Discrete Probability Distribution

1. Probability and Statistics
Week 2 – Mean and Standard Deviation, Probability Distribution
Dr. Ferdin Joe John Joseph

2. Joint Probability
• Probability of events A and B denoted by P(A and B) or P(A ∩ B) is the
probability that events A and B both occur.
• P(A ∩ B) = P(A). P(B)
• This only applies if A and B are independent, which means that if A
occurred, that doesn’t change the probability of B, and vice versa.
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3. Conditional Probability
• A and B are not independent
• When A and B are not independent, it is often useful to compute the
conditional probability, P (A|B)
• The probability of A given that B occurred: P(A|B) =
P(A ∩ B)
P(B)
• Similarly, P(B|A) =
P(A ∩ B)
P(A)
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4. • Joint probability of A and B can be denoted as
• P(A ∩ B)= p(A).P(B|A)
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5. Bayes Theorem
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6. Bayes Theorem
• Used in Naïve Bayes Classifier (Supervised Learning)
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7. Probability Distribution
• A probability distribution is a list of all of the possible outcomes of a
random variable along with their corresponding probability values.
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8. Discrete Probability Distribution
• If we consider 1 and 2 as outcomes of rolling a six-sided die, then we
can’t have an outcome in between that (e.g. We can’t have an
outcome of 1.5).
• This is called probability mass function
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9. Mean
• Let the data points be 600, 470, 170, 430, 300
• Total elements n = 5
• Mean =
600 +470+170+430+300
5
=
1970
5
= 394
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10. Variance
• Let the data points be 600, 470, 170, 430, 300
• Total elements n = 5
• Variance =
600 −𝑚𝑒𝑎𝑛 2
+ 470 −𝑚𝑒𝑎𝑛 2
+ 170 −𝑚𝑒𝑎𝑛 2
+ 430 −𝑚𝑒𝑎𝑛 2
+ 300 −𝑚𝑒𝑎𝑛 2
5
=
600 −394 2
+ 470 −394 2
+ 170 −394 2
+ 430 −394 2
+ 300 −394 2
5
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11. Variance σ2
σ2= 108520/5
σ2= 21704
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12. Standard Deviation
• Standard Deviation (SD) or σ = 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒
• σ = 21704 = 147.32
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13. Probability Distribution
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14. Example
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15. Example
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16. Discrete Probability Distributions
• A discrete random variable assumes each of its values with a certain
probability.
• In the case of tossing a coin three times, the variable X, representing
the number of heads, assumes the value 2 with probability 3/8, since
3 of the 8 equally likely sample points result in two heads and one
tail.
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17. Discrete Probability Distribution
• the probability that no employee gets back the right helmet, that is,
the probability that M assumes the value 0, is 1/3. The possible
values m of M and their probabilities are
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18. Discrete Probability Distribution
• Frequently, it is convenient to represent all the probabilities of a
random variable X by a formula. Such a formula would necessarily be
a function of the numerical values x that we shall denote by f(x), g(x),
r(x), and so forth. Therefore, we write f(x) = P(X = x); that is, f(3) = P(X
= 3).
• The set of ordered pairs (x, f(x)) is called the probability function,
probability mass function, or probability distribution of the discrete
random variable X.
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19. Discrete Probability Distribution
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20. Example
• A shipment of 20 similar laptop computers to a retail outlet contains 3
that are defective. If a school makes a random purchase of 2 of these
computers, find the probability distribution for the number of
defectives.
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21. Solution
• Let X be a random variable whose values x are the possible numbers
of defective computers purchased by the school. Then x can only take
the numbers 0, 1, and 2.
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22. Exercise
• If a car agency sells 50% of its inventory of a certain foreign car
equipped with side airbags, find a formula for the probability
distribution of the number of cars with side airbags among the next 4
cars sold by the agency.
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23. Solution
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24. Example
• 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.
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25. Solution
a) We first construct a tree diagram to represent all possible
distributions of boys and girls in the family.
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26. Solution
• 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
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27. Solution
• The discrete probability distribution of X is given by
X P(X)
0 1 / 4
1 1 / 2
2 1 / 4
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28. Solution
• 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
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29. Solution
• The standard deviation σ of the random variable X is defined by
SD = Square Root [ ∑ (X- µ) 2 P(X) ]
= (0 − 1) 2 ∗ (1/4) + (1 − 1) 2 ∗ (1/2) + (2 − 1) 2 ∗ (1/4)
= 1 / 2
= 1/1.414
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30. Exercise
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.
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31. Solution
a) When the two balanced dice are rolled, there are 36 equally likely
possible outcomes as shown below.
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32. Solution
• 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
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33. Solution
• 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 33

34. Solution
• 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
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35. Solution
• The standard deviation of is given by
SD = √ [ ∑ (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
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36. Exercise
1. 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.
2. A fair coin is tossed twice. Let X be the number of heads that are observed.
• Construct the probability distribution of X.
• Find the probability that at least one head is observed.
3. A service organization in a large town organizes a raffle each month. One thousand raffle tickets
are sold for $1 each. Each has an equal chance of winning. First prize is $300, second prize is $200,
and third prize is $100. Let X denote the net gain from the purchase of one ticket.
• Construct the probability distribution of X.
• Find the probability of winning any money in the purchase of one ticket.
• Find the expected value of X, and interpret its meaning.
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37. Solutions
• Click Here
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