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Three unbiased coins are tossed. What is the probability of getting at most two heads?
Answer & Solution
Correct 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}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{8}$$
Let E = event of getting at most two heads
Then, E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{8}$$
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