midterm review solutions math 210g-04, spring 2011

Midterm Review SolutionsMath 210G-04, Spring 2011

The barbers paradox asks the following:

Zeno’s paradox of achilles and the tortoise asserts that

Russell’s paradox implies that

A tautology is a statement that is

The Pythagoreans were divided into “mathematikoi” and

“akousmatikoi”

Dice are descendents of

A fair coin is defined as one that has a probability of ½ of coming up

heads

The statement ((pq)∧p)q is

What is the intended conclusion of the following

Solution:

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

Explain why at least one of the following pictures proves the

Pythagorean theorem

• 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?

• 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

• 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

Bayes rule, part II

Fill in the next row of Pascal’s triangle

How many ways are there to choose a subset of 4 elements

from a set of six elements?

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.

• 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?

• 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

• 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