0
$\begingroup$

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 $A$?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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

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.