We have two independent events, event A which occurs with $p=1/9$ and event B which occurs with $p=1/15$ (and can only happen after event A has already happened). What is the probability that after $n$ trials both $A$ and $B$ have occurred at least once given that event $B$ can only happen after event $A$?

  • $\begingroup$ How can they be independent if $B$ can not occur before $A$? $\endgroup$ – lulu Oct 4 '17 at 18:21
  • $\begingroup$ What does "can only occur after $A$ occured" mean ? Speaking of "after $n$ trials" is a nice intuitive way of seeing things but does not reflect anything in the "maths world" $\endgroup$ – Junkyards Oct 4 '17 at 18:22
  • $\begingroup$ Well, I guess not no. That was somewhat stupid of me. $\endgroup$ – John Smith Oct 4 '17 at 18:22
  • $\begingroup$ It means A occurs with probability 1/9 and once A does occur then B has probability 1/15. $\endgroup$ – John Smith Oct 4 '17 at 18:22
  • $\begingroup$ So...letting $X$ denote the complement of $A\cup B$, we have that prior to the first occurrence of $A$, $p_A=\frac 19$ and $p_X=\frac 89$ and after the first occurrence of $A$ we have $p_A=\frac 19, p_B=\frac 1{15},p_X=\frac {37}{45}$? $\endgroup$ – lulu Oct 4 '17 at 18:24

there are two cases.

Case I: $A$ never occurs. The probability of that is $\left(\frac 89\right)^n$

Case II: $A$ occurs but $B$ does not. That one is trickier. We need to consider the various places $A$ might have occurred first. Let $i\in \{1,n\}$ denote the first occurrence of $A$ and let $P_i$ denote the probability that $B$ does not occur at all, given that the first occurrence of $A$ is in slot $i$. Then $$P_i=\left( \frac 89\right)^{i-1}\times \frac 19\times \left(\frac {14}{15}\right)^{n-i}$$ The probability of Case II occuring is then $$\sum_{i=1}^nP_i= \left(\frac 52\right)^{1 - n} \,3^{-2 n}\, (21^n- 20^n) $$

Here I have included the closed formula more as a curiosity than any thing else. (Full disclosure: I let Wolfram Alpha compute the sum, which may account for it's somewhat odd presentation. It is really just a geometric series with some multiplicative factors tossed in.)

  • $\begingroup$ How did you change the equation into a closed form? You say you used Wolfram-Alpha, what program did you use? $\endgroup$ – John Smith Oct 5 '17 at 23:41
  • $\begingroup$ @JohnSmith Well I just typed "Sum[(8/9)^(i - 1) (1/9) (14/15)^(n - i), {i, 1, n}]" Though when I do that know I get a different result, so perhaps I mistyped it before or perhaps the two forms are equivalent. Now it is telling me "5/2 9^(-n) ((42/5)^n - 8^n)". I will look into it later. $\endgroup$ – lulu Oct 6 '17 at 11:47

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