3
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ "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
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.