Three unbiased coins are tossed. What is the probability of getting at most two heads?

A. $$\frac{3}{4}$$

B. $$\frac{1}{4}$$

C. $$\frac{3}{8}$$

D. $$\frac{7}{8}$$

Answer: Option D

Solution(By Examveda Team)

Here S = [TTT, TTH, THT, HTT, THH, HTH, HHT, HHH]
Let E = event of getting at most two heads
Then, E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{8}$$