1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?
    (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

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook