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2005 11 15
Homework
• Homework due now.
• Reading: relations
Counting, Using Permutations and
Combinations
PC13. In a competition among architects on emergency
housing, a panel votes on the four top designs (labeled poptop, cocoon, box, and dome). How many outcomes are
possible? (You must consider ties, including multi-way ties.)
Without ties, it’s P(4, 4)
With ties, there are five cases: Select Order
1) No ties:
P(4,4)
2) Two tie, other two don’t: C(4,2) * P(3,3)
3) Two sets of two tie:
C(4,2)
4) Three houses tie:
C(4,3) * P(2,2)
5) Four houses tie:
1
+
Binomial Coefficients
Pascal’s Triangle
Chu Shih-chieh’s Triangle?
Binomial Coefficients
Theorem:
Binomial Coefficients
BC1. Find the coefficients for the following terms
a. the fifth term of (2x + 3y)7
b. the coefficient of x4y7 in (x + y)11
c. the coefficient of a62b92 in (2a – b)154
C(11, 4) or
11
4
( )
Binomial Coefficients
BC6. Show that if p is a prime and k is an integer
p
such that 1  k  p-1 then p divides ( k ).
Binomial Coefficients
BC10. Prove that in any row of Pascal’s triangle,
the sum of the entries in the odd numbered
positions equals the sum of the entries in the
even numbered positions.
Beware! Coincidences Often Aren’t;
The Obvious Often Isn’t.
How many people must be in a room to make it likely
two or more of them have the same birthday?
385
How many people must be in a room to make it likely
at least one of them has the same birthday you do?
183
Monty Hall: N (>2) doors, one has what you want
behind it. You choose a door. Before opening it,
Monty opens one you didn’t choose to reveal it’s not
the one you want. Now, should you stay with the
door you selected, or switch? Or doesn’t it matter?
Doesn’t matter
Discrete Probability
• p(E) = |E| / |S|
– Events, E, equally likely; sample space, S, finite.
• If two dice are rolled once, what is the probability their sum is odd?
– (Outcomes: 2, 4, 6, 8, 10, 12;
3, 5, 7, 9, 11)
• Using counting technique to find probabilities
– Problem DP11 demonstrates well
• # of 5-card hands that have four of the same kind
• # of 5-card hands that are a full house
• Thm: p(E) = 1 - p(E)
• Thm: p(E1 E2) = p(E1) + p(E2) – p(E1 E2)
Sum of Two Dice
If two dice are rolled once, what is the probability their
sum is odd? (Since the smallest sum is two and the largest
is twelve, perhaps there are more even sums?)
Odd: (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6),
(4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)
Even: (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5),
(4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)
p(E) = |E| / |S|
=
18 / (18 + 18) =
1/2
Cards
“Suit”

“Kind”
Ace
2
3
4
5
6
7
8
9
10
Jack
Queen
King



Discrete Probability
DP7. What is the probability that a card dealt from the deck is
a. a spade or a heart?
b. a face card?
c. a jack or a spade?
16 cards satisfy: “jack or spade”, so probability is:
C(16,1) / C(52,1)
Alternatively.
4/52 + 13/52 – 1/52
Discrete Probability
DP8. Assume you are dealt a 13 card hand. What is the
probability that you have
a. the ace of spades?
b. at least two aces?
c. exactly two aces?
There are C(4,2) ways to choose the two aces.
The number of ways to choose the remaining three cards
from 48 is C(48,3), so the probability is:
C(4,2) C(48,3) / C(52,5)
where C(52,5) is the number of all five card hands.
Flush
DP11. Compute the probability of each of the following
types of poker hands:
a. flush
The number of ways to create a flush is:
C(4,1) * C(13,5)
So the requested probability is:
C(4,1) * C(13,5) / C(52,5)
Flush – an alluring incorrect answer
DP11. Compute the probability of each of the following
types of poker hands:
a. flush
What about:
C(52,1) * C(12,4) / C(52,5) ???
Choose 4 more to match its suit
Choose a first card
Answer is off by a factor equal to the number of cards in the hand.
Why?
