probability of getting at-least 2 heads Let assume, I through a fair coin three times, and I want to get at least two heads. I want to find the the probability.
To find that, 
At first I evaluated the probability of getting 1 head which is: 
$${3\choose1} \left(\frac 12\right)^{1} \times \left(\frac 12\right)^{3-1}= \frac 38 $$
the probability of getting 2 heads:
 $${3\choose 2} \left(\frac 12\right)^{2} \times \left(\frac 12\right)^{3-2}= \frac 38 $$
The total probability of getting at least two heads is equal: $\frac 38 +\frac 38 =\frac 68 = \frac 34$
But the problem is that, in a text book they  did it in different way and got the result $\left(\frac 12\right)$

At least two heads means "two or three heads". Let $X$ denote the number of heads in three tosses. In particular, $X\sim Binom(3, 1/2)$,  hence  
\begin{align}
P(X\ge 2) = P(X=2) + P(X=3) = \binom{3}{2}\frac{1}{2^3} + \binom{3}{3}\frac{1}{2^3} = \frac{1}{2} .
\end{align}
