# Probability question: blue and cyan balls in an urn

1. Nov 7, 2009

### alex07966

An urn contains b blue balls and c cyan balls. A ball is drawn at random, its colour is noted, and it is returned to the urn together with d further balls of the same colour. This procedure is repeated indefinitely.

i) What is the probability that the second ball drawn is cyan?

ii) What is the probability that the first ball drawn is cyan given that the second ball is cyan?

iii) Let Cn denote the event that the nth ball drawn is cyan. Show that P(Cn) = P(C1), for all n ≥ 1.

iv) Find the probability that the first drawn ball is cyan given that the nth drawn ball is cyan.

v) Find the probability that the first drawn ball is cyan given that the following n drawn balls are all cyan.
What is the limit of this probability as n → ∞?

Answer: Parts i and ii I can do: part i is c/(b+c) using the theorem of total prob. part ii is (c+d)/(b+c+d) using bayes theorem.
Now for part iii I have tried to prove this using induction:
True for n=1 and 2 so assume true for n = k-1, then using total prob theorem:

P(C_k) = P(C_k | C_k-1)*P(C_k-1) + P(C_k | B_k-1)*P(B_k-1)

where B_n is the event the nth ball drawn is blue.

I'm not sure if the above equation is actually correct as i dont know if C_k-1 and B_k-1 partition the sample space??

I think this gives:

= [(c+d)/(b+c+d)]*[c/(b+c)] + [c/(b+c+d)][1 - c/(b+c)] = c/(b+c) = P(C_1)

So this gives the correct answer but i get the feeling my working out is wrong!?

Now part iv: I have got P(C_1 | C_n) = P(C_n | C_1) but then I have no idea what to do from there....

Part v: = P(C_1 | C_2 intersect C_3 intersect ..... intersect C_n intersect C_n+1). Then what???

Any help would be greatly appreciated! :D