Four fair coins are tossed simultaneously. Given that at least two heads appear, what is the probability that exactly three heads occur?
4/11
With 4 coins, the total number of outcomes is \(2^4 = 16\).
The number of ways to get exactly \(k\) heads is \(\binom{4}{k}\): for 2 heads it is \(\binom{4}{2} = 6\), for 3 heads \(\binom{4}{3} = 4\), and for 4 heads \(\binom{4}{4} = 1\).
So the event "at least two heads" has \(6 + 4 + 1 = 11\) outcomes.
Using conditional probability, \(P(\text{exactly 3} \mid \text{at least 2}) = \dfrac{4}{11}\).
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