Suppose a fair coin is toss four times consecutively.
Find the PMF for random variable of number of heads following the first tail.
My take:
Let random variable $X$ be the number of heads in this experiment.
Then $X\sim Binomial(4, 0.5)$.
Hence, $P(X=k) = \binom{4}{k}{0.5}^{k}{0.5}^{4-k}$, for $k = 0, 1, 2, 3, 4$.
Let A denote the event which the first toss is tail in this experiment.
Then we know that $P(A)=0.5$.
Let random variable $Y$ be number of heads following the first tail.
Then $P(Y=y)=P(X=k\mid A)$, for $y = 0, 1, 2, 3$.
However, I think my reasoning is not valid since $k=4$ is impossible, but it's my only intuition.
Can someone please enlighten me on this question or this type of question in general?
Many thanks,
Sebastian
Best Answer
Let $Y$ be the number of heads following the first tail. We calculate $\Pr(Y=0)$, $\Pr(Y=1)$, $\Pr(Y=2)$, and $\Pr(Y=3)$. It will be done by a patient examination of cases. (If we were dealing with say $10$ tosses, then a more general approach would be worthwhile.)
The easiest one is $\Pr(Y=3)$. For $Y=3$ can only happen if we get a tail and then $3$ heads. The probability of this is $\frac{1}{2^4}$.
Now let's go to the other end, and find $\Pr(Y=0)$. This can happen in several ways. We could get TTTT, probability $\frac{1}{2^4}$. Or we could get HTTT, probability $\frac{1}{2^4}$. Other possibilities are HHTT,HHHT, HHHH. Add up. We get $\frac{5}{2^4}$.
Next we find $\Pr(Y=1)$. Here we use some machinery, not much. We could have $T$, then exactly one H. This has probability $\frac{1}{2}\cdot \binom{3}{1}\cdot \frac{1}{2^3}$, that is, $\frac{3}{2^4}$. Or else we could get HT, then exactly $1$ H. The probability is $\frac{2}{2^4}$. Or else we could have HHTH, probability $\frac{1}{2^4}$. Add up.
Finally, for $Y=2$ we can do a similar calculation, or use the fact that probabilities sum to $1$. It is best to do both, as a partial check on the correctness of our calculations.