(B) = P(\text{first girl, second boy}) + P(\text{first boy, second boy}) = \frac{12}{23} \times \frac{11}{22} + \frac{11}{23} \times \frac{10}{22} = \frac{132 + 110}{506} = \frac{242}{506} = \frac{11}{23}$

Step6: Calculate $P(A \cap B)$ (both boys)

$P(A \cap B) = \frac{5}{23}$ (from part a)

Step7: Solve conditional probability

$P(A|B) = \frac{\frac{5}{23}}{\frac{11}{23}} = \frac{5}{11}$

Answer:

a. $\frac{5}{23}$
b. $\frac{5}{11}$