Relating probabilities between different probability spaces

Question

I am trying to relate the following random experiments. The first experiment involves tossing a coin until $n$ heads have appeared. The second experiment involves tossing a coin $N$ times. Let $H$ be the random variable representing the number of heads obtained in the second experiment. My claim is that the probability that you get at least $n$ heads in the $N$ trials is equal to the probability that the number of trials in the first experiment, $X$, is less than or equal to $N$ i.e. $$\mathbb{P}(H \geq n) = \mathbb{P}(X\leq N)$$ Intuitively this makes sense to me however I do not know how to justify this claim, considering that the random variables $H, X$ are defined on different probability spaces altogether.

Answer 1

Your claim is correct. The probability that we get at least $n$ number of heads when we toss $N$ times is $$\mathbb P(H\ge n)=\sum_{h=n}^N{N\choose h}p^h(1-p)^{N-h}\,.$$ Instead of calculating $\mathbb P(X\le N)$ it is easier to calculate the probability of $\mathbb P(X>N)$ which is the one that in the first $N$ tosses we have only $n-1$ heads at most: $$\mathbb P(X>N)=\sum_{h=0}^{n-1}{N\choose h}p^h(1-p)^{N-h}$$ Obviously it is $\mathbb P(H\ge n)+\mathbb P(X>N)=1$ from which $$ \mathbb P(H\ge n)=\mathbb P(X\le N) $$ follows.

A joint probability space $\Omega$ on which this all works formally could be the space of infinite sequences of zeroes and ones: $\Omega=\{0,1\}^\mathbb N\,.$ Each $\omega\in\Omega$ is then a sequence $\omega_1,\omega_2,...$ of zeroes and ones and $$ H(\omega)=\sum_{i=1}^N\omega_i\,,\quad X(\omega)=\min_{k\in\mathbb N}\Big\{\sum_{i=1}^k\omega_i=n\Big\}\,. $$