There are two flowers, $A$ and $B$.
The probability that each one is pollinated is $0.8$.
The probability that $B$ is pollinated given $A$ is pollinated is $0.9$.
What is the probability that:
a) both flowers are pollinated?
b) one or the other or both is pollinated?
c) A is pollinated given that B is?
d) A is pollinated but B is not?
for a), my rationale is that $P(A) = 0.8$, and $P(B) = 0.8$, so $P(A \cap B) = P(A)P(B) = 0.64$. should I be taking into account the conditional statement somehow?
for b), I'm thinking the statement is literally just the identity of a union of two events, so $P(A \cup B)$, which would be $P(A) + P(B) – P(A \cap B) = 0.8 + 0.8 – 0.64 = 0.96$?
for c), $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.64}{0.8} = 0.8$. Is this right?
for d), $P(A \cap B') = P(A)P(B') = (0.8)(0.2) = 0.16$. is this right?
I know that many of my answers hinge on whether my thinking for a) is right, so I expect that a lot of this is wrong.
Any help is appreciated!
Best Answer
Note that $A$ and $B$ are not independent so $P(A\cap B)\not=P(A) P(B)$.
Rather, $P(A\cap B)=P(A) P(B\vert A)$.
This should give you (a) and then (b) and (c) just need to be corrected accordingly. The same reasoning applies to (d).