Six coins are tossed simultaneously. The probability of getting at least 4 heads is

Let $X$ be the random variabe

Getting head is success in tossing a coin 

The number of coins tossed is $n=6$

The range of the random variable is $\left\{0,1,2,3,4,5,6\right\}$

Let $p$ be the probability of success, and q be the probability of failure

here $p=\frac{1}{2},q=\frac{1}{2}$

The probability of getting at least 4 heads is $P\left(X\ge 4\right)$

Use binomial distribution, $P\left(X=r\right)=n_{C_r}{p}^r{q}^{n-r}$

Hence, 

$\begin{array}{rcl}P\left(x\ge 4\right)& =& P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)\\ & =& 6_{C_4}{p}^2{q}^4+6_{C_5}{p}^1{q}^5+6_{C_6}{q}^6\\ & =& 15{\left(\frac{1}{2}\right)}^6+6{\left(\frac{1}{2}\right)}^6+{\left(\frac{1}{2}\right)}^6\\ & =& {\left(\frac{1}{2}\right)}^6\left[15+6+1\right]\\ & =& \frac{22}{64}\\ & =& \boxed{\frac{11}{32}}\end{array}$