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Three unbiased coins are tossed. What is the probability of getting at least 2 heads?
Answer & Solution
Correct Answer:
Option
B
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{8} = \frac{1}{2}$$
Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{8} = \frac{1}{2}$$
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