The problem:

If event $A$ happens with probability $p$, what is the probability $P[K=1]$ that, after two independent trials, $A$ happened exactly once?

Is the answer:

  • $P[K=1] = p(1-p)$, since we need $A$ to happen and then we need $A$ not to happen, or
  • $P[K=1] = 2p(1-p) = p(1-p) + (1-p)p$, since there are two cases $A\overline{A}$ and $\overline{A}A$ for which $A$ happen exactly once?
  • $\begingroup$ This sounds like homework so I won't definitively answer the question but based on the phrasing, it doesn't matter whether the 'success' occurs on the first or second trial, so $P({\rm exactly \ one \ success})$ is the probability of having a success and then a failure OR having a failure and then a success. Given that, which of your options seems to be correct? $\endgroup$
    – Macro
    May 5 '12 at 3:03
  • $\begingroup$ It's actually a made-up problem to understand this basic concept, so that I could do a more complex homework problem. :) Your hints could go both ways in my mind: "it doesn't matter" could be in the sense that "we should account for both" (2nd option) or "we are already accounting for both (1st option). $\endgroup$ May 5 '12 at 3:08
  • $\begingroup$ @AndresRiofrio In response to your request, I have added one of the tag we generally use when dealing with independent events in probability. $\endgroup$
    – chl
    May 5 '12 at 18:57

Ok, well there are 4 possible outcomes

$$ {\rm success/success}, \ {\rm success/failure}, \ {\rm failure/success},\ {\rm failure/failure}, $$

which occur with probabilities

$$p^2, \ \ \ p(1-p), \ \ \ (1-p)p \ \ {\rm and} \ \ (1-p)^2$$

respectively, assuming the trials are independent. There is exactly one success in the cases ${\rm success/failure}$ and ${\rm failure/success}$. Since order doesn't matter,

$$ P({\rm exactly \ one \ success}) = P({\rm success/failure \ OR \ failure/success}) $$

In general, for two events $A,B$

$$P(A \ \ {\rm OR} \ \ B) = P(A) + P(B) - P(A \ \ {\rm AND} \ \ B) $$

In your example $$P(A) = P({\rm success/failure}) = p(1-p)$$ $$P(B) = P({\rm failure/success}) = p(1-p)$$

since ${\rm success/failure}$ (event A) and ${\rm failure/success}$ (event B) are clearly disjoint events
(it's impossible for events A and B to occur together), their intersection is empty:

$$P(A \ \ {\rm AND} \ \ B) = 0 $$


$P(A \ \ {\rm OR} \ \ B) = P(A) + P(B) - P(A \ \ {\rm AND} \ \ B)$
$P(A \ \ {\rm OR} \ \ B) = p(1-p) + (1-p)p - 0$
$P(A \ \ {\rm OR} \ \ B) = 2p(1-p) $

which was your second option.

  • $\begingroup$ Great answer :) As someone wondering the same as the OP, this really explained how to proof the solution I also expected. I've just added a little edit to simplify the math aspect even more. This is indeed really basic probability, I feel everyone should be able to understand it. $\endgroup$
    – T_D
    Jun 5 '16 at 18:23

The first option is wrong because the question does not state which trial $A$ occurs. If you say what is the probability that $A$ occurs on the first trial but not on the second, then the first option is right. Now if you say what is the probability that if $A$ occurs on the first trial it will not repeat. That probability would be $1-p$. This is really a very elementary problem in probability.

  • $\begingroup$ I was studying elemental probability. :-) $\endgroup$ Aug 21 '13 at 4:39

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