Results 1 to 6 of 6

Math Help - 10 throws of coin 0-4 heads how many possibilities?

  1. #1
    Newbie
    Joined
    Jan 2008
    From
    West Dublin
    Posts
    9

    10 throws of coin 0-4 heads how many possibilities?

    If one tosses a coin ten times, how many possible outcomes are there (assuming order is important) if of the ten throws there must be between 0 -4 heads?

    Not sure how to calculate that...

    Eg obviously total possibilities for 10 throws in general is 2^10=1024 possibilities
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1
    \sum\limits_{k = 0}^4 {{ 10 \choose k} \left( {\frac{1}{2}} \right)^{10} } is the probability of no more than four heads out of ten.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2008
    From
    West Dublin
    Posts
    9
    REally sorry but I'm still stuck/I don't fully follow?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1
    “HHHTTTTTT” that string represents tosses of a coin where we get four heads and six tails.
    That particular string has the probability of \left( {\frac{1}{2}} \right)^{10} of occurring.
    If we rearrange that string in anyway, say “THTHTTHHT” it still represents four heads and six tails. There are {{10} \choose 4} = \left( {\frac{10!}{(4!)(6!)}} \right) way to rearrange that string.
    So there are that many ways to get four heads and six tails.
    In the sum we did it for 0,1,2,3, & 4.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Plato View Post
    “HHHTTTTTT” that string represents tosses of a coin where we get four heads and six tails.
    That particular string has the probability of \left( {\frac{1}{2}} \right)^{10} of occurring.
    If we rearrange that string in anyway, say “THTHTTHHT” it still represents four heads and six tails. There are {{10} \choose 4} = \left( {\frac{10!}{(4!)(6!)}} \right) way to rearrange that string.
    So there are that many ways to get four heads and six tails.
    In the sum we did it for 0,1,2,3, & 4.
    There is probably no simpler way to explain it. Good one Plato!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, Joanna!

    Everyone is giving you probabilities . . .


    If one tosses a coin ten times, how many possible outcomes are there
    (assuming order is important) if there must be between 0 -4 heads?

    We must consider the five cases: 0 heads, 1 head, 2 heads, 3 heads, 4 heads
    . . and add the results.


    For 0 heads, there is one way (all Tails): 1

    For 1 head, there are: . {10\choose1} = 10 ways.

    For 2 heads, there are: . {10\choose2} \:=\:45 ways.

    For 3 heads, there are: . {10\choose3} \:=\:120 ways.

    For 4 heads, there are: . {10\choose4} = 210 ways.


    Therefore, there are: . 1 + 10 + 45 + 120 + 210 \:=\:\boxed{386\text{ ways}}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Consecutive Heads/Tails in weighted coin toss
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: April 14th 2013, 05:52 AM
  2. Replies: 1
    Last Post: October 17th 2011, 12:08 PM
  3. Replies: 2
    Last Post: July 7th 2010, 11:54 PM
  4. proof of the identity for a coin having even number of heads
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 26th 2009, 07:23 AM
  5. P(H heads in a row) given N coin flips
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 3rd 2009, 09:31 PM

Search Tags


/mathhelpforum @mathhelpforum