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}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{8}$$
This Question Belongs to Arithmetic Ability >> Probability
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