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STAT 111 Introductory Statistics
Lecture 7: More on Random Variables,
Probability, and Sampling
May 27, 2004
Today’s Topics
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Finishing up mean and variance
Conditional Probability
Multiplication Rule and Independence
Tree Diagrams
Bayes’s Rule
Sampling distributions
– Binomial distribution for sample counts
Recall: Rules for Means
• Let X and Y be (not necessarily independent)
random variables and let a and b be constants.
• Rule 1
E( a + bX ) = a + b E(X)
• Rule 2
E( a X + b Y ) = a E(X) + b E(Y)
• If X and Y are independent, then
E(XY) = E(X) E(Y)
Recall: Rules for Variances
• Let X and Y be random variables, and let a and b
again be constants.
• Rule 1
Var(a + b X) = b2 Var(X)
• Rule 2 If X and Y are independent, then
Var(X ± Y) = Var(X) + Var(Y)
• Rule 3 If X and Y have correlation ρ, then
Var(X ± Y) = Var(X) + Var(Y) ± 2ρ σX σY
Example: Pick 3 Ticket
• Suppose you buy a $1 Pick 3 ticket on each of
two different days. The payoffs X and Y on the
two tickets are independent. Let X + Y be the
total payoff. Calculate
– Expected value of the total payoff
– Variance of the total payoff
– Standard deviation of the total payoff
Example: Heights of Women
• The height of young women between 18 and 24
in America is approximately normally distributed
with mean µ = 64.5 and s.d. σ = 2.5.
• Two women are randomly chosen from this age
group.
• What are the mean and s.d. of the difference in
their heights?
• What is the probability that one is at least 5”
taller than the other?
• What is the IQR of heights in this age group?
Conditional Probability
• The probability of an event can change if we
know some other event has occurred.
• The conditional probability of an event gives us
the probability of one event under the condition
that we know the outcome of another event.
• Let A and B be any two events such that P(B) > 0.
The conditional probability of A assuming that B
has already occurred is written P(A | B):
P( A  B)
P( A | B) 
P( B)
Example: Rolling Dice
• Let A be the event that a 4 appears on a single
roll of a fair 6-sided die, and let B be the event
that an even number appears.
– Find P(A | B) and P(B | A).
• Suppose we add another (different-colored so we
can distinguish between the two) die to the mix,
and let C be the event that the sum of the two
dice is greater than 8.
– Find P(A | C) and P(C | A).
Example: Gender of Children
• Suppose we have a family with two children.
Assume all four possible outcomes ({older boy,
younger boy},…) are equally likely. What is the
probability that both are girls given that at least
one is a girl?
• Suppose instead that we ignored the age of the
children and distinguished only three family
types. How would this change the above
probability?
Example: Drawing Cards
• Draw 2 cards off the top of a well-shuffled deck.
• What is the probability that the second card is an
Ace, given that the first card was an Ace?
• On the other hand, consider only the first card for
a minute. Suppose you do not see what the card
is, and your friend tells you the card is a King.
What is the probability that the card is a
diamond?
Multiplication Rule
• The probability that both event A and event B
occur is given by
P(A and B) = P(A) P(B | A) = P(B) P(A | B)
• Here, P(A | B) and P(B | A) have the usual
meaning of being conditional probabilities.
Example: Home Security
• House security experts estimate that an untrained
house dog has a 70% probability of detecting an
intruder – and, given detection, a 50% chance of
scaring the intruder away.
• What is the probability that Fido successfully
thwarts a burglar? (The probability of a trained
watchdog detecting and running off an intruder is
estimated to be around 0.75)
Example: Drawing Chips from an Urn
• An urn contains 5 white chips and 4 blue chips.
Two chips are drawn sequentially and without
replacement. What is the probability of obtaining
the sequence (W, B)?
• The multiplication rule can be extended to higherorder intersections. For example, suppose we
