What is the probability of obtaining exactly four tails in five flips of a coin if at least three are tails?

Answer 1

If at least three flips are tails, there are two scenarios where we can obtain exactly four tails in five flips. Either the first four flips are tails and the last flip is heads, or the first flip is heads and the next four flips are tails. Each scenario has a probability of 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32. Therefore, the probability of obtaining exactly four tails in five flips if at least three are tails is 1/32 + 1/32 = 1/16.

Answer 2

We need to determine the separate event.

Let A = obtaining four tails in five flips of coin

Let B = obtaining at least three tails in five flips of coin

Apply Binomial Theorem for this problem, and we have:

P(A | B) = P(A ∩ B) / P(B)

P(A | B) means the probability of "given event B, or if event B occurs, then event A occurs."

P(A ∩ B) means the probability in which both event B and event A occur at a same time.

P(B) means the probability of event B occurs.

Work out each term...

P(B) = (5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0

It's obvious that P(A ∩ B) = (5 choose 4)(½)4(½) since A ∩ B represents events A and B occurring at the same time, so there must be four tails occurring in five flips of coin.

Hence, you should get:

P(A | B) = P(A ∩ B) / P(B)

= ((5 choose 4)(½)4(½))/((5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0)