SOLUTION: 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 b

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Question 1161040: 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 have tried and got 41/81 and 89/216. What am I doing wrong???
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Honestly, now.... How are we supposed to know what you are doing wrong if all you do is show us your (apparently wrong) answers?

The rules for posting questions on this forum say that you should show any work you have done on the problem. If you had done that, we might be able to tell you what you did wrong....

Scenario 1: draw two black on first draw

P(B,B) = (4/9)(3/8) = 12/72 = 1/6

The probability of ending up with two black by scenario 1 is 1/6.

Scenario 2: draw one black and one white; put back the white, draw a black

P(B,W) = (4/9)(5/8) = 20/72 = 5/18
P(W,B) = (5/9)(4/8) = 20/72 = 5/18
P(one black and one white on first draw) = 5/18+5/18 = 5/9

When you put the white back, there are now 5 white and 3 black; the probability of drawing a black on the second draw is 3/8.

The probability of ending up with two black by scenario 2 is (5/9)(3/8) = 15/72. = 5/24

Scenario 3: draw two white; put both back and draw again, getting two black

P(W,W) = (5/9)(4/8) = 20/72 = 5/18

(As a check of these calculations, note that the sum of the probabilities for all scenarios for the first draw is 1: 12/72+40/72+20/72 = 72/72 = 1.)

Since you put back both whites, the bag now contains the same balls it did at the beginning. The probability of drawing two black on the second draw is then 12/72 = 1/6, as before.

The probability of ending up with two black by scenario 3 is (5/18)(1/6) = 5/108.

ANSWERS:

The probability of having two black balls after the first draw is 1/6.

The probability of having two black balls after the second draw is the sum of the probabilities for the three scenarios: 1/6+5/24+5/108 = 36/216+45/216+10/216 = 91/216.