I'm trying to understand the following question and solution:
Let A be the event that it rains tomorrow and suppose $P(A) = 1/3$. Also supposed that a fair coin is tossed; let B be the event it lands heads up. We have $P(B)=1/2$.
Now I ask what is $P(A|B)$? Due to independence $P(A|B)=P(A)=1/3$. The result of the coin toss had nothing to do with tomorrow's weather. Two events are independent if $P(A \cap B) = P(A)P(B)$
Now I have tried to recreate this problem so I can use the set theoretic definitions. It was suggested that $P(\{(r,h)\}) = 1/6, P(\{(r,t)\}) = 1/6$ be chosen for $P(Rain_{tomorrow})$ if independence of a fair coin toss is assumed. I'm having challenges understanding how this assumption is made. Does anyone know?
$S = Rain \times CoinToss$, $|S| = |Rain| \times |CoinToss| = 2 \times 2$
$Rain_{tomorrow} = \{ (r,h), (r,t) \}$ $P(\{(r,h)\}) = 1/6, P(\{(r,t)\}) = 1/6, P(Rain_{tomorrow}) = 1/3$
$Heads_{tomorrow} = \{ (r,h), (\lnot r,h) \}$ $P(\{(r,h)\}) = 1/6, P(\{(\lnot r,h)\}) = 2/6, P(Heads_{tomorrow}) = 1/2$
$P(Rain_{tomorrow}|Heads_{tomorrow}) = P(Rain_{tomorrow} \cap Heads_{tomorrow})/P(Heads_{tomorrow}) = (1/6)/(1/2) = 2/6 = 1/3$
Now to check for independence: $P(A \cap B) = P(A)P(B)$ $P(A \cap B) = P(\{(r,h\}) = 1/6$ $P(A)P(B) = (1/2) * (1/3) = 1/6$
So it would appear the events are independent.
Appreciate your guidance.