An urn contains 4 white balls and 6 black balls. A ball is chosen at random and

Question

An urn contains 4 white balls and 6 black balls. A ball is chosen at random and its color noted. The ball is then replaced, along with 3 more balls of the same color (so that there are now 13 balls in the urn). Then another ball is drawn at random from the urn. (a) Find the chance that the second ball drawn is white. (b) Given that the second ball drawn is white, what is the probability that the first ball drawn is black? (c) Suppose the original contents of the urn are w white and b black balls, and that after a ball is drawn from the urn, it is replaced along with d more balls of the same color. In part (a), w was 4, b was 6, and d was 3. Show that the chance that the second ball drawn is white is w/(w + b). Note that the answer does not depend on the value of d. Why is that?

Explanation / Answer

a) There is 4/10 chance you pick the white ball, and if you do, there will now be 7 white balls out of 13 so the chance of you picking a white the second time is 7/13.

There is 6/10 of you picking a black one, and if you do there will still be 4 whites out of 13 balls and so the chance of you picking a white one will be 4/13 the second time.

so the chance of you getting a white ball is the probability of the 2 situations added together:
(4/10)(7/13) + (6/10)(4/13) = 28/130 + 24/130 = 52/130 = 2/5

b)  There is (4/10)*(7/13)=28/130 chance u get a white then a white,

and (6/10)*(4/13) = 24/130 chance u get a black then a white,

Hence the probability that the first draw is black is 24/(24+28) = 6/13

c) There is w/(w+b) chance of u drawing a white ball, and if you do, there will now be w+d white balls and w+b+d balls in total. Hence the chance of u drawing white the second time will be (w+d)/(w+b+d)

There is b/(w+b) chance of u drawing a black ball, and if you do, there will still be w white balls and w+b+d balls in total. Hence the chance of you drawing white the second time will be w/(w+b+d)

Hence the chance of you drawing a white then a white is:

[w/(w+b)] [(w+d)/(w+b+d)] = [ (w^2 + wd) / (w+b)(w+b+d) ]

The chance of you drawing a black then a white is:

[b/(w+b)] [w/(w+b+d)] = [ wb / (w+b)(w+b+d) ]

Hence the required probability is

[ (w^2 + wd + wb) / (w+b)(w+b+d) ] =[(w(w+d+b))/(w+b)(w+b+d)] = w/(w+b)

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