Question 7 of 12

Question 7 (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) Two independent rolls of a fair die, sample space size $36$. (b) $P(\text{male})=0.48$, $P(\text{female})=0.52$; $P(\text{blond}\mid\text{male})=0.45$, $P(\text{blond}\mid\text{female})=0.60$; dark hair is the complement of blond within each sex.

Find. (a) $P(\text{first}\ge\text{second})$. (b) $P(\text{male}\mid\text{dark hair})$.

Approach. (a) count favorable outcomes out of 36 equally-likely pairs (or use symmetry with the tie count); (b) apply Bayes' theorem with the law of total probability for the denominator.

1. (a) $P(\text{first}\ge\text{second})$. Of the 36 equally likely ordered pairs $(i,j)$ with $i,j\in\{1,\ldots,6\}$, exactly 6 are ties ($i=j$). By symmetry, the remaining $30$ split evenly between $i\gt j$ and $i\lt j$, so $15$ pairs have $i\gt j$. The event $i\ge j$ is the union of "ties" and "$i>j$": $$P(\text{first}\ge\text{second}) = \dfrac{6+15}{36} = \dfrac{21}{36}$$ $\boxed{P=\dfrac{7}{12}\approx 0.583}$
2. (b) $P(\text{male}\mid\text{dark hair})$ via Bayes' theorem. Dark-hair rates are the complements of the blond rates: $P(\text{dark}\mid M)=1-0.45=0.55$, $P(\text{dark}\mid F)=1-0.60=0.40$. Total probability of dark hair: $$P(\text{dark}) = P(M)P(\text{dark}\mid M)+P(F)P(\text{dark}\mid F) = (0.48)(0.55)+(0.52)(0.40) = 0.264+0.208 = 0.472$$ Bayes' theorem: $$P(M\mid\text{dark}) = \dfrac{P(M)P(\text{dark}\mid M)}{P(\text{dark})} = \dfrac{0.264}{0.472}$$ $\boxed{P(M\mid\text{dark}) = \dfrac{33}{59}\approx 0.559}$