In a simultaneous throw of two coins, the probability of getting at least one head is-
A. $$\frac{1}{2}$$
B. $$\frac{1}{3}$$
C. $$\frac{2}{3}$$
D. $$\frac{3}{4}$$
Answer: Option D
Solution(By Examveda Team)
Here S = {HH, HT, TH, TT}Let E = event of getting at least one head = {HT, TH, HH}
$$\therefore P\left( E \right) = \frac{{n(E)}}{{n(S)}} = \frac{3}{4}$$
This Question Belongs to Arithmetic Ability >> Probability
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