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Left handed probability.

  1. Oct 25, 2008 #1
    4 out of 5 recent US Presidents have been left handers? Given that 1 in 10 of the population is left handed what is the probability of this? Say it was taking socks out of draw with the light off.. 1 White to 9 Black. What are chances of having 4 White and 1 Black.. I get 0.00045 ??

    More accurately what is the probability that a continuous sequence of 5 presidents out of 43 has 4 left handers?
  2. jcsd
  3. Oct 25, 2008 #2


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    Dearly Missed

    There is no particular reason, or improbability against, for a stronly skewed tendency within such a tiny sample.
  4. Oct 25, 2008 #3


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    The probability of 4 out of 5 people being left handed is 5C4(.14)(.9)= 5(.0004)(.9)= .00045 as you have. The "out of 43" is irrelevant.
  5. Oct 25, 2008 #4
    I don't think it is irrelevant.

    The more presidents there has been the more likley you will get a sequence of 4 out of 5.

    If there had been thousands or millions then the chances are that there would be a lot of '4 out of 5's

    You would have to look at all the possible conbinations of L and R you could have with 43 presidents and for those that have at least one '4 out 5' work out those combimations' probabilities and add them up. There are maybe short cuts or at least estimates.
  6. Oct 25, 2008 #5


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    You could look at this as a problem involving a number of runs in a sequence of Bernoulli trials (president is either left- or right-handed). The sequence is 43 long, and you want the probability of a run of length 4, given p = 1/10.
    The question is whether independence would be verified, because of presidents having served multiple terms - but this would apply to the binomial calculation as well.
    You could reduce the problem to only the individual presidents and lessen the dependence issue.
    At the end of the day, I too would simply write this off as "interesting, but so what".
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