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May 29, 2020

#1
+732
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May 29, 2020
#2
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Oh wait. I meant this:

A bag contains five white balls and four black balls. Your goal is to draw two black balls.

You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)

Guest May 29, 2020
#3
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and

Guest May 29, 2020
#4
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But can you help me with this first?

A bag contains five white balls and four black balls. Your goal is to draw two black balls.

You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)

May 29, 2020
#5
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I think he'll help both of us if he sees our replies.

Guest May 29, 2020
#6
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ok gimme some time, i have to wash the dishes

May 29, 2020
#7
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i'll try to help when i get back :)

#8
+1

Thank you so much!!

My questions are the two links that I attached on one of the replies above.

Guest May 29, 2020
#9
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ok so the answer to this problem:

A bag contains five white balls and four black balls. Your goal is to draw two black balls.

You draw two balls at random. Once you have drawn two balls, you put back any white balls, and redraw so that you again have two drawn balls. What is the probability that you now have two black balls? (Include the probability that you chose two black balls on the first draw.)

case 1: getting 2 black balls on 1st try = 4/9*3/8= 1/6

case 2: getting 1 black and 1 white: the prob that you chose one black ball and one white ball is $$\dfrac{4 \cdot 5}{\binom{9}{2}} = \dfrac{5}{9}$$.  but then, you'd have to put back the white ball, so the bag currently contains three black balls and five white balls. so the prob that you draw a black from there is $$\frac{3}{8}$$.

case 3: getting 2 white balls: the probability that you chose two white balls is $$\dfrac{\binom{5}{2}}{\binom{9}{2}}= \dfrac{5}{18}$$. like in the last case, you'd have to put back both white balls, so the bag contains four black balls and five white balls. The probability that you draw two black balls is then $$\frac{1}{6}$$.

so, the answer is $$\frac{1}{6} + \frac{5}{9} \cdot \frac{3}{8} + \frac{5}{18} \cdot \frac{1}{6} = \boxed{\frac{91}{216}}.$$

May 29, 2020
#10
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Guest May 29, 2020
edited by Guest  May 29, 2020
#11
+732
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edited by lokiisnotdead  May 29, 2020
#12
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No, if you read carefully, the question is a bit different.

Can you answer my other question at least then?

Guest May 29, 2020
#13
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sure i'll try!

just out of curiosity where is the 2nd question from? i feel like I've seen it b4

#14
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i posted it on ur original question!

May 29, 2020
edited by lokiisnotdead  May 29, 2020
#15
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Thank you so much!!!

Guest May 29, 2020