# One coin chosen between a biased coin and a fair coin, and is tossed n times. Find probability of having gotten the biased coin.

In a different question, I had asked for clarification on the following problem where I wanted to just understand the problem. Now, I have attempted it and wish to know if my solution is right.

Problem statement:

A drawer contains two coins. One is an unbiased coin, which when tossed, is equally likely to turn up heads or tails. The other is a biased coin, which will turn up heads with probability $$p$$ and tails with probability $$1 − p$$. One coin is selected (uniformly) at random from the drawer. Two experiments are performed:

a) The selected coin is tossed $$n$$ times. Given that the coin turns up heads $$k$$ times and tails $$n − k$$ times, what is the probability that the coin is biased?

b) The selected coin is tossed repeatedly until it turns up heads $$k$$ times. Given that the coin is tossed $$n$$ times in total, what is the probability that the coin is biased?

My attempt:

a) Let $$F$$ be the set of outcomes where I have chosen the fair coin, and $$B$$ be the set of outcomes where I have chosen the biased coin. Let $$A_k$$ be the set of outcomes where I tossed $$n$$ times and got $$k$$ heads. I need to find $$P(B|A_k)$$.

$$P(A_k \cap F) = \frac{1}{2} {{n}\choose{k}}\frac{1}{2^n}$$ $$P(A_k \cap B) = \frac{1}{2} {{n}\choose{k}}p^k (1-p)^{n-k}$$ Since $$F$$ and $$B$$ partition the sample space, we have

$$P(A_k) = \frac{1}{2}{{n}\choose{k}} \left \{ p^k (1-p)^{n-k}+\frac{1}{2^n} \right \}$$

We from Bayes' theorem know that

$$P(B|A_k)=\frac{P(A_k|B)}{P(A_k)}$$

Hence we get

$$P(B|A_k)=\frac{p^k (1-p)^{n-k}}{p^k (1-p)^{n-k}+\frac{1}{2^n}}$$

b) In this case, we keep tossing till we get $$k$$ heads. Now this means that the last toss is a head. Now we know that we had to toss $$n$$ times to get $$k$$ heads.

As in (a) above, let $$F$$ be the set of outcomes where I have chosen the fair coin, and $$B$$ be the set of outcomes where I have chosen the biased coin. Let $$C_n$$ be the event that I had to toss $$n$$ times to get $$k$$ heads. I need $$P(B|C_n)$$.

$$P(C_n \cap F) = \frac{1}{2} \times \frac{1}{2} \times {{n-1}\choose{k-1}}\frac{1}{2^{n-1}}$$ $$P(C_n \cap B) = \frac{1}{2} \times p \times {{n-1}\choose{k-1}}p^{k-1} (1-p)^{n-k}$$

Again using Bayes' theorem along the lines of what was done in (a), we get

$$P(B|C_n)=\frac{p^k (1-p)^{n-k}}{p^k (1-p)^{n-k}+\frac{1}{2^n}}$$

I am a bit skeptic about my answer as the answers to (a) and (b) are turning out to be the same. I cannot find an intuitive explanation as to why that is.

Please provide feedback and let me know if I have solved this question correctly. In case there is a mistake, please point me to it.

$$P(A_k|B)$$
$$P(A_k\cap B)$$.
That both results are the same is a bit suprising, but then the ending state of both a) and b) are very similar: You have thrown $$n$$ coins and seen $$k$$ heads. The only difference is that in b) you know that the last coin is heads. But that doesn't change the general fact that the possible selections of which of those $$n$$ coints is heads or tails is the same for both the fair and unfair coin, and that this value cancel's out when yo do calulate the quotient.
• Thanks a lot for the feedback. I will not edit and leave it as $P(A_k|B)$ just so that readers can locate my mistake. I suppose I left out $P(B)$ so that when you have $P(A_k|B) \times P(B)$, then the equation would have been correct. Thanks a lot once again! – TryingHardToBecomeAGoodPrSlvr Mar 20 at 21:54