A coin is tossed 7 times. The probability of getting at least 4 heads is:

$\frac{5}{8}$ $\frac{3}{4}$ $\frac{1}{4}$ $\frac{1}{2}$

$\frac{1}{2}$

Coin tossed 7 times P(atleast 4 heads) = P(4 heads) + P(5 heads) + P(6 heads) + P(7 heads) $=\frac{{^7C}_4+{^7C}_5+{^7C}_6+{^7C}_7}{2^7}=\frac{35+21+7+1}{128}=\frac{64}{128}$ $=\frac{1}{2}$