Probability of getting heads given that first flip was a head?

Solution 1:

Yeah, that is right. You can also use a concept called independence; if the two coin tosses are independent, then knowing that the first one is heads does not change at all the probability of heads for the second one.

Solution 2:

If we are give the information that "the first coin was a head" then, from, HH, HT, TH, TT, would remove both TT and TH. That leaves only HH and HT so that the probability that the second flip is also a head is 1/2.

Solution 3:

the two events are unrelated, the outcome of the second is (as mentioned) independent of the first. So, the odds of the second being heads is 1/2.

The odds of both being heads is 1/4.

If you did 49 flips - and they all came up heads - the odds of the next one being heads is still 1/2.

Solution 4:

In general, for two events $A, B$ $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. $$ Let $A$ be the event of heads on the second toss, and $B$ the event of first toss being a head. Then $$ P(A\mid B)=\frac{1/4}{1/2}=1/2. $$ The numerator is explained by noting that of the four possible sequences of two tosses (all equiprobable), we want $HH$.