Whats the probability you lose money? Suppose you are playing a game that costs $ $8 to play.  You  flip 10 coins and, for every head, you win $2.  Whats the probability you lose money ?
$$
\begin{array}{c|cccccccccc}
             x& 1 & 2 & 3 & 4 & 5 &  6 & 7 &8 &9&10       \\ \hline
         p(x)
          &   5/512 &   45/1024 &  15/128 &   63/256 &   105/512 &   
\end{array}
$$
Since the loosing money means getting less than 4 heads I just did an addition of the $$P_X(1)+P_X(2)+P_X(3)=0.1708$$

With pmf of X being $P_X(x)=\binom{10}{x}(0.5)^x(0.5)^{10-x}$ 

And g(X)=Y=-8+2X the equation that define the net losses or winnings.
can this be a viable solution the way I presented it ?
If yes do you know any good shortcut for this exercise ?

Don't forget about the possibility of a super bad luck event, that is you might get all tails.
probability of losing money is equal to 
$$\frac{1}{2^{10}}\sum_{i=0}^3 \binom{10}{i}= \frac{1+10+45+120}{1024}\approx 0.1719$$
