Question:
A coin is tossed two times. Find the probability of getting atmost one head.
Solution:
The possible outcomes,if a coin is tossed 2 times is
$S=\{(H H),(T T),(H T),(T H)\}$
$\therefore$ $n(\mathrm{~S})=4$
I et $F=$ Fvent of chetting atmost one head
$=\{(T T),(H T),(T H)\}$
$\therefore$ $n(E)=3$
Hence, required probability $=\frac{n(E)}{n(S)}=\frac{3}{4}$