2.7 Independence and the Union of Events — Answers and Rubrics

$P(A)=2/6=1/3$, $P(B)=1/2$, and independence gives $P(A\cap B)=1/6$. Then $P(A\mid B)=(1/6)/(1/2)=1/3=P(A)$, and $P(B\mid A)=(1/6)/(1/3)=1/2=P(B)$. Finally $P(A\cup B)=1/3+1/2-1/6=2/3$.

$P(S\text{ union }H)=P(S\cup H)=0.60+0.50-0.40=0.70$. They are not independent because $P(S\cap H)=0.40\neq0.60(0.50)=0.30$; equivalently $P(H\mid S)=0.40/0.60\neq0.50$. Adding 0.60 and 0.50 double-counts the 40% who visited both.

$P(R)=P(L)=1/3$, $P(R\cap L)=0$, and $P(R\cup L)=2/3$, where L labels Blue. Given Red, Blue has conditional probability 0 instead of its marginal 1/3, so the events are dependent despite being disjoint.