In the polya urn scheme what is the probability that the 3rd ball is red given exactly one of the first four balls is red?
so P(R3|exactly one of the first four balls is red) =
P(R3 and exactly one of the first four balls is red)/(exactly one of the first four balls is red)
P(R3) = P(R1)P(R2|R1)P(R3|R2 andR1) + P(B1)P(R2|B1)P(R3|B1 and R2) + P(B1)P(B2|B1)+P(R3|B1 and B2) + P(R1)P(B2|R1)P(R3|R1 andB2).
This is equal to (r/b+r)(r+c/b+r+c)(r+2c/b+r+2c) + (b/b+r)(b+c/b+r+c)(r/b+r+2c) + (b/b+r)(r/b+r+c)(r+c/b+r+2c)+(r/b+r)(b/b+r+c)(r/b+r+2c) = r/b+r I think.
P(exactly 1 is red out of 4) = (4 choose 1)(r/b+r)(b/b+r)^3
So do you divide P(R3)= r/b+r by P(exactly 1 is red out of 4)?
I am stuck here.
Also how would you find the probability that the 2nd ball is red and the nth ball is red? The probability second ball is red is still r/b+r. The probability that the nth ball is red is 1. So would it just be r/b+r?