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

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?

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 $$P(F')=P(F',D)+P(F',D')=P(F',D)=0.5\times0.01$$.