A3 - conditional probabilities

conditional probability

$P (A | B) = \frac{P (A \cap B)}{P (B)}$

P (both colts | at least one colt) = \frac{P (both colts and atleast one colt)}{P (at least one colt)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}

P (both colts | first born is a colt) = \frac{P (both colts and first born is a colt)}{P (first born is a colt)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}

now, given that $10 %$ horses produce only colts, $10 %$ only fillies, and $80 %$ are equally likely to produce both gender

\begin{matrix} P (both colts) = 0.1 \times 1 + 0.1 \times 0 + 0.8 \times 0.25 = 0.3 \\ P (at least one colt) = 0.1 \times 1 + 0.1 \times 0 + 0.8 \times 0.75 = 0.7 \\ ∴ P (both colts|at least one colt) = \frac{3}{7} \end{matrix}

product rule

P (E_{1} \cap E_{2}) = P (E_{1}) P (E_{2} | E_{1}) = P (E_{2}) P (E_{1} | E_{2})

chain rule for conditioning

let $E_{1}, E_{2}, \dots, E_{n}$ be events, where $P (\cap_{i = 1}^{n} E_{i}) > 0$ , then

P (\cap_{i = 1}^{n} E_{1}) = P (E_{i}) \times P (E_{2} | E_{1}) \times P (E_{3} | E_{1} \cap E_{2}) \dots P (E_{n} | \cap_{i = 1}^{n - 1} E_{i})