Conditional Probability Coin and Urn Question

Hi,

Quick question (hopefully) on conditional probability. The question says, Consider a Biased coin, where the chance of heads is 2/3 and tails is 1/3. If the coin shows tails, a chip is drawn from urn 1, and if heads, drawn from urn 2. Urn 1 has 3 white chips and four red chips. Urn 2 has 6 white and 3 red chips. Chips drawn from the urns are not replaced. What is the probability that a white chip will be drawn on the second toss?

We have stared at this question for ages and so far have only figured that we could be able to use a tree diagram to figure this out, any help would be appreciated. Thanks!!

Re: Conditional Probability Coin and Urn Question

Hey Cotty.

Are you allowed to use known results of distributions? If so, take a look at this distribution for sampling without replacement:

Hypergeometric distribution - Wikipedia, the free encyclopedia

Re: Conditional Probability Coin and Urn Question

Hey Chiro, thanks for getting back to me! Ahh well i don't remember ever hearing anything about it, and I think they would rather we do it how they have taught us. Looks interesting though!!

Re: Conditional Probability Coin and Urn Question

The hyper-geometric distn. isn't really of much use here. I would say a tree diagram might be the easier thing to do visually. Mathematically what you have is P(W2) = P(W2|HH,R1)+P(W2|HH,W1)+P(W2|TT,R1)+P(W2|TT,W1)+P( W2|TH,R1)+P(W2|TH,W1)+P(W2|HT,R1)+P(W2|HT,W1).