Tutorial on discrete probability distributions with examples and detailed solutions.

The probabilities P(X) are such that

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

b) Find the mean and standard deviation of X.

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

- 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 µ of the random variable X is defined by

**μ**=**∑**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**= √ [**∑**(X- µ)^{ 2}P(X) ]

= √ [ (0 - 1)^{ 2}* (1/4) + (1 - 1)^{ 2}* (1/2) + (2 - 1)^{ 2}* (1/4) ]

= 1 / √ (2)

a) Obtain the probability distribution of X.

b) Find the mean and standard deviation of X.

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

- 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

The graph of the probability of the sum of two dice is shown below.

- As an exercise, check that

**∑**P(X) = 1

- b) The mean of X is given by

**µ**=**∑**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

**µ**=**σ**√ [**∑**(X- µ)^{ 2}P(X) ]

= √ [ (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

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

- 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 µ of the random variable X as follows

**µ**=**∑**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**= √ [**∑**(X- µ)^{ 2}P(X) ]

= √ [ (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)