Download Discrete Probability Distributions

Discrete Probability
Distributions
Quantitative Methods II
Plan for Today
• Discrete and continuous random variables
• Probability distributions
• Histograms for discrete probability
distributions
• Examples
• The expected value, the variance, and the
standard deviation
• Lotteries (example: Lotto 6/49)
• Probability of winning, expected payoff,
payoff rate, expected winnings.
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Random Variables
• We say that x is a random (quantitative)
variable if the value of x is determined by an
outcome of an experiment or an observation
and cannot be predicted precisely.
• A discrete random variable can have finitely
many or a countable number of values.
• A continuous random variable can take on
countless number of values on a given interval.
• Examples: 1. Number of passengers on a bus.
2. Weight of a watermelon.
Probability Distributions
• A probability distribution is an assignment of a
probability to each value or to each interval of
values of a random variable.
• For a discrete probability distribution (DPD),
each distinct value of a random variable has its
own probability.
• The sum of all probabilities equals 1.0000
• Let us consider an example:
Example: it is known that 25% of German adults
are smokers. A sample of 4 German adults is
selected at random.
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Example: smokers
• We can have 0, 1, 2, 3, or 4 smokers in the
sample. The respective probabilities can be
computed (recall how!) and put in a table:
x
P(x)
0
1
2
3
4
0.3164 0.4219 0.2109 0.0469 0.0039
Check that σ 𝑃(𝑥) = 1 .
This table is an example of a discrete probability
distribution (DPD).
A histogram for a DPD
• The data in the table can be better visualized
using a histogram.
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The expected value
• The mean of the probability distribution is
also called the expected value and is computed
using the following formula:
𝜇 = ෍ 𝑥 ∙ 𝑃(𝑥)
In the previous example,
𝜇 = 0 ∙ 0.3164 + 1 ∙ 0.4219 + 2 ∙ 0.2109 +
+ 3 ∙ 0.0469 + 4 ∙ 0.0039 = 1.0000
Why does it make sense to expect one smoker
out of four randomly selected Germans?
The variance and standard deviation
These quantities are computed as follows:
𝜎2 = σ 𝑥 − 𝜇
2
∙ 𝑃(𝑥)
and
𝜎 = 𝜎2
In the previous example,
𝜎 2 = 0 − 1.0 2 ∙ 0.3164 + 1 − 1.0 2 ∙ 0.4219
+ 2 − 1.0 2 ∙ 0.2109 + 3 − 1.0 2 ∙ 0.0469 +
+ 4 − 1.0 2 ∙ 0.0039 = 0.7500
and
𝜎 = 𝜎 2 = 0.8660
Why is it a reasonable value?
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Example: marbles in a bag
There are 16 marbles in a bag: 6 yellow and 10
red. Four marbles are chosen at random and
the number of red marbles, x , is counted.
We have the following probability table
(explain how the numbers are obtained)
x
P(x)
0
1
2
3
4
0.0082 0.1099 0.3709 0.3956 0.1154
Again, check that σ 𝑃(𝑥) = 1.0000 (or close)
The histogram
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Example (red and yellow balls): continued
• Compute the expected value, the variance,
and the standard deviation of this probability
distribution.
• μ = 2.5000
• Why does this number make perfect sense?
• 𝜎 2 = 0.7496
• 𝜎 = 0.8658
• Does this number seem reasonable?
Example: lottery 6/49
• For $3 a player chooses 6 numbers from 1 to 49.
• Six winning numbers plus 1 complimentary
number are drawn at random by the lottery
officials.
