Tutorial on Discrete Probability Distributions

Tutorial on discrete probability distributions with examples and detailed solutions.

Let X be a random variable that takes the numerical values X1, X2, ..., Xn with probablities p(X1), p(X2), ..., p(Xn) respectively. A discrete probability distribution consists of the values of the random variable X and their corresponding probabilities P(X).

The probabilities P(X) are such that

P(X) = 1

Example 1:Let the random variable X represents the number of boys in a family.

a) Construct the probability distribution for a family of two children.

b) Find the mean and standard deviation of X.

Solution to Example 1:

  • a) We first construct a tree diagram to represent all possible distributions of boys and girls in the family.

    tree diagram for boys and girls distribution


  • Assuming that all the above possibilities are equally likely, the probabilities are:

    P(X=2) = P(BB) = 1 / 4
    P(X=1) = P(BG) + P(GB) = 1 / 4 + 1 / 4 = 1 / 2
    P(X=0) = P(GG) = 1 / 4

  • The discrete probability distribution of X is given by

    X P(X)
    0 1 / 4
    1 1 / 2
    2 1 / 4


  • Note that

    P(X) = 1


  • b) The mean of the random variable X is defined by

    = X P(X)


    = 0 * (1/4) + 1 * (1/2) + 2 * (1/4) = 1

  • The standard deviation σ of the random variable X is defined by

    σ = Square Root [ (X- ) 2 P(X) ]


    = Square Root [ (0 - 1) 2 * (1/4) + (1 - 1) 2 * (1/2) + (2 - 1) 2 * (1/4) ]

    = 1 / square root (2)

Example 2:Two balanced dice are rolled. Let X be the sum of the two dice.

a) Obtain the probability distribution of X.

b) Find the mean and standard deviation of X.

Solution to Example 2:

  • a) When the two balanced dice are rolled, there are 36 equally likely possible outcomes as shown below .

    possible outcomes when two dice are rolled


  • The possible values of X are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

  • The possible outcomes are equally likely hence the probabilities P(X) are given by

    P(2) = P(1,1) = 1 / 36
    P(3) = P(1,2) + P(2,1) = 2 / 36 = 1 / 18
    P(4) = P(1,3) + P(2,2) + P(3,1) = 3 / 36 = 1 / 12
    P(5) = P(1,4) + P(2,3) + P(3,2) + P(4,1) = 4 / 36 = 1 / 9
    P(6) = P(1,5) + P(2,4) + P(3,3) + P(4,2) + P(5,1)= 5 / 36
    P(7) = P(1,6) + P(2,5) + P(3,4) + P(4,3) + P(5,2) + P(6,1)
    = 6 / 36 = 1 / 6
    P(8) = P(2,6) + P(3,5) + P(4,4) + P(5,3) + P(6,2) = 5 / 36
    P(9) = P(3,6) + P(4,5) + P(5,4) + P(6,3) = 4 / 36 = 1 / 9
    P(10) = P(4,6) + P(5,5) + P(6,4) = 3 / 36 = 1 / 12
    P(11) = P(5,6) + P(6,5) 2 / 36 = 1 / 18
    P(12) = P(6,6) = 1 / 36

  • The discrete probability distribution of X is given by

    X P(X)
    2 1 / 36
    3 1 / 18
    4 1 / 12
    5 1 / 9
    6 5 / 36
    7 1 / 6
    8 5 / 36
    9 1 / 9
    10 1 / 12
    11 1 / 18
    12 1 / 36


  • As an exercise, check that

    P(X) = 1


  • b) The mean of X is given by

    = X P(X)
    = 2*(1/36)+3*(1/18)+4*(1/12)+5*(1/9)+6*(5/36)
    +7*(1/6)+8*(5/36)+9*(1/9)+10*(1/12)
    +11*(1/18)+12*(1/36)
    = 7

  • The standard deviation of is given by

    = σ Square Root [ (X- ) 2 P(X) ]
    = Square Root [ (2-7)2*(1/36)+(3-7)2*(1/18)
    +(4-7)2*(1/12)+(5-7)2*(1/9)+(6-7)2*(5/36)
    +(7-7)2*(1/6)+(8-7)2*(5/36)+(9-7)2*(1/9)
    +(10-7)2*(1/12)+(11-7)2*(1/18)+(12-7)2*(1/36) ]
    = 2.41

Example 3:Three coins are tossed. Let X be the number of heads obtained. Construct a probability distribution for X and find its mean and standard deviation.

Solution to Example 3:

  • The tree diagram representing all possible outcomes when three coins are tossed is shown below.

    tree diagram for all possible outcomes when three coins are tossed.


  • Assuming that all three coins are indentical and all possible outcomes are equally likely, the probabilities are:

    P(X=0) = P(TTT) = 1 / 8
    P(X=1) = P(HTT) + P(THT) + P(TTH)
    = 1 / 8 + 1 / 8 + 1 / 8
    = 3 / 8
    P(X=2) = P(HHT) + P(HTH) + P(THH)
    = 1 / 8 + 1 / 8 + 1 / 8
    = 3 / 8
    P(X=3) = P(HHH) = 1 / 8

  • The discrete probability distribution of X is given by

    X P(X)
    0 1 / 8
    1 3 / 8
    2 3 / 8
    3 1 / 8


  • Note that

    P(X) = 1


  • We now compute the mean of the random variable X as follows

    = X P(X)


    = 0 * (1/8) + 1 * (3/8) + 2 * (3/8) + 3 * (1/8) = 1.5

  • We now compute the standard deviation σ of the random variable X as follows

    σ = Square Root [ (X- ) 2 P(X) ]


    = Square Root [ (0 - 1.5) 2 * (1/8) + (1 - 1.5) 2 * (3/8) + (2 - 1.5) 2 * (3/8) + (3 - 1.5) 2 * (1/8) ]

    = 0.87 (rounded to 2 decimal places)



More references on
elementary statistics and probabilities.


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Updated: 2 April 2013

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