Download Probability Models

7.1: Discrete and
Continuous Random
Variables
Game of Craps
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Rules
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If the sum is 7 or 11, the player WINS immediately.
If the sum is 2,3, or 12, the player LOSES immediately.
If the sum is anything else, the player continues rolling
the dice until she obtains the original sum (WIN) or rolls
a 7 (LOSS).
Our Mission
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Estimate the probability of a player winning at craps.
Game of Craps
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Activity
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Every pair plays a total of 20 games.
One player rolls the dice while the other records wins
and losses. (You may switch jobs after 10 games).
Calculate the relative frequency (percent) of wins.
Combine class results.
Game of Craps
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Compute the Exact Probability of Winning
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P(win) = P(win on the first roll) + P(win on some roll after
the first)
P(win) = P(roll a 7 or 11) + P(roll a 4,5,6,8,9, or 10 on the
first roll and then roll “point” before rolling a 7)
P(win) = 8/36 + 2([P(4) x P(4 before 7)] + [P(5) x P(5
before 7)] + [P(6) x
P(6 before 7)])
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Note: P(4) = P(10), P(5) = P(9), and P(6) = P(8)
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8
3 3 4 4 5 5

 2(      )
36
36 9 36 10 36 11
 0.4929
Discrete Random Variables
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Random Variable – variable whose value is a
numerical outcome of a random phenomenon.
A discrete random variable X has a countable
number of possible values.
Probability Distribution of a discrete random
variable X:
Value of X: x1
x2
x3
… xk
Probability: p1 p2 p3 … pk
Discrete Random Variables

Probability Distribution of a discrete random
variable X:
Value of X: x1
x2
x3
… xk
Probability: p1 p2 p3 … pk
1.
Every probability pi is a number between 0 and 1.
The sum of the probabilities is 1.
2.
Ex) Stat at State

Students in Statistics 101 at NC State earned the
following: 21% A’s, 43% B’s, 30% C’s, 5% D’s,
and 1% F’s. (If A = 4 on a four-point scale)
Value of X: 0
1
2
3
4
Probability: .01 .05 .30 .43 .21

What is the probability that the student earned a
B or better?
P(X>3) = P(X = 3) + P(X = 4)
= 0.43 + 0.21 = 0.64
Probability Histograms
Random
Digits
0 to 9
Notice that the
heights of the
probabilities add
up to 1.
Histograms allow
us to easily
compare
distributions.
Benford’s
Law
Ex) Tossing Coins

List all
What is the probability distribution of the discrete
possible
random variable X that counts the numberoutcomes
of
(HHHH,
heads in four tosses of a coin?
etc.)
X=0 X=1
X=2
X=3
X=4
# of Heads:
0
1
2
3
4
Probability: 0.0625 0.25 0.375 0.25 0.0625
Ex) Tossing Coins
# of Heads:
0
1
2
3
4
The probability distribution is
exactly
symmetric. It is 0.0625
an
Probability:
0.25 0.375 0.25 0.0625
idealization of the relative
frequency distribution.
What is the probability of
at least one head?
P(X>1) = 1 – P(X=0)
= 1 – 0.0625
= 0.9375
Continuous Random Variables
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Now, consider the following spinner. How can we
assign probabilities to the following events?
0.3 < x < 0.7
0
•Because the spinner can come to rest
anywhere:
2.5
7.5
S = {all numbers x such that 0 < x < 10}
•We CANNOT assign probabilities to
each x and then sum because there are
infinitely many possible values.
5
•Therefore, we must use the Density
Curve…
Density Curves
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Ex) Assign probabilities for generating a random
number between 0 and 1.
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The probability of any interval of numbers is the area
above the interval and under the curve.
Density Curves
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A continuous random variable has values
that are not isolated numbers but an entire
interval of numbers.
A probability distribution for a continuous
random variable is defined in terms of a
density curve.
A density curve is a nonnegative function that
has area = 1 between it and the horizontal
axis.
Density Curves
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The probability model for a continuous random
variable assigns probabilities to intervals of
outcomes.
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NOT to individual outcomes.
In fact, all continuous probability distributions
assign probability 0 to every individual
outcome.
Normal Distributions
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Remember that N(μ,σ) is our Normal Distribution
notation with mean μ and standard deviation σ.
When looking at random variables, if X has the N(μ,σ)
distribution, then the standardized variable:
Z
X 

is a standard Normal random variable having the
distribution N(0,1)
Ex) Honor Code
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Using a SRS of 400 college students, we learn that
12% of students would report a case of cheating.
(p=0.12)
If we used another SRS the parameter would most
likely change. Therefore, we will use p̂ (random
variable).
In this case, assume N(0.12, 0.016)
What is the probability that the survey result differs
from the truth about the population by more than 2
percentage points?
Ex) Honor Code
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What is the probability that the survey result differs
from the truth about the population by more than 2
percentage points?
P( pˆ  0.10 or pˆ  0.14)  1  P(0.10  pˆ  0.14)
0.10  0.12 pˆ  0.12 0.14  0.12
 1  P(


)
0.016
0.016
0.016
 1  P (1.25  Z  1.25)
About 21% of
 1  (0.8944  0.1056)
sample
Results will be off  1  (0.7888)
by more than two
percentage points. 0.2112