A bag contains one counter, known to be either white or black.
A white counter is put in, the bag shaken, and a counter drawn out,
which proves to be white.
What is now the chance of drawing a white
counter?
The following assumes it's equally likely to contain a black or a white counter before placing the new white counter.
Without replacement:
P(original was white and white was drawn) = 1/2
P(white would be drawn somehow) = 1/2 + (1/2)*(1/2) = 3/4
P(original was white given the known outcome of white) = 1/2 / (3/4) = 4/6 = 2/3
If the drawn counter is not placed back into the bag then that is the probability that the lone counter remaining in the bag is white as that would happen if and only if the original counter were white.
With replacement:
If the drawn counter were placed back in the bag then the chance of drawing another white counter is 2/3 + (1/3)*(1/2) = 2/3 + 1/6 = 5/6.
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Posted by Charlie
on 2017-11-07 14:19:37 |
2 scenarios: with and without replacement
