Conditional probability or Bayes' theorem I'm trying to do a question in probability:
''we flip three coins'' 
''What is the probability that the second coin landed tails, given that two coins
(exactly) landed head?''
I have set out the sample space - S= {HHH, HHT, THH, HTH, HTT, TTH, THT, TTT} 
E1 - tails on the second coin - {HTH}
E2 - outcomes feat. 2 heads exactly - {HHT, HTH, THH}
P(E1 + E2) = 1/8
P(E2) = 3/8
P(E1/E2) = P(E1+E2)/P(E2) =  [(1/8)]/[(3/8)] = 1/3
? Is this correct or will I have to use Bayes' theorem for this?
 A: $\color{green}{\checkmark}$ The calculations using Conditional Probability are correct. $\mathsf P(A\cap B)=\mathsf P(A)\mathsf P(B\mid A)$


*

*To use Bayes' Theorem you'd need the other conditional probability. $\mathsf P(A\mid B)\;\mathsf P(B)= \mathsf P(B\mid A)\;\mathsf P(A)$



Your labeling of the events is a bit misleading.
You have:

the sample space - S= {HHH, HHT, THH, HTH, HTT, TTH, THT, TTT} 
E1 - tails on the second coin - {HTH} 

No, tails on the second coin is: {HTH, HTT, TTH, TTT} 

E2 - outcomes feat. 2 heads exactly - {HHT, HTH, THH}

And that leads to: E1 $\cap$ E2 : {HTH}  , tails on the second coin and exactly two heads.
Which, despite the labelling, you then have used correctly in the conditional probability calculation. 
$$\begin{align}
\mathsf P(E_1\mid E_2) & = \frac{\mathsf P(E_1\cap E_2)}{\mathsf P(E_2)}
\\ & = \frac{1/8}{3/8}
\\ & = \tfrac{1}{3}
\end{align}$$
Showing that it was just an error in labelling not understanding. But still, try to avoid doing that on an exam.
