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Counting problem, exactly 5 heads obtained, coin

  1. Oct 25, 2006 #1
    Hello everyone, i'm having some issues on this problem:
    A coin is tossed ten times. In each case the outcome H (for heads) or T (for tails) is recorded. (One possible outcome of the ten tossings is denoted THHTTTHTTH.)

    I got a, d right i believe.
    but I need someone to check if i did the others ones right.


    a. what is the total number of possible outcomes of the coin-tossing experiemnt?

    I said: 2^10 = 1,024

    b. In how many of the possible outcomes are exactly 5 heads obtained?
    [# of outcomes with exactly 5 heads] = [total # of outcomes] - (10 choose 5)

    This one I'm not sure what i'm suppose to do, when i say 10 choose 5, i mean you have a set of 10, and i'm wanting to choose 5 elements out of that set that are H but i'm not sure how to represeent this...

    c. In how many of the possible outcomes is at least 8 heads obtained?
    For this one would i do:
    [# of outcomes with at least 8 heads obtained] = [total number of outcomes] - [number of outcomes that contain 8 heads]

    = 1024 - (10 choose 8)
    = 1024 - 10!/[8!(2!)]
    = 1024 - 45 = 979


    d. In how many of the possible outcomes is at least one head obtained?
    [# of outcomes w/ at least 1 head] = [total # of outcomes] - [# of outcomes with no heads] = 1024 - 1 = 1023.

    e. In how many of the possible outcomes is at most one head obtained?

    I think i got this one, but the # of outcomes with 1 head i might have got wrong.

    [# of outcomes with at most 1 head] = [# of outcomes with no heads] + [# outcomes with 1 head]
    = 1 + (10 choose 1)
    = 1 + 10!/[1!(10-1)!]
    = 1 + 10!/9! = 11



    Any help would be great, i'm having issues figuring out how to choose once i figure out the method it seems.
     
  2. jcsd
  3. Oct 25, 2006 #2
    I am pretty sure (b) and (c) are wrong, but the rest are right.

    One thing you could notice is that [tex]1024 = 2^{10} = \binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} + \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10}[/tex]

    What this means is that you can partition the 1024 outcomes according to how many heads there are in each possible outcome (you could also replace heads with tails here).

    [tex]\binom{10}{i}[/tex] would be the number of outcomes with exactly i heads.

    So [tex]\binom{10}{3}[/tex] would be the number of outcomes with exactly 3 heads.

    So you should be able to figure out (b) from this.

    In your explanation you keep subtracting from 1024, I am not sure why you are doing this (well I have an idea why) but you should not be doing this. Remember X - A = A' (Where A,A' are subsets of X, and A' is the complement of A). That is, 1024 - number of sets with 5 heads = number of sets that do not have 5 heads.

    In part (c) you are doing [total] - [number that have exactly 8], as I said above this would be the [number that do not have 8], not the number that have at least 8. The negation of [at least 8] is [less than 8], so you would have to say [number with at least 8] = [total] - [number with less than 8] but you would not want to do it this way since you would have more things to compute than if you just do it directly. So, I would just compute each case directly (that is, the number of outcomes for 8 heads, 9 heads, and 10 heads, and add these).

    edit.. the LaTeX is not working, just click the "images" to see the code I guess.
     
    Last edited: Oct 26, 2006
  4. Oct 27, 2006 #3
    Thanks matt i'll take another look at it once we get our homework back
     
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