Assume we have $n$ cards, indexed successively by the integers $1$ to $n$. Now each card is either marked or not. Suppose that card $i$ is marked with probability $p_i$, independently of the others.

Let the r.v. $X = 0, 1,...$ be the highest index among the marked cards (with $X = 0$ if all cards are unmarked).

What is $P(X=k)$ for $0 \leq k \leq n$ , $P(X \geq k)$ and $P(X \geq k| k \geq 1)$ for $1 \leq k \leq n$ ?

Does $X$ follow the distribution of the maximum?

For $P(X=k)$ when I compute $P(X=0)$ get $P(X=0) = \Pi_i (1-p_i)$ then $P(X=1) = \sum_{i=1}^n p_i(\Pi_{j \neq i} (1-p_j))$, but I think I am doing this wrong?


You can think iteratively from the card numbered $n$. This one is marked with probability $p_n$. So we have the probability $$P(X=n)=p_n.$$ Now think about the case $X=n-1$ in this case, the $n^\text{th}$ card is unmarked and the $(n-1)^\text{th}$ is marked. This has probability $$P(X=n-1)=(1-p_n)p_{n-1}.$$ You probably can take it from here.

  • $\begingroup$ Thank you. I just needed a fresh look at it. $\endgroup$ – Note Nov 17 '18 at 2:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.