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Discrete Random Variables – Outline
1.Two kinds of random variables
–Discrete
–Continuous
2. Describing a DRV
3. Expected value of a DRV
4. Variance of a DRV
5. Examples
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Two kinds of random variables
A. Discrete (DRV)
• Outcomes have countable
values (can be listed)
• E.g., # of people in this room
– Possible values can be
listed: might be …51 or
52 or 53…
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Two kinds of random variables
A. Discrete
B. Continuous (CRV)
• Not countable
• Consists of points in an
interval
• E.g., time till coffee break
• CRVs are the subject of next
week’s lecture
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Describing a DRV
• We begin by defining the DRV
of interest
• For example, if our
experiment is to toss a fair coin
twice, we could define the
variable X as the # of heads
that occurs in those two
tosses.
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Describing a DRV
• To describe a DRV, we specify
the possible values and their
respective probabilities.
• For the coin-toss experiment,
we have:
Possible
values
0
1
2
Notes:
P(x) ≥ 0, ∀x.
∑p(x) = 1.0
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Probability
.25
.5
.25
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Expected value of a DRV
• We can compute the average
or “expected value” of a DRV
• This is the average value of X
that we would find over many
repetitions of the experiment
• It is a theoretical quantity
based on the possible values
and their probabilities
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Expected value of a DRV
• Example experiment: two
tosses of a fair coin
• DRV: X = # of heads that
occur
•Expected value of X: the mean
# of heads that would occur
over many repetitions of the
experiment.
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Expected value of a DRV
Expected value:
Variance:
μ = E(x) = ∑[x * p(x)]
σ2 = ∑[(x – μ)2 * p(x)]
Note: σ = √σ2
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Expected Value of a DRV
For the coin toss:
Possible
values
0
1
2
μ = (0 * ¼) + (1 * ½) + (2 * ¼)
=1
Probability
¼
½
¼
σ2 = [(0-1)2 * ¼] + [(1-1)2 * ½]
+ [(2-1)2 * ¼]
=½
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Example 1
12 candidates apply for positions in a local law firm (8
men and 4 women). The firm decides that they only
have enough money to hire 3 of the applicants. They
decide to give each applicant an aptitude test and hire
the 3 with the highest scores. Assume that law aptitude
is randomly distributed in the population and that, on
average, there are no differences between women and
men. Let the random variable X be the number of
women who score in the top 3 on the aptitude test.
What is the distribution of X?
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Example 1
First, what is the DRV?
• X = # of women who score in
the top 3 on the law aptitude
test.
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Example 1
Second, what are the possible
outcomes?
0, 1, 2, and 3
Do you see why?
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Example 1
Next, what are the probabilities of these outcomes?
The question says that law aptitude is randomly
distributed with no sex difference. Therefore, the
probability that we get 0 women in the top 3 is:
8 4
(3 ) ( 0 )
12
(3 )
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Example 1
# of ways of choosing 0
women out of 4
8 4
( 3) ( 0 )
12
(3)
Total # of ways of
choosing 3 candidates
out of 12
# of ways of choosing 3
men out of 8
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Example 1
The question says that law aptitude is randomly
distributed with no sex difference. Therefore, the
probability that we get 0 women in the top 3 is:
8 4
(3 ) (0 )
12
(3 )
=
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56
220
= .254
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Example 1
We compute the probabilities of other outcomes
similarly. For P(x =1):
8 4
(2) (1 )
12
(3)
=
112
220
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= .509
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Example 1
P(X = 2):
8 4
( 1 ) ( 2)
12
( 3)
=
48
220
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= .218
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Example 1
P(X = 3):
8 4
(0) ( 3)
12
(3 )
=
4
220
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= .018
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Example 1
X
0
1
2
3
P(X)
These are the values
that X can take
.254
.509
.218
.018
These are the
probabilities for each
value of X
This is how we present the distribution.
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Example 2
In the game GAMBLE, a player flips a coin and the game ends
when the first head occurs or after the fifth flip. However, the
nature of the coin’s bias changes over successive flips, such that
the true ratio of heads to tails is n:1 for a given flip, where n is
the ordinal position of a given flip in the series of flips (for
example, the ratio of heads to tails for the third flip would be
3:1). The player loses $20 if the game ends after the first flip. The
player wins (n X $10) if the game ends after any of flips 2 through
5. Let X be the amount of money that can be won or lost playing
the game one time (losses are represented by negative
amounts). What is the Expected Value of X ?
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Example 2
First, what are the possible outcomes?
X = amount of money that can be won or lost when
playing the game once. Recall that the player loses
$20 (X = -20) if the game ends on the first flip.
