# Logic behind the coin flip problem, and conditional probability in general

The problem is the following:

Three coins are in a bag. The first coin flips heads with probability $$50$$%, the second coin flips heads with probability $$60$$%, and the third coin flips heads with probability $$70$$%. I pull out a coin and flip heads. If I flip this coin again, what is the probability I will get heads?

According to the solution key, the correct answer is: $$\mathsf P(HH\mid H)=\frac{\mathsf P(HH)}{\mathsf P(H)} = \frac{\frac{1}{3}\cdot(0.5^2+0.6^2+0.7^2)}{\frac{1}{3}\cdot(0.5+0.6+0.7)} = \frac{11}{18}$$

I can't really follow the logic here: why do we use the conditional probability and (the same question actually) why does the fact "the first flip is a head" matter? Aren't the first and second flips independent with each other?

Broadly, I still struggle with conditional probability, in particular I have difficulty in distinguishing independent and dependent activities. When should we use conditional probability, and when should we not?

• The first flip gives you information about which coin was flipped. If the first flip gives head, it is more likely that this was coin 3 than one of the other coins. Do you know Bayes' theorem? – Gerhard S. Mar 3 at 17:29
• @GerhardS. Thanks for the comment. Yes, I know Bayes' theorem, but I still struggle with using it under different contexts. The problem above is an example: I have a hard time judging whether the "conditions" matter or not. – Robert Mar 3 at 18:37