Probability at least 2 flips are tails I would like to do this the direct ways

a fair coin is tossed 3 times, find $P(\text{at least 2 flips were tails})$

Doing this directly,
$$P(\text{at least 2 flips were tails}) = P(\text{2 tails}) + P(\text{3 tails})$$
Is the answer:

$$ = (0.5)^2 + (0.5)^3$$

But the text says the probability should be $(1/2)$ which isnt in agreement with my answer?
 A: We use the concept of Binomial distribution here. Let $n=3$ be the number of trials, $k $ be the number of successful trials (here, getting a tail), $p=0.5$, the probability of getting a tail and $X $, a random variables determining the number of successes.
We thus need to find $P (X=2) + P (X=3)$. Thus, $$P_{\text {req}} = P (X=2)+P (X=3) $$ $$= \binom{3}{2}(0.5)^2 (0.5)^{3-2} + \binom {3}{3}(0.5)^3$$ $$ =\frac {3}{8}+\frac {1}{8} = 0.5$$ Hope it helps. 
A: Here is the sample space:
$\Omega = \left\{HHH, HHT, HTH, \color{red}{HTT}, THH, \color{red}{THT},\color{red}{TTH}, \color{red}{TTT}\right\}$
All the eight outcomes are equally likely to occur. In four out of eight of them (the ones that are colored red), we see at least two tails. Therefore, the probability of getting at least two tails is $\frac{4}{8} = 0.5$.
A: From $3$ fair coin flips, if you don't have at least two tails, you have at least two heads. Since these two states are clearly equal in probability and exclude any other possibility, they must each have probability of $\frac 12$.
A: Probability (exactly one tail) $=\dbinom{3}{1}\left(\dfrac{1}{2}\right)^1\left(\dfrac{1}{2}\right)^2=\dfrac{3}{8}$
Probability (no tail) $=\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}$
Required probability 
$=1-\dfrac{3}{8}-\dfrac{1}{8}=\dfrac{1}{2}$