Possible values (in dollars):
X = -20, 20, 30, 40, or 50
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Example 2
Next, what has to happen to produce each of these
outcomes? We need to specify that in order to work
out the probabilities.
X
What has to happen
-20
H
20
T, H
30
T, T, H
40
T, T, T, H
50
T, T, T, T, H or T, T, T, T, T
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Example 2
• Now, what are the
probabilities?
• To answer, we have to specify
the probabilities of H on each
flip – because the question
tells us these probabilities keep
changing.
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Example 2
Flip #Ratio H:T
1
1:1
2
2:1
3
3:1
4
4:1
5
5:1
P(H)
.5
.67
.75
.80
.83
Remember that the ratio of
heads to tails is n:1 for the nth
trial
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2:1 means that out of
every 3 flips, 2 will be
heads and 1 will be
tails – so we get
P(heads) = 2/3 (= .67)
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Example 2
We can now work out the probabilities:
X P(X)
This is P(heads) for flip #1
-20 .5
This is P(tails) for flip #2
20 .5*.67
30 .5*.33*.75
40 .5*.33*.25*.80
50 .5*.33*.25*.20*.83 + .5*.33*.25*.20*.17
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Example 2
X
-20
20
30
40
50
P(X)
.5
.335
.12375
.033
.00825
ΣP(X) = 1.00
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Example 2
We’re not finished yet – the question asks:
“What is the Expected Value of X?”
µ = (-20)*.5 + (20)*.335 + (30)*.12375)
+ (40)*.033 + (50)*(.00825)
= 2.145
(Now we’re finished!)
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Example 3
3. In a local basketball league, when a person is
fouled they get to shoot two free throws. Over
the years of playing in the league, Beth has
calculated that every time she takes two free
throws, she makes her first free throw 60% of
the time. If she makes her first free throw, she
makes the second one 70% of the time. If she
misses the first one, she makes the second
one only 40% of the time.
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Example 3
a) Let X be the # of free throws that Beth makes
when she goes to the free throw line to shoot
two free throws. What is the probability
distribution for X?
First, what are the possible outcomes? Beth
could make 0, 1, or 2 free throws.
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Example 3
Secondly, how might these outcomes be achieved?
Outcome
First throw
Second throw
0
1
1
2
Miss
Hit
Miss
Hit
Miss
Miss
Hit
Hit
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Example 3
Let’s define:
Event A = Beth makes the free throw on the first
attempt
Event B = Beth makes the free throw on the
second attempt
We are given
P(A) = .60
P(B│A) = .70
P(B’│A) = .30
P(B│A’) = .40
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Example 3
Compute:
For X = 0 P(A’B’) = P(A’) * P(B’│A’)
= .60 * .40
= .24
For X = 1 P(AB’) = P(A) * P(B’│A)
= .60 * .30
= .18
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Example 3
For X = 1 P(A’B) = P(A’) * P(B│A’)
= .40 * .40
Thus, P(X = 1) = .18 + .16
= .16
= .34
For X = 2 P(AB) = P(A) * P(B│A)
= .60 * .70
= .42
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Example 3
X
P(X)
0
1
2
.24
.34
.42
μ = Σ(x* p(x))
= 0 *.24 + 1*.34 + 2*.42
= 1.18
σ2 = Σ[(x-μ)2 * p(x)]
= [(0-1.18)2 * .24] +
[(1-1.18)2 * .34] +
[(2-1.18)2 * .42]
= .627
σ = √.627 = .792
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Example 3
b) In the next game, Beth gets fouled twice. On both
occasions, she gets to shoot two free throws. This is
equivalent to taking a sample of size 2 from the
distribution you created in part a) above. Let Y = the
total # of free throws Beth makes in those 2
opportunities to shoot two free throws. What is the
probability distribution of Y?
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Example 3
First, what are the possible outcomes?
“a sample of size 2” means 2 trips to the free throw
line, with two throws on each trip. Therefore, possible
outcomes for Y are 0, 1, 2, 3, and 4.
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Example 3
Secondly, what the respective probabilities of these
outcomes?
To answer this question, we have to think about how
each outcome might arise…
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Example 3
Y
0
1
1
2
2
2
3
3
4
# hits first trip # hits second trip
0
0
0
1
1
0
2
0
0
2
1
1
2
1
1
2
2
2
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P(Y = 1) will
have two
components
P(Y = 2) will
have three
components
Example 3 – the probability
distribution
Y Component probabilities
0
1
2
3
4
P(Y)
.24 * .24
(.24*.34) + (.34*.24)
(.42*.24) + (.24*.42) + (.34*.34)
(.34*.42) + (.42*.34)
(.42*.42)
.0576
.1632
.3172
.2856
.1764
Σ = 1.000
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