# Midterm Review, Math 210G-07, Spring 2009 - PowerPoint

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```					Midterm Review Solutions
Math 210G-04, Spring 2011
following:

B) In a town, one barber cannot stay in business
but two barbers can
C) If every man except the barber shaves himself,
who shaves the barber?
D) If every man except the barber shaves
himself, how can the barber stay in business?
Zeno’s paradox of achilles and the
tortoise asserts that
A) In a race between Achilles and the tortoise,
Achilles can never pass the tortoise
B) In a race between Achilles and the tortoise,
Achilles will never win
C) In a race between Achilles and the tortoise,
the race ends as soon as Achilles and the
tortoise are tied
D) Achilles is afraid to race the tortoise
A) There cannot be a set that contains all other
sets
B) A set that contains only apples cannot contain
any oranges
C) A set cannot contain itself
D) This sentence is false
(NB: both A and C are correct in this case)
A tautology is a statement that is
A) Always true
B) Always tight
C) redundant
The Pythagoreans were divided
into “mathematikoi” and
“akousmatikoi”
• A) True
• B) False
Dice are descendents of
•   A) pottery
•   B) rubiks cube
•   C) dodecahedrons
•   D) buckminsterfullereen
•   E) bones
A fair coin is defined as one that
has a probability of ½ of coming up
• A) True
• B) False
The statement
((pq)∧p)q is
A) Always true
B) True only when p is true
C) True only if q is true
D) True only if both p and q are true
In the Monte Hall problem, if a car is behind one
of three doors and, after you choose one door
Monte shows a goat behind one of the others,
then you should always switch to the other
door.
• A) True
• B) False
What is the intended conclusion of the
following
• All writers, who understand human nature,
are clever. (W -> C)
• No one is a true poet unless he can stir the
hearts of men. (~S -> ~ P)
• Shakespeare wrote “Hamlet”. (SH->WH)
• No writer, who does not understand human
nature, can stir the hearts of men. (~W->~S)
• None but a true poet could have written
“Hamlet”. (~P->~WH)
Solution:
The statements are W -> C; ~S -> ~ P; WH->SH;
~W->~S; and ~P->~WH. The symbols C and SH
each only appear once so:
~C->~W->~S->~P->~WH->~SH or SH->C.
The intended conclusion is Shakespeare was
clever.
Determine the intended implication of the following collection of
statements:
• (a) No interesting poems are unpopular among people of real taste.
• (b) No modern poetry is free from affectation.
• (c) All your poems are on the subject of soap-bubbles.
• (d) No affected poetry is popular among people of real taste.
• (e) No ancient poem is on the subject of soap-bubbles.
Solution
• (a) ~P->~I (b) ~A->~M (c) Y->S (d) P->~A (e)
~M->~S
• Altogether,
• Y->S->M->A->~P->~I
• Your poems are not interesting
Explain why at least one of the
following pictures proves the
Pythagorean theorem
Solution

• The picture on the right says that
(a+b)*(a+b)=c*c+2*a*b or that a*a+b*b=c*c
which is the same as the Pythagorean
theorem.
Propose a solution to the following
problem:
A king decides to give 100 of his prisoners a test. If they pass,
they can go free. Otherwise, the king will execute all of them.
The test goes as follows: the prisoners stand in a line, all
facing forward. The king puts either a black or a white hat on
each prisoner. The prisoners can only see the colors of the
hats in front of them. Then, in any order they want, each one
guesses the color of the hat on their head. Other than that,
the prisoners can not speak. To pass, no more than 1 of them
may guess incorrectly. If they can make their strategy before
hand, how can they be assured that they will survive?
Solution
• The last prisoner has 99 prisoners in front.
Since 99 is odd, either the number of black
hats or of white hats in front is even in
number, but not both. The prisoner at back
states the color of the hats in front that are
even in color. Now Prisoner 99 can infer the
color of his hat, from this , prisoner 98 can
infer the color of his, etc.
Compute the probabilities of the
following outcomes for rolling a pair of
dice
•   The sum of the dice is odd
•   The sum of the dice is larger than 6
•   The sum of the dice is less than 1
•   The sum of the dice is less than 6
Solution
• The probability that the sum of the dice is
even is ½
• The probability that the sum is larger than 6 is
21/36=7/12
• The probability that the sum is less than one is
zero.
• The probability that the sum is less than 6 is
10/36=5/18.
Bayes rule states that

Find P(A|B) if P(B|A)=1/2, P(A)=1/3 and
P(B)=2/3.
SOLUTION: P(A|B) =(1/2)*(1/3)/(2/3)=1/4
Bayes rule, part II
•   Professor Doolittle conducted a survey and discovered that 80% of the students
who got an A on his exam studied for it for 4 or more hours.
•   He also learned that half of his students studied for 4 or more hours for the exam.
•   20 % of Doctor Doolittle’s students get an A on the exam.
•   Use Bayes’ rule to determine the probability that someone who studies for 4 or
more hours the day before will get an A on his exam.

•   SOLUTION: P(S|A)=0.8; P(S)=1/2; P(A)=0.2 so
•   P(A|S)=P(S|A)*P(A)/P(S)=0.8*0.2/(1/2)=8/25
Fill in the next row of Pascal’s triangle
1
1         2         1
1        3         3         1
1       4         6         4        1
1       5        10        10        5       1
1         6       15        20        15       6       1
How many ways are there to
choose a subset of 4 elements
from a set of six elements?
•   A) 30
•   B) 15
•   C) 4
•   D) 6
•   E) None of the above
Poker hands
• Compute the number of possible ways of
choosing 5 cards from a deck of 52 cards.
• Compute the number of possible ways of
getting 4 of a kind in a five card poker hand.
Explain your result and its probability of
happening.
Solution
• 52 choose 5 equals 52*51*50*49*48 /
(5*4*3*2) =2,598,960
• 4 of a kind: 13 (ranks) * 48 (remaining choices)
= 624
• Probability of 4 of a kind is
624/2598960=0.00024… or about 24 out of
100,000
3 card guts
• Three card cuts is a version of poker in which
each player gets three cards. Straights and
flushes are not allowed. The best possible
hand is three aces.
• What is the number of distinct hands in 3 card
guts?
• How many of these hands allow three of a
kind?
• How many allow a pair (but not three of a
kind)?
• How many allow no pairs?
Solution
• The number of 3 card guts hands is 52 choose 3, or
52*51*50/(3*2*1)=17*26*50=22,100.
• For each given rank (e.g. “aces” there are “4 choose 3” ways
of getting three of a kind of that rank, ie 4. There are 13 ranks.
Multiply these to get 52 ways of getting 3 of a kind.
• There are 4 choose 2 or six ways of getting a pair in each rank,
there are 48 choices for the third card of a different rank.
Multiply these times the 13 ranks to get 6*48*13=3744 guts
hands having a pair but not three of a kind. There is less than
a one in five chance of getting a pair or better.
• The remaining hands have no pairs: subtract 3744+52 from
22,100 to get 18304.
Find the average of the following
numbers and their standard deviation
• The variance of numbers x1 ,…, xNis the sum of
the squares of their differences from their
mean, divided by N-1.
• The sample deviation is the square root of the
variance.
• The numbers are: 72, 66, 70, 54, 60, 78, 72,
64, 66, 56, 82
Solution
• Their mean is
• M=(72+66+70+54+60+78+72+64+66+56+82)/
11 =67.27
• Their variance is
• V= [(72-67.27)2 +(55-67.27)2+…+(82-
67.27)2]/10
• =75.41

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