Question

Probability: I need help.
So there are 6 colored balls and 3 are black. Ramona picks on and keeps it. Carlos picks one up too. What is the probability they with both pick a black ball.

So its 3/6. Ramona has a 1/2 chance to pick a black ball. But then when she keeps it Carlos has a 2/5 chance to pick a black ball.
I think im supposed to multiply but Im not too sure.

Answer 1

The two events are independent, therefore the two probabilities should be multiplied.

Answer 2

It's not independent since "Ramona picks one and keeps it. "

Answer 3

there's no replacement

Answer 4

So what do I do? multiply?

Answer 5

you still multiply though

P(both pick black) = P(Ramona picks black)*P(Carlos picks black based off Ramonas pick)

P(both pick black) = (3/6)*(2/5)

P(both pick black) = (3*2)/(6*5)

P(both pick black) = 6/30

P(both pick black) = 1/5

This is assuming no replacement is made after Ramona picks a ball

Answer 6

The events are independent, the reason being that the occurrence of one has absolutely no effect on the occurrence of the other. @jim_thompson5910 you are confusing occurrence with probability.

Answer 7

thank you @jim_thompson5910 !! you're amazing :)

Answer 8

No the events cannot be independent

What Ramona picks changes what Carlos will pick

Answer 9

they can only be independent if the ball Ramona picks is replaced or put back

Answer 10

Sorry @jim_thompson5910 the events are independent and the correct result is obtained when the two probabilities found by @airrbae are multiplied.

Answer 11

yes multiplication does occur, but they're still not independent

if one event affects the probability of the other, then one event is dependent on the other (and they are not independent)

Answer 12

When the probability is calculated using the hypergeometric distribution (sampling without replacement) the value obtained is consistent with the proof of independence which is
\[P(A\cap B)=P(A)\times P(B)\]

Answer 13

The formula used would be

P(A and B) = P(A)*P(B | A)

Now if A and B were independent, then P(B | A) = P(B) since event A wouldn't change P(B) at all which is why you can only say

P(A and B) = P(A)*P(B)

if A and B were independent events

Answer 14

but they are not independent events

Carlos has 1 less ball to choose from when Ramona picks out her ball, so the probability will be altered

Answer 15

Whether or not Ramona has picked a ball has no effect on whether or not Carlos picks a ball. Independence relates to the occurrence of an event (picking a ball) and not to whether the probability of a certain outcome from an event (picking a ball) is affected by the outcome of a previous event.
In any case the proof of independence that I gave should settle the matter.

Answer 16

you have to define your events carefully

I'm talking about the event of picking a black ball

you're just talking about picking any ball in general

Answer 17

we're focused on the probability of picking a black ball for both people

so that's why

A = event of Ramona picking a black ball

B = event of Carlos picking a black ball

Answer 18

@jim_thompson5910 Thank you very much. I accept your explanation that conditional probability is the key to understanding the correct theory. Thank you for your help :)

Answer 19

you're welcome