You flip a coin four times. First flip is heads. What are the chances that at least three of the four flips will be heads?

Question

Wondering if my work is correct.

2^4 = 16

Successful attempts to get at least three heads: HHHH, HHHT, THHH, HTHH, HHTH = 5

5/16 is my answer. I just wanted to make sure this was correct. Thank you.

Answer 1

No. Your answer is not correct. First, note that you included THHH as a "success," but that case violates your condition that "First flip is heads."

Given that the first flip is heads, there are only $2^3 = 8$ equally likely scenarios:

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, and HTTT.

Now do your counting...

Answer 2

Given that the first flip is heads, you wind up with at least three heads in all if and only if the majority of the final three flips come up heads. For any set of odd flips (such as three), the majority is either heads or tails with equal probability, namely $1/2$.