1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Your calculations and arguments are correct. There is only a clerical error in a) in the line after you mention Bayes's theorem: You wrote

$$P(A_k|B)$$

in the enumerator when you meant (and had calculated before)

$$P(A_k\cap B)$$.

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – TryingHardToBecomeAGoodPrSlvr Mar 20 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.