Assume a jar has 7 red marbles and 5 black marbles. Draw out 3 marbles with and without replacement. Find the requested probabilities.
(c) P(one red and two black marbles)
With replacement , without replacement .
(d) P(red on the first draw and black on the second draw and black on the third draw)
With replacement , without replacement .

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- anonymous

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- anonymous

What does with replacement mean?

- anonymous

putting back in the jar

- anonymous

Oh. I dunno about that, but without replacement is pretty easy.

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- anonymous

i understand more of the with replacement but if you can help with the without replacement that would be good

- anonymous

For the first question of one red and two black we just want to know how many ways we can choose one red, and two black divided by how many ways we can choose 3 marbles from 12.
So:
\[\frac{{7 \choose 1} * {5 \choose 2}}{12 \choose 3}\]

- anonymous

so 12/3?

- anonymous

No,
7 choose 1 = 7
5 choose 2 = 10
12 choose 3 = 220
So it is 70/220.
Are you familiar with the choose function (and it's notation)?

- anonymous

kind of

- anonymous

was that without replacement?

- anonymous

Yes

- anonymous

do you know how to do d?

- anonymous

For the other one it's just
(Probability of choosing a red)*(Probability of choosing a black)*(Probability of choosing a black)
So (7/12) * (5*/11) * (4/10)

- anonymous

140/1320....can be reduced though right?

- anonymous

14/132 = 7/66

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