A man flips a fair coin with sides heads and tails five times. Given that the man receives heads on at least two of the coin flips, what is the probability that he receives tails exactly twice after the five flips of the coin?

Please provide the solution for this problem

At least 2 heads:

All arrangements of 2H & 3T = \frac{5!}{2! * 3!} = 10

All arrangements of 3H & 2T = \frac{5!}{3! * 2!} = 10`

All arrangements of 4H & 1T = \frac{5!}{4! * 1!} = 5

All arrangements of 5H = \frac{5!}{5! * 0!} = 1

P(T = 2) = N(T = 2)/N(All) = 10/26 or 5/13

The questions asks for the probability of receiving 2 tails after the 5 flips of the coins. Doesn’t that mean that we’re to flip the coin 2 times again after the first 5 times? There seems to be a problem with this question’s answer.

Ah, I misinterpreted the problem:

I interpreted it as “Probability that he received Tails twice in the five flips”

The answer should be \frac{1}{4}

@vidishas99 The video solution gives the same solution as you did in the first go. I think the language of this problem is iffy.