[Guest]

An urn contains seven blue balls and six green balls. A series of drawings of one ball at a time is made such that the ball removed is returned immediately after the taking of the next following ball.

Determine the probability that the eighth ball drawn is blue, when:

(A) No information is available about the preceding drawings.

(B) The third ball drawn is known to be blue.

[Guest]

I came up with a recurrence formula for P(B)_n, the chance of taking a blue ball on the nth draw where P(B)₁ = 7/13

P(B)_n+1 = (6/12)P(B)_n+1 + (7/12)(1 - P(B)_n+1)

which produced a chance of 7/13 for each draw. So P(B)₈=7/13

Starting with P(B)₃ = 1, P(B)₈ = 0.5385...... (very close to 7/13)

Interestingly this formula will always tend towards 7/13.