Probability Basics. Combinatorics. Bayes Theorem. Part 1

Published 9/27/2025

\text{In a family with 2 children, at least one is a girl. What is the probability that both are girls?}\\[8pt] \textbf{Answer: }\frac{1}{3}.

\text{A family has 2 children. One child chosen uniformly at random from the two is a boy. What is the probability that both children are boys?}\\[8pt] \textbf{Answer: }\frac{1}{2}.

\text{In a family with 3 children, at least one child is a boy. What is the probability that all three children are boys?}\\[8pt] \textbf{Answer: }\frac{1}{7}.

\text{Two independent servers A and B are up with probabilities }p_A=0.9,\ p_B=0.8.\ \text{The system is up iff at least one server is up.}\\ \text{Given the system is up, what is the probability that both servers are up?}\\[8pt] \textbf{Answer: }\frac{P(A\cap B)}{P(A\cup B)}=\frac{0.72}{0.9+0.8-0.72}=\frac{36}{49}.

\text{A die is selected: fair with probability } \tfrac{2}{3},\ \text{loaded with probability } \tfrac{1}{3}.\\ \text{Loaded die rolls }6\text{ with probability } \tfrac{1}{2}\ \text{and each of }1\text{--}5\text{ with probability } \tfrac{1}{10}.\\ \text{A single roll shows }6.\ \text{What is the probability the selected die was loaded?}\\[8pt] \textbf{Answer: }\frac{\tfrac{1}{3}\cdot\tfrac{1}{2}}{\tfrac{1}{3}\cdot\tfrac{1}{2}+\tfrac{2}{3}\cdot\tfrac{1}{6}}=\frac{3}{5}.

\text{Factory }F_1\text{ makes }40\%\text{ of chips with defect rate }1\%.\ \text{Factory }F_2\text{ makes }60\%\text{ with defect rate }3\%.\\ \text{A randomly chosen chip is defective. What is the probability it came from }F_1\text{?}\\[8pt] \textbf{Answer: }\frac{0.4\cdot0.01}{0.4\cdot0.01+0.6\cdot0.03}=\frac{2}{11}.

\text{A family has 2 children. It is known that at least one child is a boy born on a Tuesday. What is the probability that both children are boys?}\\[8pt] \textbf{Answer: }\frac{13}{27}.

\text{A family has 2 children. The older child is a boy. What is the probability that both children are boys?}\\[8pt] \textbf{Answer: }\frac{1}{2}.

\text{In a two-child family, each birth is a boy with probability }p\in(0,1)\text{ independently. Given that at least one child is a boy, what is the probability that both are boys?}\\[8pt] \textbf{Answer: }\displaystyle \frac{p^2}{1-(1-p)^2}=\frac{p}{2-p}.