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, \(P(\text{exactly 3} \mid \text{at least 2}) = \dfrac{4}{11}\).
Important Questions from Probability of Random Experiments
A, B, C and D are mutually exclusive and exhaustive events.
If 2P(A) = 3P(B) = 4P(C) = 5P(D), then what is 77P(A) equal to ?
A fair coin is tossed 6 times. What is the probability of getting a result in the 6t h toss which is different from those obtained in the first five tosses ?
Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is
A biased coin with the probability of getting head equal to \(\frac{1}{4}\) is tossed five times. What is the probability of getting tail in all the first four tosses followed by head ?
Three dice are thrown. What is the probability that each face shows only multiples of 3 ?
