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an urn contains 100 balls. 60 of the balls are white and 40 of them are black. We take out one ball and put it aside without looking at its color. We then take out a second ball. What is the probability that the second ball is white?

What I have tried so far is I calculate the overall number of outcomes which is 9900. if the first ball taken out is white, then the probability of the second ball being white is 59/99. if the first ball taken out is black, then the probability of the second ball being black is 60/99. I was thinking about should I take the average of these two numbers or not. But now I figured I should multiply the results by their weights. 59/990.6 + 60/990.4 = 3/5.

serendipity0217
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2 Answers2

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Hint: Every individual ball has an identical chance for being the second drawn and sixty among the one hundred are white.

Graham Kemp
  • 129,094
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There are 2 possible scenarios before you draw the white ball on the second try:

  1. You draw a white ball the first time
  2. You draw a black ball the second time

Scenario 1: $P(white, white) = \frac{60}{100} * \frac{59}{99} = \frac{59}{165}$

Scenario 2: $P(black, white) = \frac{40}{100} * \frac{60}{99} = \frac{8}{33}$

Since both scenarios led to you drawing a white ball the second time, you add them together, to get the following probability: $P(\text{2nd draw is white}) = \frac{3}{5}$

SoInstant
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