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A fair coin is tossed independently four times. The probability of the event that the number of heads is greater than the number of tails is

1. $1/16$
2. $1/8$
3. $1/4$
4. $5/16$

Correct answer: $5/16$

Solution

The event requires more heads than tails in 4 tosses, so the number of heads must be 3 or 4. Thus the probability is $\binom{4}{3}/16 + \binom{4}{4}/16 = 4/16 + 1/16 = 5/16$.

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