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 $\left(\frac 12\right)$
 A: Option #$1$:
Add up the following:


*

*The probability of getting exactly $\color\red2$ heads, which is $\dfrac{\binom{3}{\color\red2}}{2^3}=\dfrac38$

*The probability of getting exactly $\color\red3$ heads, which is $\dfrac{\binom{3}{\color\red3}}{2^3}=\dfrac18$
Hence the probability of getting at least $2$ heads is $\dfrac38+\dfrac18=\dfrac12$.

Option #$2$:
First, split it into disjoint events of equal probability:
 C 1 | C 2 | C 3 | At least 2 heads
-----|-----|-----|------------------
  H  |  H  |  H  | Yes
  H  |  H  |  T  | Yes
  H  |  T  |  H  | Yes
  H  |  T  |  T  | No
  T  |  H  |  H  | Yes
  T  |  H  |  T  | No
  T  |  T  |  H  | No
  T  |  T  |  T  | No

Then, count the number of combinations with at least $2$ heads, which is $4$.
Finally, divide this number by the total number of combinations, which is $8$.
Hence the probability of getting at least $2$ heads is $\dfrac48=\dfrac12$.
A: 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}
A: You misread the question. The question asks for how many ways can you get two or three coins that are heads, not how many ways you can have up to 2 heads. 
Here's how I would approach the problem:
First we need to find out how many possibilities there are. If we think of flipping a coin 3 times as 3 binary digits, where 0 and 1 are heads and tails respectively, then the number of possibilities must be $2^3$ or 8.
Next we need to figure out the probability of each event and add them together. The probability of getting 3 heads is easy since it can only happen one way $(000)$, so it must be $\frac {1}{8}$. The probability of getting two heads is: $$\frac {3 \choose 2}{8} = \frac{3}{8}$$
Add those together and you get $\frac{4}{8}$ which equals $\frac{1}{2}$.