throw 3 red chips and 5 yellow chips into our urn.
Five chips are drawn sequentially and without
replacement. What is the probability of obtaining
the sequence (W, R, W, B, Y)?
Independence
• Recall that two events A and B are independent if
knowing one occurs does not change the
probability that the other occurs.
• When two events are independent, we have that
P(B | A) = P(B) and P(A | B) = P(A)
• Recall our example about the probability of a
single card draw being a diamond given that we
are told it is a King.
Many Independent Events
• Suppose we have n independent events
A1, A2, …, An. Then the multiplication rule is
P( A1  A2  ...  An )  P( A1 ) P( A2 ) ... P( An )
Example: Ten Rolls of a Die
• Roll a die ten times.
• What is the probability that we roll a 2 10 times?
• What is the probability that we roll at least one
2?
Example: Height of Women
• Randomly select 8 American young women aged
18 to 24.
• What is the probability that all 8 women are
more than 65 inches tall?
• What is the probability at least one of the women
is between 63 and 67 inches tall?
Tree Diagrams
• A tree diagram is often helpful for solving more
elaborate calculations, and in particular,
problems that have several stages.
• In a tree diagram, each segment in the tree
represents one stage of the problem.
• Each complete branch shows a possible path.
• Tree diagrams combine both the addition and
multiplication rules.
Sample Tree Diagram
Probability
of outcome
in Stage 1
Stage 2
Stage 1
H
0.5
First flip
0.5
T
H
0.5
Second flip
0.5
T
0.5
H
Second flip
0.5
T
Conditional
probability of
outcome in
Stage 2 given
the outcome in
Stage 1
Example: Tossing a Coin Twice
P(HH) = P(H)P(H)
= (0.5)(0.5) = 0.25
H
H HH
Second flip
T HT
H TH
First flip
Second flip
T
T
P(HT) = P(H)P(T)
= (0.5)(0.5) = 0.25
P(TH) = P(T)P(H)
= (0.5)(0.5) = 0.25
TT
P(TT) = P(T)P(T)
= (0.5)(0.5) = 0.25
Example: Dependent Coin Flips
• The previous example has independent coins
flips, but we can imagine situations where coin
flips will be dependent.
• Consider the following situation. We have two
coins, one fair (P(H) = 0.5 = P(T)) and one
biased (P(H) = 0.75, P(T) = 0.25).
– Flip fair coin; if H, use biased coin next; otherwise,
use fair coin again.
– Flip biased coin after every H, fair coin after every T.
• Coin flips now are no longer independent.
Example: Chips in an Urn
• As a slightly different example, consider an
urn with 5 white chips and 4 blue chips. We
draw three chips sequentially and without
replacement. What is the probability of
obtaining the sequence (W, B, B)? Calculate
this using a tree diagram.
Example: Lab Testing
• A lab test can yield either a positive or negative
result. For people with a particular disease, it
will produce a positive result 90% of the time.
But it will also produce a positive result in 0.1%
of all healthy people. Suppose that 0.01% of the
population actually has the disease.
– What is the probability that an individual is healthy?
– That a sick individual produces a negative test result?
– That an individual is healthy and has a negative test
result?
Tree Diagram for Lab Testing
Stage 2
Stage 1
0.90
0.0001
?
ill
+
Ill and +
+
Ill but -
Test
Individual ?
0.001
healthy
?
Test
-
Healthy but +
Health and -
More on Lab Testing
• We know from our initial conditions that if a
person chosen randomly has the disease, we get a
positive test with probability 0.90.
• What we’re more interested in usually is
diagnosing individuals with the disease. In other
words, we want to know what the probability is
that an individual has the disease given that his
test result is positive.
Bayes’s Rule
• What we need is a method that allows us to use
our known conditional probabilities to compute
the conditional probabilities “in the other
direction.”
• The formula we use is called Bayes’s Rule and
can be stated as follows: if A and B are any two
events whose probabilities are not 0 or 1, then
P( B | A) P( A)
P( A | B) 
P( B | A) P( A)  P( B | Ac ) P( Ac )
Derivation of Bayes’s Rule
P( A  B)
P( A | B) 
P( B)
P ( B  A)

