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

You simultaneously 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 choose two black balls on the first draw.)

 

I know that the probability of getting two black balls on the first draw is 3/28. I need help with the rest of it though. 

 Mar 29, 2023
 #1
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There are two ways to get two black balls: either you draw two black balls on the first attempt, or you draw one black ball and one white ball on the first attempt, then draw two black balls on the second attempt.

The probability of drawing two black balls on the first attempt is:

P(draw two black balls on the first attempt) = (3/8) * (2/7) = 6/56 = 3/28

The probability of drawing one black ball and one white ball on the first attempt is:

P(draw one black ball and one white ball on the first attempt) = (3/8) * (5/7) + (5/8) * (3/7) = 30/56 = 15/28

If you draw one black ball and one white ball on the first attempt, then the probability of drawing two black balls on the second attempt is:

P(draw two black balls on the second attempt | one black ball and one white ball on the first attempt) = (3/8) * (2/7) = 6/56 = 3/28

To calculate the probability of drawing two black balls, we need to add the probability of the two cases:

P(draw two black balls) = P(draw two black balls on the first attempt) + P(draw two black balls on the second attempt | one black ball and one white ball on the first attempt)

P(draw two black balls) = (3/28) + (15/28)*(3/28) = 54/784 = 27/392

 Mar 29, 2023
 #2
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Sorry, but on my end, it said wrong.

Guest Mar 29, 2023

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