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.)

I know there are three cases: draw two black balls immediately, draw one black and one white, and draw two white. The probability to draw two blacks is 1/6, then I'm stuck.

Guest Apr 23, 2020

#1**+1 **

Probability of getting two black balls: 4/8 x 3/7 = 12/56

Probability of getting the first ball to be black and the second ball to be white: 4/8 x 4/7 = 16/56

Now, put back the white ball and draw; the probability of getting a black ball on this draw is 3/7.

Final probability for this case: 16/56 x 3/7 = 48/392.

Probability of getting the first ball to be white and the second ball to be black: 4/8 x 4/7 = 16/56

Now, put back the white ball and draw; the probability of getting a black ball on this draw is 3/7.

Final probability for this case: 16/56 x 3/7 = 48/392.

Probability of getting two white balls: 4/8 x 3/7 = 12/56

Now, put back the two white balls and draw; the probability of getting two white balls: 4/8 x 3/7 = 12/56.

Final probability for this case: 12/56 x 12/56 = 144/3136.

To find the total probability, we need to add these answers: 12/56 + 48/392 + 48/392 + 144/3136 = 99/196.

geno3141 Apr 23, 2020