- #1

ms. confused

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

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

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- Thread starter ms. confused
- Start date

- #1

ms. confused

- 91

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

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

- #2

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.

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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- #3

The Bob

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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.ms. confused said:

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

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

The Bob (2004 ©)

- #4

Pyrrhus

Homework Helper

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- #5

HallsofIvy

Science Advisor

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Cyclovenom said:

On the contrary, this is an example of

That's why Silverwing and The Bob use

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.

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