Results 1 to 3 of 3

Math Help - Probability Distrubtion - ANSWER CHECK!

  1. #1
    Member Last_Singularity's Avatar
    Joined
    Dec 2008
    Posts
    157
    Quote Originally Posted by cnmath16 View Post
    48. A coin is tossed 8 times. Calculate the probability of tossing

    a) 6 heads and 2 tails

    b) 7 heads and 1 tail

    c) at least 6 heads
    The coin question is binomially distributed with p = 0.5 and n = 8. The problem of getting k successes out of n trials is calculated by
    \frac{n!}{(n-k)!k!} p^k (1-p)^{n-k}

    For more properties of the distribution, see: Binomial distribution - Wikipedia, the free encyclopedia
    Last edited by mr fantastic; February 8th 2009 at 02:16 AM. Reason: Post #2 comes first - a consequence of moving a reply to a duplicate question (and yes, I had my reasons for moving it here).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2008
    Posts
    53

    Probability Distrubtion - ANSWER CHECK!

    48. A coin is tossed 8 times. Calculate the probability of tossing..

    a) 6 heads and 2 tails

    My answer:
    Let p=P(heads) = 1/2
    Let q= P (tails) =1/2
    From the expansion of (p+q)^8, use the term containing p^6.

    Thus, P(6 heads) = c (8,2)(1/2)^6 (1/2)^2
    = 8! 6! 2! (1/2)^8
    = (56/2)(1/256)
    = 28/256
    = 14/128
    = 7/64 is the probability of 6 heads and 2 tails occuring

    b) 7 heads and 1 tail
    My answer:
    From the expansion of (p+q)^8, use the term containing p^7

    Thus, P(7 heads) = c (8,1) (1/2)^7(1/2)
    = 8! 7! 1! (1/2)^8
    = 8/256
    = 1/32 is the probability of 7 heads and 1 tail

    c) at least 6 heads
    My answer:
    P (at least 6 heads) = P(6 heads) + P (7 heads) + P (8 heads)
    P (6 heads) = 28/256
    P (7 heads) = 8/256
    P(8 heads) = 1/256

    Thus, the total probability of 6 heads is equal to 28/256 + 8/256 + 1/256 which is equal to P= 37/256
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by cnmath16 View Post
    48. A coin is tossed 8 times. Calculate the probability of tossing..

    a) 6 heads and 2 tails

    My answer:
    Let p=P(heads) = 1/2
    Let q= P (tails) =1/2
    From the expansion of (p+q)^8, use the term containing p^6.

    Thus, P(6 heads) = c (8,2)(1/2)^6 (1/2)^2
    = 8! 6! 2! (1/2)^8
    = (56/2)(1/256)
    = 28/256
    = 14/128
    = 7/64 is the probability of 6 heads and 2 tails occuring

    b) 7 heads and 1 tail
    My answer:
    From the expansion of (p+q)^8, use the term containing p^7

    Thus, P(7 heads) = c (8,1) (1/2)^7(1/2)
    = 8! 7! 1! (1/2)^8
    = 8/256
    = 1/32 is the probability of 7 heads and 1 tail

    c) at least 6 heads
    My answer:
    P (at least 6 heads) = P(6 heads) + P (7 heads) + P (8 heads)
    P (6 heads) = 28/256
    P (7 heads) = 8/256
    P(8 heads) = 1/256

    Thus, the total probability of 6 heads is equal to 28/256 + 8/256 + 1/256 which is equal to P= 37/256
    All correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Check my probability answer?
    Posted in the Statistics Forum
    Replies: 5
    Last Post: July 26th 2009, 02:39 PM
  2. Probability -- Check my answer please!
    Posted in the Statistics Forum
    Replies: 2
    Last Post: January 10th 2009, 10:44 AM
  3. Answer Check - Probability
    Posted in the Statistics Forum
    Replies: 3
    Last Post: December 29th 2008, 10:16 AM
  4. Probability Answer Check
    Posted in the Statistics Forum
    Replies: 3
    Last Post: May 4th 2008, 01:23 PM
  5. check probability answer please
    Posted in the Statistics Forum
    Replies: 2
    Last Post: August 7th 2006, 09:14 PM

Search Tags


/mathhelpforum @mathhelpforum