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The probability of an event occurring at least x times?

  1. Nov 17, 2014 #1
    Assuming that an event has 80% chance of occurrence and 20% of not occurring per "round", what is the formula to calculate the likelihood that it will occur at least n times assuming that there are x rounds?

    i.e.
    If you flip a coin 3 times, what are the chances that at least 2 times will turn up heads?

    This example would be simple enough to map out all of the occurrences and find out which proportion meet those criteria, but the examples i will need to work with are too complex to do so.

    I tried to find the formula online but didnt know the name and couldnt describe it in such a way that a google search yielded results.

    Thanks!
     
  2. jcsd
  3. Nov 17, 2014 #2

    phinds

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    Forget formulas. If you don't understand the fundamentals, you may well apply a formula incorrectly.

    Just think about the basics. If an event has a 80%/20% chance of occurring/not occuring, what's the likelihood that it will occur/not occur a second time?
     
  4. Nov 17, 2014 #3

    Simon Bridge

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    What he said
     
  5. Nov 17, 2014 #4
    I'm really not sure how that relates to what I asked. Im well aware that combined probability of multiple independent events is multiplicative. But I'm asking about the odds (for example) of an event occurring at least 5 times out of 8 chances.

    Not exactly 5 times, at least 5 times, and in no particular required order of successes vs failures.

    For that I would need to sum the aggregate probability of all possible sequence combinations where there are at least 5 successes. This includes:

    YYYYYNNN
    YYNNNYYY
    YYYYYYYY
    NNYNYYYY
    etc.

    So unless im blatantly missing an obvious concept, im not sure which fundamentals in your post will help me derive a formula to solve what i asked efficiently.
     
  6. Nov 17, 2014 #5

    phinds

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    It was not clear from your post that you understand the fundamentals so well. It sounded as though you were simply searching for a formula to plug number into.

    I don't know if there's an equation per se. I would find a generic expression for "exactly n times" and extrapolate it as a sum of terms and see if that looks like it would yield a single expression or simply have to be such a sum of terms.
     
  7. Nov 17, 2014 #6

    Simon Bridge

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    Well if you don't follow suggestions you'll never know ;) Here, I'll start you off...

    For 2 independent trials, probability of success being p, the probability of exactly
    2 successes is ##P(2)=p^2##
    1 success is ##P(1)=2p(1-p)## ... because that requires one success and one failure, and there are two ways this can happen
    0 successes is ##P(0)=(1-p)^2##
    ... so the probability of at least one success is ##P(1)+P(2) = p^2 + 2p(1-p) = 2p-p^2##

    Repeat for three independent trials and spot the pattern.
    Mind you - you'd have to recognize the binomial coefficients.

    You can just look up "binomial probability", which I see you found while I was typing, and memorize the equation.
     
  8. Nov 17, 2014 #7

    That turned out to be exactly the right approach. It's a sum of terms using exactly n times, and I found that equation pretty easily online

    thanks!
     
  9. Nov 17, 2014 #8
    Great, thanks for the detailed response!
     
  10. Nov 18, 2014 #9

    Stephen Tashi

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    Try the general topic of "generating functions" - often mentioned in connection with problems involving how many ways a certain sum of money can be made using coins of various types.

    The problem you described may just amount to a sum of terms each of which is a "binomial coefficient" times probabilities raised to powers. (For example, the probability of "at least 7 heads" in 10 independent tosses of a fair coin is the sum: probability of exactly 7 heads + probability of exactly 8 heads + probability of excactly 9 heads + probability of exactly 10 heads. So if you can solve the "probability of exactly..." kind of problem, you can solve the "probability of at least" sort of problem.
     
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