Question

Imagine an urn with two balls, each of which may be either white or black. One of these balls is drawn and is put back before a new one is drawn. Suppose that in the first two draws white balls have been drawn. What is the probability of drawing a white ball on the third draw?

Answer 1

Answer:

Accepting it is similarly likely for each ball to white or dark, the likelihood is 1/4 that the two balls are white, 1/2 that one ball is white and one ball is dark, and 1/4 that the two balls are dark.

At that point, if the two balls are white, the likelihood that the two balls drawn are white is 1.

In the event that one ball is white and one is dark, the likelihood that the two balls drawn are white is 1/4

In the event that the two balls are dark, the likelihood of drawing 2 white balls is 0

Consequently,

P(both balls are white and the two balls drawn are white) = 1/4 * 1

= 1/4

P(one ball is white and one is dark and both drawn balls are white = 1/2 * 1/4

= 1/8

At that point,

P(2 white balls|both drawn are white) = (1/4) / (1/4 + 1/8)

= 2/3

P(1 dark and 1 white ball|both drawn are white) = (1/8) / (1/4 + 1/8)

= 1/3

At that point,

P(next ball is white) = P(2 white balls|both drawn are white)*P(next white given 2 white balls) + P(1 white and 1 dark ball|both drawn are white)*P(next white given 1 white and 1 renounce)

= (2/3)*1 + (1/3)*(1/2)

= 2/3 + 1/6

= 5/6