Five Card Hand with Four Same Rank
DP11. Compute the probability of each of the following
types of poker hands:
d. 4 of a kind
Choose a kind: C(13,1)
Choose four out of four of a kind: C(4,4)
Any fifth card: C(48,1)
Number of possible five card hands: C(52,5)
Desired probability: C(13,1) * C(4,4) * C(48,1) ~ 0.00024
C(52,5)
Full House
DP11. Compute the probability of each of the following types
of poker hands:
f. full house (3 of a kind + one pair)
Arrange 2 out of 13 kinds: P(13,2)
Choose 3 out of four of the first kind: C(4,3)
Choose 2 out of four of second kind: C(4,2)
Desired probability: P(13,2) * C(4,3) * C(4,2) ~ 0.0014
C(52,5)
Lottery
DP15. To play the Big Bang lottery, the contestant guesses 7 numbers
out of the integers from 1 to 77, no repetitions allowed. To win, the set
of guessed numbers must be a subset of the 12 integers from the same
range drawn by a lottery official. (Assume on each draw, any number
in the range has an equal probability of being selected.)
a. What is the probability that a contestant who plays once wins the
Big Bang lottery?
Number of ways to select 7 of 12 numbers:
divided by number of ways to select 7:
- OR Number of ways for lottery to select 12 when drawing our 7:
divided by number of ways to select 12:
C(12,7)
C(77,7)
C(70,5)
C(80,12)
Defining Independence
Two events, E1 and E2, are called independent if
p(E1  E2) = p(E1) * p(E2).
It’s very important to know whether two events are
independent.
- If events are treated as independent, and they’re
not, it can seriously degrade the quality of any
analysis that assumes they are.
Analysis in which events can be assumed
independent is often simplified considerably.
Evaluating Independence
DP18. Let E1, E2, … En be events in the sample space S. For each of
the following propositions, restate it in English, say whether it is
true or false, and give an argument supporting your answer.
a. E1  E2  P(E1)  P(E2)
b. P(E1 E1) = 1
c. If E1, E2, and E3 are independent events, then P(E1  E2  E3) = P(E1) P(E2) P(E3).
d. P(E1  E2) = P(E1) + P(E2)
e. P(E1  E2)  P(E1) + P(E2)
E1  E2  P(E1)  P(E2)
TRUE.
Why?
E1  E2  |E1|  |E2|  |E1| / |S|  |E2| / |S|  P(E1)  P(E2)
Evaluating Independence
DP18. Let E1, E2, … En be events in the sample space S. For each of
the following propositions, restate it in English, say whether it is
true or false, and give an argument supporting your answer.
a. E1  E2  P(E1)  P(E2)
b. P(E1 E1) = 1
c. If E1, E2, and E3 are independent events, then P(E1  E2  E3) = P(E1) P(E2) P(E3).
d. P(E1  E2) = P(E1) + P(E2)
e. P(E1  E2)  P(E1) + P(E2)
P(E1  E2)  P(E1) + P(E2)
TRUE.
|E1  E2|  |E1| + |E2|
 |E1  E2| / |S|  |E1| / |S| + |E2| / |S|
 P(E1  E2)  P(E1) + P(E2)
Why?
by def of 
In a room with 82
people in it, what are
the odds two or more
of them have the same
birthday?
In a room with 82
people in it, what are
the odds one of them
has the same birthday
you do?
- different question!
Look here
???
Stick, or switch?
X
Monty Hall with N Doors
DP17. Generalize the Monty Hall problem to n  4 doors. After
you’ve selected a door and Monty opens one with a valueless
prize behind it, should you switch doors if offered the
opportunity? Compute the probability that you win if you
stick with your original choice and the probability that you
win if you switch to another unopened door.
¼ chance of winning initially, and if you do not change.
Thus, ¾ chance it’s one of the other doors. Since there are only
two equally likely choices, chances if you change are 3/8.
Beware! Coincidences Often Aren’t;
The obvious often isn’t.
How many people must be in a room to make it likely
two or more of them have the same birthday?
385
How many people must be in a room to make it likely
at least one of them has the same birthday you do?
183
Monty Hall: N (>2) doors, one has what you want
behind it. You choose a door. Before opening it,
Monty opens one you didn’t choose to reveal it’s not
the one you want. Now, should you stay with the
door you selected, or switch? Or doesn’t it matter?
Doesn’t matter
An Empirical Test of Our Theory
Everybody write down a number: 1, 2, or 3
Loop until my point is made
Student a: Tell us your number (pick a door)
Student b: Tell us a number that is not the number
“a” chose, and not the number you wrote down (open
a door that is not the one “a” selected and not the one with the prize)
Student a: Pick the remaining number (switch!)
Student b: Is this the number you wrote down?
yes -> switching paid off
no -> didn’t pay off
Transition
Match Some
DP16. Consider a lottery in which the contestant selects 6
numbers from among 40 possible, and 6 numbers are also
chosen randomly by the state. What is the probability that
exactly 5 (i.e. not 6) of the numbers chosen by the contestant
match 5 of the 6 chosen by the state?
Ways to select 5 out of 6: C(6,5)
Ways to select remaining number:
* C(34,1)
divided by number of ways to select 6:
C(40,6)