A little help with what seems to be conditional probability

I want to make sure that I'm working this question correctly. Here goes:

A jar has 5 white balls. A ball is taken out randomly - if it's white, a black ball is inserted instead, and we continue to take balls out. If it's black, we stop.

Whats the probability that we will take 6 balls out until we stop?

So I actually drew it out and treated the event of '6 balls out' as 'we take out all the white ones' and ended up multiplying each probability of taking a white out at each stage.
So - at first we have 5/5 chance of getting a white one out, multiplied by 4/5 chance, 3/5... etc.

Is this a correct way to solve this? I don't know why but it seems iffy really.

Thanks

Muffins said:

I want to make sure that I'm working this question correctly. Here goes:

A jar has 5 white balls. A ball is taken out randomly - if it's white, a black ball is inserted instead, and we continue to take balls out. If it's black, we stop.

Whats the probability that we will take 6 balls out until we stop?

So I actually drew it out and treated the event of '6 balls out' as 'we take out all the white ones' and ended up multiplying each probability of taking a white out at each stage.
So - at first we have 5/5 chance of getting a white one out, multiplied by 4/5 chance, 3/5... etc.

Is this a correct way to solve this? I don't know why but it seems iffy really.

Thanks

Click to expand...

Drawing a tree is what I would do - if that happened to show me an algebraic way to do it, that would be fine, too.

Ball #1, P(white) = 5/5 = 1 --> insert black

Ball #2, P(white) = 4/5 --> insert black
...........P(black) = 1/5 stop

Ball #3, P(white) = (4/5)(3/5) --> insert black
...........P(black) = (4/5)(2/5) stop

. . .

"Clearly" the probability for having to draw out 6 balls is (5/5)(4/5)(3/5)(2/5)(1/5)

That confirms what I did.

Thank you!