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Math Help - little problems

  1. #1
    Member
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    little problems

    Hello,
    Question no:1
    A coin is biased such that a head is thrice as likely to occur as a tail. Find the probability distribution of heads and also find the mean and variance of the distribution when it is tossed 4 times.

    (i am confused in this question can somebody explain this.according to point of view"head is thrice "may be we can take as 0.75".
    i solve this question by following method it is correct or there is another method plz explain it.)
    P(no heads) = 4C0 (.75)0(.25)4 = .004

    P(1 head, 3 tails) = 4C1(.75)1(.25)3 = .047

    P(2 head, 2 tail) = 4C2(.75)2(.25)2 = .211

    P(3 head, 1 tail) = 4C3(.75)3(.25)1 = .423

    P(4 head, no tail)= 4C4(.75)4(.25)0 =.316


    Mean= E(X) = 0(.004)+1(.047) +2(.211)+3(.423)+4(.316) = 3 heads


    Variance = E(X2) -[E(X)]2
    E(X2) = 0(.004)+1(.047)+4(.211)+9(.423)+16(.316) = 9.754

    Variance = 9.754 - 9 = .754


    Question no:2
    FIND THE CEILING OF X, AND THE FLOOR OF X, OF THE FOLLOWING VALUE
    (1)-2.07
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by m777 View Post
    Hello,
    Question no:1
    A coin is biased such that a head is thrice as likely to occur as a tail. Find the probability distribution of heads and also find the mean and variance of the distribution when it is tossed 4 times.

    (i am confused in this question can somebody explain this.according to point of view"head is thrice "may be we can take as 0.75".
    i solve this question by following method it is correct or there is another method plz explain it.)
    P(no heads) = 4C0 (.75)0(.25)4 = .004

    P(1 head, 3 tails) = 4C1(.75)1(.25)3 = .047

    P(2 head, 2 tail) = 4C2(.75)2(.25)2 = .211

    P(3 head, 1 tail) = 4C3(.75)3(.25)1 = .423

    P(4 head, no tail)= 4C4(.75)4(.25)0 =.316


    Mean= E(X) = 0(.004)+1(.047) +2(.211)+3(.423)+4(.316) = 3 heads


    Variance = E(X2) -[E(X)]2
    E(X2) = 0(.004)+1(.047)+4(.211)+9(.423)+16(.316) = 9.754

    Variance = 9.754 - 9 = .754
    This looks OK to me

    Question no:2
    FIND THE CEILING OF X, AND THE FLOOR OF X, OF THE FOLLOWING VALUE
    (1)-2.07
    Ceiling is the smallest integer greater than the number so ceil(-2.07)=-2.

    Floor is the largest integer less than the number so floor(-2.07)=-3.

    RonL
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