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!

  • $\begingroup$ "should I be taking into account the conditional statement somehow?" Well, YES, definitely. $\endgroup$
    – Did
    Jun 1 '14 at 15:56
  • 1
    $\begingroup$ That's what I was worried about. I guess I'm confused by when to consider a conditional statement, because what we've learned so far in my class is everything but conditional statements. I'm assuming that the presence of a conditional statement in a question means that you can't just use the simple multiplication rule to find a probability, and that you must use, for example $P(A \cap B) = P(A)P(B|A) = 0.72$. Is this right? $\endgroup$
    – Kestrel
    Jun 1 '14 at 16:08

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).


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