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Probability problem

  1. Dec 10, 2013 #1
    You flip a count 10 times what is the probability of getting exactly four tails?

    My solution.
    out of ten flips you can get

    0 tails out of 10 flips
    1 tails out of 10 flips
    2 tails out of ten flips
    3 tails out of ten flips
    4 tails out of ten flips
    5 tails out of ten flips
    6 tails out of ten flips
    7 tails out of ten flips
    8 tails out of ten flips
    9 tails out of ten flips
    or
    10 tails out of ten flips

    So there are 11 different cases. Therefore would the answer be 1/11?
     
  2. jcsd
  3. Dec 10, 2013 #2

    Dick

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    Do you really think those cases are equally probable??
     
  4. Dec 10, 2013 #3

    phinds

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    As Dick suggests, you need to rethink the problem. What is the probability of NOT getting a tail on any single flip?
     
  5. Dec 10, 2013 #4

    Ray Vickson

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    You are missing some basics, so to clarify, start with the simpler case of 2 flips. You can write down all the possible ways to obtain 0 tails, 1 tail and 2 tails, and from there you can get the probabilities.

    You need to assume something about the flips: (i) they are fair (equal chance of H or T each time); and (ii) the flips are independent (that is, the results on one flip do not affect the probabilities of H or T on any other flip---the coin does not "remember" anything).
     
    Last edited: Dec 10, 2013
  6. Dec 10, 2013 #5
    0 tails:
    Flip 1: H
    Flip 2: H

    1 tails:
    Flip 1: H
    Flip 2: T

    OR

    Flip 1: T
    Flip 2: H

    2 tails:
    Flip 1: T
    Flip 2: T

    There are 4 cases.. is that correct?
     
  7. Dec 10, 2013 #6

    Dick

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    Yes, so for two coins that shows the probability of 0 tails is 1/4, 1 tail is 2/4 and 2 tails is 1/4. It's going to be tedious to write out cases that way with 10 coins. Haven't you learned to use combinatorial arguments or the binomial distribution to calculate it without writing cases out?
     
  8. Dec 10, 2013 #7
    So then it would be

    10 choose 0
    10 choose 1
    10 choose 2
    .
    .
    .
    10 choose 10
    Then would the answer be

    10 choose 4/(Ʃ(10 choose i)) i from 0 to 10
     
  9. Dec 10, 2013 #8

    Dick

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    Yes, it would. But there's an easy way to evaluate the sum in the denominator. It's all total ways of choosing heads or tails. It's 2^10.
     
  10. Dec 10, 2013 #9
    Thank you!
     
  11. Dec 31, 2014 #10
    Let p=Probability of getting Heads=0.5
    q=Probability of getting Tails=0.5
    assuming all flips are independent, probability of getting 4 heads=p*p*p*p*q*q*q*q*q*q
    you are doing the mistake of assuming getting 1, 2, 3 ...heads(or tails) is equally likely. They clearly aren't. This is a typical example of bernoulli trials.
     
  12. Dec 31, 2014 #11

    haruspex

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    Well, no, that's the probability of getting four heads first then six tails. And it's the same as any other specific sequence of outcomes.
     
  13. Dec 31, 2014 #12
    Okay i should have multiplied by 10!/4!6!
     
  14. Dec 31, 2014 #13
    You also have to consider the order in which you get the heads and tails, and this problem would clearly become tedious without the binomial distribution, or without applying permutations. What I mean is that you'll have to consider the probability of ALL the different ways tails in the combination T,T,T,T,H,H,H,H,H,H could be arranged. (Like having H,T,T,H,T,TH,H,H,H is different and its probability also has to be included in the final answer.)
     
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