I have 3 mutually exclusive events, probability of each one happening is $\frac13$. What's the probability none of the events happens 3 times out of 3 tries.

My approach is this : $(\frac23*(1+\frac13+\frac19))^3=0.893$

Simulation shows 0.889, That 0.004 makes me think I'm incorrect.

Note:I'm looking for general formula, these numbers in my real problem aren't equal and even change during process.

  • $\begingroup$ What was your final answer to the question? $\endgroup$ – Greenparker Aug 18 '17 at 11:00
  • 2
    $\begingroup$ I find the wording of your question somewhat confusing. I think Lukasz Derylo probably interpreted it correctly given the result he came up with, but I'd just like to point out that in that interpretation, A, B and C are not independent events; rather, they are mutually exclusive. $\endgroup$ – Ruben van Bergen Aug 18 '17 at 11:59
  • $\begingroup$ Yep you are right. $\endgroup$ – luka25 Aug 18 '17 at 12:09

Let's call your events A, B and C. You have $3^3=27$ possible results here (AAA, AAB, AAC, ABA and so on). 3 of them are not of your interest (e.g. these are situations where one of the events happens 3 times: AAA, BBB, CCC). So you have 24 possible "good results". $24/27=0.888888...$

  • $\begingroup$ This time it's simple because every result have similar chances of happening, I'm more interested in formula(when numbers aren't so good) $\endgroup$ – luka25 Aug 18 '17 at 12:08
  • $\begingroup$ A slightly different way of getting there would be: the probability of none of these events happening 3 (out of 3) times, is complementary to the probability that any one event happens exactly 3 times. For each one of your 3 events, the probability of occuring 3 times is $\frac{1}{3}^3=\frac{1}{27}$. Thus, the probability that any one of your 3 events is repeated three 3 is $3\times\frac{1}{27}=\frac{3}{27}=\frac{1}{9}$. So the probability that none of the events is repeated 3 times is $1-\frac{1}{9}=\frac{8}{9}=0.888888...$. But this is fundamentally the same as Lukasz Derylo's answer. $\endgroup$ – Ruben van Bergen Aug 18 '17 at 12:49
  • $\begingroup$ Nah, the probability that any one of your 3 events is repeated three is (1-1/27)^3 $\endgroup$ – luka25 Aug 18 '17 at 12:57
  • $\begingroup$ No I'm afraid that's wrong. We're considering a disjunction of three possible outcomes: event 1 is repeated three times (AAA), OR event 2 is repeated three times (BBB), OR event 3 is repeated three times (CCC). Thus the probability of any of these events happening is simply the sum of the individual probabilities of these events (that is, $P(AAA \vee BBB\vee CCC)=P(AAA)+P(BBB)+P(CCC)$). $\endgroup$ – Ruben van Bergen Aug 18 '17 at 13:06
  • $\begingroup$ Ah, you're right $\endgroup$ – luka25 Aug 18 '17 at 13:18

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.