I have an unfair coin that lands as head with probability of $\dfrac{2}{3}$. If I flip the coin 5 times, what is the probability that I get

Yes that is the answer!

$${\mathtt{10}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}^{{\mathtt{3}}} = {\frac{{\mathtt{40}}}{{\mathtt{243}}}} = {\mathtt{0.164\: \!609\: \!053\: \!497\: \!942\: \!4}}$$

If the probability of getting a head is 2/3, then the probability of getting a tail is 1/3.

Wanting 2 heads on 5 tosses is a combinations problem:  ₅C₂·(2/3)²·(1/3) ³   =  10·(2/3)²·(1/3) ³ 

then, what would be the solution sorry?