• There are seven winning categories and prizes:
W6 (Jackpot)
W5 + C
W5
W4
W3
W2 + C
W2
$ 10 000 000
(typical prizes listed)
$ 100 000
$ 2 000
$ 100
$ 10
$5
$ 3 (free play)
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Computing probabilities for Lotto 6/49
For the category W2, where two winning
numbers are guessed correctly:
1
8
15
22
29
36
43
2
9
16
23
30
37
44
3
10
17
24
31
38
45
4
11
18
25
32
39
46
5
12
19
26
33
40
47
6
13
20
27
34
41
48
7
14
21
28
35
42
49
Computing probabilities for Lotto 6/49
The total number of possible choices is
49𝐶6 = 13 983 816
The number of choices for two winning numbers
are 6𝐶2 = 15
The number of choices for four losing numbers
are 42𝐶4 = 111 930
Thus, the probability of winning in this category:
𝑃 2𝑊 =
6𝐶2 ∙ 42𝐶4
= 0.1201
49𝐶6
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Computing probabilities for Lotto 6/49
For the category W2+C, where 2 winning and the
complimentary number are guessed correctly:
1
8
15
22
29
36
43
2
9
16
23
30
37
44
3
10
17
24
31
38
45
4
11
18
25
32
39
46
5
12
19
26
33
40
47
6
13
20
27
34
41
48
7
14
21
28
35
42
49
Computing probabilities for Lotto 6/49
The number of choices for two winning numbers
are 6𝐶2 = 15
The number of choices for the complimentary
number: 1𝐶1 = 1
The number of choices for three losing numbers
are 42𝐶3 = 11 480
Thus, the probability of winning in this category:
𝑃 2𝑊 + 𝐶 =
6𝐶2 ∙ 1𝐶1 ∙ 42𝐶3
= 0.01231
49𝐶6
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Probabilities for other categories:
6𝐶3 ∙ 43𝐶3
= 0.01765
49𝐶6
6𝐶4 ∙ 43𝐶2
𝑃 4𝑊 =
= 0.0009686
49𝐶6
6𝐶5 ∙ 42𝐶1
𝑃 5𝑊 =
= 0.00001802
49𝐶6
6𝐶5 ∙ 1𝐶1
𝑃 5𝑊 + 𝐶 =
= 0.000000429
49𝐶6
6𝐶6
𝑃 Jackpot =
= 0.000000071
49𝐶6
𝑃 3𝑊 =
Another way of formulating probabilities
One chance in 1 / P(x). Consider Lotto 6/49:
Category
2W
2W + C
3W
4W
5W
5W + C
6W
Chances
1 in 8.3
1 in 81.2
1 in 56.7
1 in 1033
1 in 55 492
1 in 2 330 636
1 in 13 983 816
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Overall probability of winning
• It equals the sum of all winning probabilities.
• In the example of Lotto 4/49, we have
𝑃 winning = 0.1201 + 0.01231 + 0.01765 +
+ 0.0009686 + 0.00001802 + 0.000000429 +
+ 0.000000071 = 0.1510
or 1 in 6.6 (so the majority of all tickets contain
losing combinations)
The expected payoff
It is defined as the sum σ 𝑥 ∙ 𝑃(𝑥), where x is the
winning amount and P(x) is the corresponding
probability. In the example of Lotto 6/49:
෍ 𝑥 ∙ 𝑃(𝑥) = 3 ∙ 0.1201 + 5 ∙ 0.01231 +
+ 10 ∙ 0.01765 + 100 ∙ 0.0009686 +
+ 2000 ∙ 0.00001802 + 100 000 ∙ 0.000000429
+ 10 000 000 ∙ 0.000000071 =
= $1.48
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The payoff rate
It is defined as follows:
The payoff rate =
Expected payoff
Price of ticket
=
1.48
3.00
= 0.49
It can also be expressed in terms of percentages
The payoff rate = 49% . This also indicates the
rate of return on your investment.
The expected winnings
They are defined as follows:
Expected winnings =
= Expected payoff – Price of ticket
In the example of Lotto 6/49,
Expected winnings = 1.48 – 3.00 = – $1.52
The negative sign indicates that the player is
actually expected to lose $1.52 with the
purchase of each ticket.
Is it profitable to play such a lottery?
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