Question
Three unbiased coins are tossed. What is the probability of getting at most two heads?
Options:
A .  $$\frac{3}{4}$$
B .  $$\frac{1}{4}$$
C .  $$\frac{3}{8}$$
D .  $$\frac{7}{8}$$
Answer: Option D
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}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{8}$$

Was this answer helpful ?
Next Question

Submit Solution

Your email address will not be published. Required fields are marked *

Latest Videos

Latest Test Papers