How is the sample space the same in the given binomial distribution?

Question

In the following video lecture of Prof. Blitzstein, Probability, we have the following: https://www.youtube.com/watch?v=k2BB0p8byGA

time stamp: 22:20-22-40 (20 seconds)

If you don't wish to watch the video, my doubt is as follows:

Respected Prof. Blitzstein says that given two random variables binomially distributed -- X Bin(n,p) and Y Bin(m,p). Then X+Y makes "sense" because the sample space of both are the same. However, the sample space of X consists only those sequences of successes and failures that are of length 'n', and the the sample space of Y consists only those sequences of successes and failures that are of length 'm'.

So, how can the sample spaces of both X and Y possibly be the same?

Answer 1

The professor is not saying that the sample space of $X$ and $Y$ is the same. What he is saying is that if you take a coin that lands on heads with probability $p$, and flip it $n + m$ times, then:

1. The number of heads in the first $n$ flips can be written as $X \sim B(n, p)$

2. The number of heads in the last $m$ flips can be written as $Y \sim B(m, p)$

3. $X$ and $Y$ are independent of each other, since the flips that determine their values do not overlap

4. The number of heads across all the flips can be written as $X + Y \sim B(n + m, p)$ (where the right-hand side comes straight from the fact that there are $n + m$ independent flips, each with probability $p$, and the left-hand side comes from the fact that the total number of heads must be the number of heads in the first $n$ flips plus the number of heads in the last $m$ flips)