How Do You Calculate the Probability of Both Picking a Red Jelly Bean?

[ms. confused]

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.

 

[The Bob]

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.

The Bob (2004 ©)

 

[Pyrrhus]

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 $P(A \cap B)$.

On the contrary, this is an example of dependent events. (it is "sampling without replacement"). If Jim goes first and takes a red jelly bean, there are 10 jelly beans left only 3 of which are red so Joan's prob of taking a red jelly bean is 3/10. If Jim takes either a yellow or green jelly bean there would 10 jelly beans left still but now 4 would be red so the probability of Joan taking a red jelly bean would be 4/10.
That's why Silverwing and The Bob use conditional[/b] probability: the probability that Joan picks a red jelly bean given that Jim picks a red jelly bean.

Of course, the problem doesn't say that Jim picked before Joan but it doesn't matter- if Joan picks first the probability that she picks a red jellybean is 4/11 and the probability that Jim picks red given that Joan picked red is 3/10 so you get exactly the same answer: (4/11)(3/10)= 6/55.