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Suppose we put eight white and two black balls into a bag and then draw forth the balls one at a time.
If we repeat this experiment many times, which draw is most likely to produce the first black ball?

Try to guess (=estimate guided by intuition), then evaluate.
You'd be surprised.

Source: A.E. Lawrence, “Playing With Probability”
Mathematical Gazette, (1969)

 No Solution Yet Submitted by Ady TZIDON No Rating

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 solution | Comment 1 of 6
Probability that the first black ball occurs at

Draw 1: 2/10 = 1/5 = 0.2 (the most likely)
Draw 2: 4/5 * 2/9 = 8/45 = 0.17777...
Draw 3: 4/5 * 7/9 * 2/8 = 7/45 = 0.155555...
Draw 4: 4/5 * 7/9 * 6/8 * 2/7 = 2/15 = 0.1333333...

continuing:

Draw 5: 1/9 = 0.11111...
Draw 6: 4/45 = 0.08888...
Draw 7: 1/15 = 0.06666...
Draw 8: 2/45 = 0.04444...
Draw 9: 1/45 = 0.02222...

So the modal case is not the mean case. I guess that's the surprise. It's less surprising if there were replacement, as then it would be obvious that each successive delay just requires more  successive white ball draws. Here, you might think the successively larger proportion of black balls might offset the requirement of larger numbers of white ball choices, but that's not the case.

 Posted by Charlie on 2015-09-02 11:45:16

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