1. Problem: A couple plans to have three children. We assume each child is equally likely to be a boy or a girl with probability $\frac{1}{2}$ independently. Formula: For $n$ independent trials with success probability $p$, the probability of exactly $k$ successes is given by the binomial formula: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ Here, "success" can be having a girl or a boy depending on the question. --- a. Probability all children are girls: - Number of children $n=3$, number of girls $k=3$, $p=\frac{1}{2}$ - $$P(3\text{ girls}) = \binom{3}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^0 = 1 \times \frac{1}{8} \times 1 = \frac{1}{8}$$ b. Probability two children are boys: - Number of boys $k=2$, $p=\frac{1}{2}$ - $$P(2\text{ boys}) = \binom{3}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^1 = 3 \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{8}$$ c. Probability at least 2 children are girls: - This means 2 or 3 girls. - Calculate $P(2\text{ girls}) + P(3\text{ girls})$. - $$P(2\text{ girls}) = \binom{3}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^1 = 3 \times \frac{1}{4} \times \frac{1}{2} = \frac{3}{8}$$ - $$P(3\text{ girls}) = \frac{1}{8}$$ - Total: $$\frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$$ d. Probability couple will have 2 boys at the most: - This means 0, 1, or 2 boys. - Calculate $P(0\text{ boys}) + P(1\text{ boy}) + P(2\text{ boys})$. - $$P(0\text{ boys}) = \binom{3}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^3 = 1 \times 1 \times \frac{1}{8} = \frac{1}{8}$$ - $$P(1\text{ boy}) = \binom{3}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^2 = 3 \times \frac{1}{2} \times \frac{1}{4} = \frac{3}{8}$$ - $$P(2\text{ boys}) = \frac{3}{8}$$ (from part b) - Total: $$\frac{1}{8} + \frac{3}{8} + \frac{3}{8} = \frac{7}{8}$$ --- 2. Problem: Rolling a pair of dice once. Each die has 6 faces, so total outcomes = $6 \times 6 = 36$. We want the probability the sum of the dice is a certain number. --- a. Probability sum is 5: - Possible pairs: (1,4), (2,3), (3,2), (4,1) - Number of favorable outcomes = 4 - $$P(\text{sum}=5) = \frac{4}{36} = \frac{1}{9}$$ b. Probability sum is 9: - Possible pairs: (3,6), (4,5), (5,4), (6,3) - Number of favorable outcomes = 4 - $$P(\text{sum}=9) = \frac{4}{36} = \frac{1}{9}$$