P( B)
P( B | A) P ( A)

c
P( B  A)  P( B  A )
P ( B | A) P( A)

c
c
P( B | A) P ( A)  P ( B | A ) P( A )
Example: Lab Testing
• In our example, A is the event that an individual
is ill, and B is the event that the test result is
positive.
• So, let’s calculate the probability of an individual
being ill given a positive test result.
Example: Coins and Urns
• A biased coin, twice as likely to come up heads as
it is tails, is tossed once.
– If heads, draw chip from urn I, which contains 3 white
chips and 4 red chips.
– If tails, draw chip from urn II, which contains 6 white
chips and 3 red chips.
• Given that a white chip was drawn, what is the
probability that the coin came up tails?
Population and Sampling Distributions
• The population distribution of a variable is the
distribution of its values for all members of the
population.
• The population distribution is also the
probability distribution of the variable when we
choose one individual from the population at
random.
Population and Sampling Distributions
• A statistic is any numeric measure that is used to
describe the data we obtain.
• If the data are obtained using random sampling, a
statistic is a random variable, and hence its value
varies from sample to sample.
• The probability distribution of the statistic is
known as its sampling distribution.
Population and Sampling Distributions
• The sampling distribution of a statistic depends
not only on the population distribution, but also
on the sample size and the method used to collect
the data from the population.
• A statistic can be used to estimate the parameter
of the population.
Example: Two Different Surveys
• Two national surveys are planned to estimate the
proportion p of people who are skeptical of news
media.
– First survey randomly selects 1000 people.
– Second survey randomly selects 10,000 people.
• Will the results be the same?
• Are the survey results biased?
• If the these two surveys are repeatedly performed,
which one will yield less variable estimates?
The Binomial Setting
• There are a fixed number n of trials.
• The n trials are all independent.
• Each trial has one of two possible outcomes,
labeled “success” and “failure.”
• The probability of success, p, remains the same
for each trial.
Example: Chips in an Urn
• An urn contains 5 white chips and 4 red chips.
Randomly draw 3 chips with replacement from
this urn.
• Let S represent the outcome where a red chip is
drawn, and F the outcome where a white chip is
drawn.
• This is a binomial experiment with p = 4/9.
• Would this still be a binomial experiment if the
chips were drawn without replacement?
The Binomial Distribution
• We briefly mentioned the binomial distribution in
passing previously.
• The distribution of the count X of successes in
the binomial setting is called the binomial
distribution with parameter n and p, where
– n is the number of trials
– p is the probability of a success on any trial
• The count X is a discrete random variable,
typically abbreviated as X ~ B(n, p).
The Binomial Distribution
• Possible values of X are the whole numbers from
0 to n.
• We know that the count of successes is a binomial
random variable.
• Is it true that the count of failure is also a
binomial random variable?
• If it is, what are the parameters of its distribution?
The Binomial Distribution
• If X ~ B(n,p), then
n x
n!
n x
x
n x


P( X  x)    p (1  p) 
p (1  p)
x!(n  x)!
 x
• Examples: Let n = 3.
 n   3
3!
3  2 1
x  0 :      

1
 x   0  0!3! (1)(3  2 1)
 n   3
3!
3  2 1
x  1 :      

3
 x   1  1!2! (1)( 2 1)
Developing Binomial Probabilities for
n=3
S1 p
p
1-p
p
1-p
F1
1-p
S2
p
S3
F2 1-p
p
S2 1-p
p
1-p
p
F3
S3
F3
S3
F3
S3
F2 1-p
F3
P(SSS) = p3
P(SSF) = p2(1 – p)
P(SFS) = p2(1 – p)
P(SFF) = p(1 – p)2
P(FSS) = p2(1 – p)
P(FSF) = p(1 – p)2
P(FFS) = p(1 – p)2
P(FFF) = (1 – p)3
Binomial Probabilities for n = 3
• Let X be the number of successes in three trials.
P(X = 0) = (1 – p)3
P(X = 1) = 3p(1 – p) 2
P(X = 2) = 3p2(1 – p)
P(X = 3) = p3
P(FFF) = (1 – p)3
X=0
P(SSF) = p2(1 – p)
P(SFS) = p2(1 – p)
X=1
P(SFF) = p(1 – p)2
P(FSS) = p2(1 – p)
X=2
P(FSF) = p(1 – p)2
P(FFS) = p(1 – p)2
X=3
P(SSS) = p3
Example: Rolling a Die
• Roll a die 4 times, let X be the number of times the
number 5 appears.
• “Success” = get a roll of 5, so P(Success) = 1/6.
X=0
X=1
X=2
X=3
X=4
4!
P( X  0) 
(1 / 6)0 (1  1 / 6) 40  0.4823
0!(4  0)!
4!
P( X  1) 
(1 / 6)1 (1  1 / 6) 41  0.3858
1!(4  1)!
4!
(1 / 6) 2 (1  1 / 6) 42  0.1157
2!(4  2)!
4!
P( X  3) 
(1 / 6)3 (1  1 / 6) 43  0.0154
3!(4  3)!
P( X  2) 
P( X  3) 
4!
(1 / 6) 4 (1  1 / 6) 40  0.0008
4!(4  0)!
Example: Rolling a Die
• Find the probability that we get at least 2 rolls of
5.
P( X  2)  1  P( X  2)
 1  P( X  0)  P( X  1)
 1  0.4823  0.3858
 0.1319
Example: Flying
• Suppose an airline operates a daily shuttle service
from Altoona to Hoboken.
– Two round-trip flights, one on a plane with two
engines, the other on a plane with four engines.
– Each engine on each plane fails independently with
probability p.
– Each plane arrives safely only if at least half of its
engines remain in working order.
• For what values of p would you prefer to fly in
the two-engine plane?