# Probability of mutually exclusive events

Let A and B be two mutually exclusive events with positive probability associated with some random experiment. The experiment is replicated until event B is observed. Find the probability that by that time event A was already observed twice.

I'm not sure where to begin with this problem, any insights?

• Are the event A and B the only outcomes of the experiment? Dec 6, 2016 at 4:52
• No. They are just two events with positive probability.
– ViC
Dec 6, 2016 at 4:54
• Exactly twice or at least twice? Dec 6, 2016 at 5:45
• Exactly twice..
– ViC
Dec 6, 2016 at 5:46

Since we aren't interested whether a relative sequence ...A...A...B occurs on the third trial or the hundreth, the events other than $A$ and $B$ are irrelevant, and we can greatly simplify the problem.

Let the odds in favor of $A$ over $B$ be $3:2$, say,
then we can take $p = 0.6, q = 0.4$,
and the required probability is $0.6^2\cdot0.4$

let A = getting a number divisible by $3$,
let B = getting $5$,
then odds in favor of A $=2:1$,
$p=2/3, q = 1/3$,
Pr ...A...A...B $= \left(\dfrac 23\right)^2\cdot\dfrac13 = \dfrac2{27}$
• To clarify. $~$ We want the probability that, of all events where either $A$ or $B$ occurs, the first $B$ will occur on the third such. $$\mathsf P(A\mid A\cup B)^2\,\mathsf P(B\mid A\cup B) =\dfrac{\mathsf P(A)^2\,\mathsf P(B)}{(\mathsf P(A)+\mathsf P(B))^3}$$ Dec 6, 2016 at 8:54
Suppose event B occurs at time $k$ with probability $q$, event A's occurrence before $k$ is a binomial distribution ~ Binomial$(k-1,p)$. So the probability is $$P(\text{A occurs twice before B} )=\sum_{k=3}^\infty {k-1\choose2}p^2q(1-p-q)^{k-3}$$