# Random biased coins problem

So the problem goes something like this: There are 3 coins in a bag (C1, C2, C3). C1 comes up heads $1/3$ of the time, C2 $1/2$ of the time, and C3 $3/4$ of the time. Now you choose a coin at random, flip it, record the outcome and without replacing the coin back in the bag, pick another coin (out of the 2 remaining) and flip it.
a) What is the probability that 1 heads and 1 tails is flipped and b) what is the probability that a heads was flipped on the 1st flip given a tails was flipped on the 2nd.

I have figured out how to go about this if we put the coin back into the bag. In that case the 1 heads and 1 tails is $P(H_1T_2) + P(T_1H_2)$ which using a probability tree I found to be 95/216 (not sure if this is correct, just what I got as an answer). Same way for b) i found the answer to be 228/432 when you put the coin back into the bag.

I am not sure how to go about finding the solution to when the coin is not put back into the bag. Any help would be appreciated.

I am fairly new to probability so if the problem isn't well stated I apologize.

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For a) you need to work out all the cases. One way you can get 1H+1T is to pull C1 first, get a head, then get a tail. The chance of this is $\frac 13$(pull C1)$\frac 13$(get head)$[\frac 12\cdot \frac 12$(pull C2 and get tail)$+\frac 12 \cdot \frac 14$(pull C3 and get tail)$]=\frac 19(\frac 14 + \frac 18)=\frac 1{24}$. If you are careful, all the possibilities will be disjoint and you can add them. Not replacing is represented by the fact that the second draw is restricted to C2 and C3 (given that you start with C1).

For b) you need to calculate the probability $\frac{H_1T_2}{H_1T_2+T_1T_2}$. $H_1T_2$ is just part of the result of a). $T_1T_2$ is calculated similarly.

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Ok, this helps a ton, thank you. – Lok Sep 18 '12 at 3:30

You can still make a tree for the version without replacement. Here’s a large part of a rough sketch of one:

The first three-way branching represents the first draw; the two-way branching that follows represents the result of the first coin flip, heads or tails; the two-branching below that represents the second draw; and the final two-way branching, which I didn’t draw in its entirety, represents the result of the second coin flip. The probabilities for the coin flips depend on which coin is being flipped; the probabilities for the second draw are $1/2$ for each of the coins remaining in the bag.

To solve the first part of the problem using this method, just complete the tree and add up the probabilities of the final states corresponding to getting a head and a tail. For the second part, add up the probabilities of the final states that had a tail on the second flip; call that total $p_2$. Then add up the probabilities of the final states that had tails on both flips; call that total $p_{12}$. The probability that the first flip was tails given that the second was tails is then $\dfrac{p_{12}}{p_2}$.

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This is also very helpful, thank you. Also just a follow up question which may not be appropriate for this forum but how did you make that tree diagram, I have been trying to learn latex and cannot figure out how to make a tree diagram on it. – Lok Sep 18 '12 at 4:18
@Lok: I did a quick and dirty drawing in Paint, actually, and uploaded it into my answer. – Brian M. Scott Sep 18 '12 at 4:24
Ah ok, thank you – Lok Sep 18 '12 at 4:25