A coin is tossed two times.


A coin is tossed two times. Find the probability of getting atmost one head.


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 $=\frac{n(E)}{n(S)}=\frac{3}{4}$

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