Bayesian with multiple conditions, conditioning on an intersection - how?

Question

We have 100 coins, one of which may be double-headed with probability P(D)=0.5 (else all coins are fair). We randomly choose one coin and flip it 7 times, all of which are heads.

Q: What is the probability that the coin we have chosen is double headed?

So:
Let D be the event that there is a double-headed coin: P(D) = 0.5
Let H be the event that the chosen coin lands heads 7 times in a row.
Let F be the event that the chosen coin is fair. (For a fair coin to land heads 7 times in a row the probability is 0.0078)

So I'm guessing what we are looking for is the sum of P(Fc|H, D) + P(Fc|H, Dc)
(Fc = F complement, Dc = D complement). But I can't even get anywhere with the first part, because:

$$P(Fc|H,D)= { P(D|Fc,H)P(Fc|H)\over P(D|H)}.$$

of which how do I get to P(D|Fc,H)? How do I get the probability that there is a dodgy coin given that 7 heads in a row and a dodgy coin have both occurred?

Answer 1

No, the asked probability is $P(F'|H)=P(F',D|H)$ (because $P(F',D'|H)=0$). Also, for your formula, $P(A|B)+P(A|B')$ can be greater than $1$, and they shouldn't be summed. We apply the usual Bayes formula:

$$P(F'|H)=\frac{P(H|F')P(F')}{P(H)}$$

We simply have $P(H|F')=1$ because if the chosen coin is not fair, all will be heads. And $P(F')$, the probability of choosing the unfair coin is $0.5\times 0.01=0.005$. Using total probability law, the denominator can be decomposed as

$$P(H)=P(H|F')P(F')+P(H|F)P(F)$$

You already have calculated $P(H|F)$ as $0.0078$ and $P(F)=1-P(F')$. I haven't mentioned event $D$ but it's already embedded in the calculation of $P(F')=P(F',D)+P(F',D')=P(F',D)=0.5\times0.01$.