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Coint toss probability question

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A coin is tossed 3 times. at least 1 head is obtained. Determine the probability that exactly 1 head is obtained



    2. Relevant equations



    3. The attempt at a solution

    brute force indicates that there 7 possible combinations:

    HHH, HHT, HTH, HTT, THH, THT, TTH (because at least one heads is obtained).

    Out of these we see that there are 3 occasions where there is exactly one heads.
    If we let S be the sample space of all possable outcomes, and let E be the event of having exactly one heads, then a basic rule of probability indicates that the probability P(E)=|E|/|S|=3/7.

    However, using another method:
    fix one coin throw's result as heads. Then the probability of the other two throws being tails is (1/2)(1/2)=1/4.
    Since we can fix heads 3 times, this indicates that the probability is 3(1/4)=3/4.

    Can anyone spot where the flaw is in either of these attempts at a solution?
    thanks
    (this isn't a hw question btw)
     
  2. jcsd
  3. May 25, 2010 #2
    3/7 is when you drop "at least 1 head is obtained" out
    1/4 is when when "at least 1 head is obtained" and that is the only head obtained
     
  4. May 25, 2010 #3

    D H

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    You can't do that. Probabilities add only if the underlying events are mutually exclusive events in the same parent population. That isn't the case here.
     
  5. May 26, 2010 #4

    Can you please explain?
     
  6. May 26, 2010 #5
    I think the flaw (in the second attempt) is the following:

    You summed the probabilities of events {HTT}, {THT} and {TTH} because they are disjoint events whose union is the event targeted. The error was to consider each of them as intersections of two equally likely, independent events whose probability is 1/2 (getting two tails).
    Rather, they are individual instances of a sample space with equally likely 7 elements as you deduced directly. So their probabilities are 1/7 and the desired probability equals 3(1/7)=3/7.

    Have you learned random variables? If you have, I guess there's an easier way to do this exercise. Let X denote the number of heads in the three throws. Then, as each of the 3 throws is independent and the probability of obtaining a head in each of them remains constant and equal to 1/2, X is Binomial with parameters n=3 and p=1/2. The desired probability is then

    [tex]P(X=1|X\geq 1)=\frac{P(X=1,X\geq 1)}{P(X\geq 1)}=\frac{P(X=1)}{1-P(X=0)}.[/tex]

    But since [tex]P(X=k)={n \choose k}p^k(1-p)^{n-k}[/tex] for a Binomial variable, we have [tex]P(X=0)=1/8[/tex] and [tex]P(X=1)=3/8[/tex]. So,

    [tex]P(X=1|X\geq 1)=\frac{3/8}{1-1/8}=\frac{3/8}{7/8}=\frac{3}{7}.[/tex]

    Notice the comma in [tex]P(X=1,X\geq 1)[/tex] means intersection.
     
  7. May 27, 2010 #6
    Thanks a lot, that made sense.
     
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