Which coin?

The probability of the fair coin getting the sequence is $\dfrac{32}{1024}$ . The probability of the unfair coin getting the sequence is $\dfrac{27}{1024}$ . Therefore the probability that the coin picked is the unfair coin is $\dfrac{27}{27+32}=\dfrac{27}{59}$ and our desired answer is $27+59=\boxed{86}$ .

Probability With the Fair Coin: $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{32}$

Probability With the Unfair Coin: $\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}=\frac{27}{1024}$

Combined Probabilities: $\frac{1}{32}+\frac{27}{1024}=\frac{59}{1024}$

Probability with Unfair Coins(Out of all the Possible Outcomes): $\frac{\frac{27}{1024}}{\frac{59}{1024}}=\frac{27}{59}$

Therefore the answer is $27+59 = \boxed{86}$

Use Bayes' Theorem:

...or you can think of it like this. There is a certain probability that this will happen assuming you choose a die at random (i.e. either fair or unfair). This is gotten by taking the probability assuming it's a fair die and taking the probability assuming it's an unfair die and multiplying by the probability that it's fair and unfair, respectively (them adding those up):

p(outcome) = p(outcome|fair) * p(fair) + p(outcome|unfair) * p(unfair)

In this case the chances of being fair or unfair would be 50/50 = 0.50 % (you don't know know which one). The probability that it is unfair is simple the portion of the probability that comes from the unfair part...that is:

p(unfair | outcome) = p(outcome | unfari) * p(unfair) / p(outcome) <-- this is Bayes' Theorem

So let's find p(outcome)...that's easy:

Assuming fair coin:

Every roll has probability 1/2, so it's just: 1/2⁵ = 1/32

Unfair coin: Three heads and two tails: (3/4)³ * (1/4)² = 27/4⁵ --> add them up to get the probability of this outcome (assuming a random choice of fair or unfair):

1/32 * 1/2 + 27/32² * 1/2 = 1/2 *(32 + 27)/32² = 1/2 * 59/32² -->

Now take the proportion of that total which was the unfair coin:

27/32² *1/2 /(1/2 *59/32²) = 27/59 27+59=86

lets say that fair coins probability of yielding H or T is 1/2 and that of unfair coin is 3/4 and 1/4 now for getting HTHTH with fair coin P(f)=1/2x1/2x1/2x1/2x1/2=1/32 with unfair coin is P(u)=3/4x1/4x3/4x1/4x3/4=27/(32x32) so probability of getting HTHTH using unfair coin is P(u)/(P(u)+P(f))=27/(27+32) so m=27 and n=32+27 and.............m+n=86

Suppose we did x trials with fair coin and another same x number of trials with unfair coin.

Using fair coin, the number of times we will get the "HTHTH" sequence is P1*x, where P1 is the probability of getting the given sequence with fair coin.

Similarly, using unfair coin, the number of time we get 'HTHTH" sequence is P2*x, where P2 is the probability of getting the given sequence with unfair coin.

Now, let's just consider the number of times we got 'HTHTH' sequence. Let's take this our sample space and find the probability that unfair coin was being used to get the 'HTHTH' sequence.

The probability will be = No of times we got "HTHTH" sequence using unfair coin/No of times we got 'HTHTH' sequence using both coins

= P2 x/(P1 x + P2*x) = P1/(P1 + P2)