Jelly-Bean Probability

A jar contains four red, four yellow, and three green jelly beans. If Joan and Jim take one jelly bean each, the probability that they both take a red jelly bean is:

4/11? (1/2)(4/11) + (1/2)(4/11)

or should I be using P(A) x P(B) for this?

Silverwing
P(A) = 4/11 (There are 4 red jelly beans for Joan to pick, out of a total possible 11)
P(B) = 3/10 (After Joan has now taken a red jelly bean, obviously there are only 3 left and a total possible 10 to pick)

Your question wants Joan AND Jim to pick a red jelly bean, so you should multiply the two. If the question wanted either Joan OR Jim to pull one, you would add the probabilities.

Not too sure where you got (1/2)(4/11) from :P Maybe if you post your reasoning, we could help you out.

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ms. confused said:
A jar contains four red, four yellow, and three green jelly beans. If Joan and Jim take one jelly bean each, the probability that they both take a red jelly bean is:

4/11? (1/2)(4/11) + (1/2)(4/11)

or should I be using P(A) x P(B) for this?
As Silverwing has said, they are both taking a bean, which means it is an 'AND' situation. If the question said they replaced it then the two events would be independent but the question suggests they are not independent of each other.

Pyrrhus
Homework Helper
This is probably a independent event probability, because in this knowing P(A) and P(B) isn't enough to determine $P(A \cap B)$.

HallsofIvy
This is probably a independent event probability, because in this knowing P(A) and P(B) isn't enough to determine $P(A \cap B)$.