Probability of picking one coin out of two tossing it twice and getting the same result? 
You have two coins to choose from. Getting a head from the 1st coin is $\frac{1}{2}$, for the 2nd coin is $\frac{1}{5}$. You pick one coin at random and flip it two times.
What is the chance that the coin lands the same way both times?

I must have messed up somewhere in the setup as I'm not getting the same answer as the solution.
The setup that I have is P(1st coin being head)=$\frac{1}{2} * \frac{1}{2}+\frac{1}{2}*\frac{1}{5}=\frac{7}{20}$ thus the chance of getting heads for both time is:
$\frac{7}{20}*\frac{7}{20}$
Similarly, the P(1st being tail) = $\frac{1}{2}*\frac{1}{2}+\frac{1}{2}*\frac{4}{5}=\frac{13}{20}$ thus the chance of getting tails both times is :
$\frac{13}{20}*\frac{13}{20}$
Lastly P(coin lands same way both times) = P(HH)+P(TT)=$\frac{7}{20}*\frac{7}{20}+\frac{13}{20}*\frac{13}{20}$
What did I do wrong here?
Thank you!
 A: Your workings assume you randomly pick the coin for both tosses, when in fact you do such picking only on the first toss. Start from there and for each coin work out the probability of getting both heads or both tails:
$$\frac12\left(\left(\frac12\right)^2+\left(\frac12\right)^2\right)+\frac12\left(\left(\frac15\right)^2+\left(\frac45\right)^2\right)=\frac{59}{100}$$
A: 
What did I do wrong here?

You are computing $P(HH)=P(H)P(H)$, which assumes the events are independent, but actually they aren't. (They only are when conditioned on the picked coin!).
The correct (and simple) way is: using conditional probabilities, adding a variable $X=1$ for the event of choosing the first coin, $X=2$ for choosing the second coin:
$$P(HH)= \sum_X P(HH \mid X) P(X) \\
= P(HH \mid X=1)P(X=1)+P(HH \mid X=2)P(X=2)
\\ = \left(\frac12\right)^2  \frac12 + \left(\frac15\right)^2 \frac12
\\=0.145$$
Similarly
$$P(TT)
 =\left(\frac12\right)^2  \frac12 + \left(\frac45\right)^2 \frac12=0.445$$
Then $P(HH)+P(TT)=0.59$
