Math 461 Homework

In how many ways can 3 boys and 3 girls sit in a row if no two people of the same sex are allowed to sit together?

2*3!3! = 72 You can have either of the patterns bgbgbg, gbgbgb. For each of these patterns, you can independently permute the set of boys, and the set of girls. # boys = 3, so 3! # girls = 3, so 3! and the factor of 2 comes from the fact that there are two patterns.

If N people, including A and B, are randomly arranged in a circle, what is the probability that A and B are next to each other?

2/(N-1)

A pair of dice is rolled until a sum of either 5 or 7 appears. Find the probability that a 5 occurs first.

2/5

A committee of 7, consisting of 2 Republicans, 2 Democrats, and 3 Independents, is to be chosen from a group of 5 Republicans, 6 Democrats, and 4 Independents. How many committees are possible?

(5 choose 2) * (6 choose 2) * (4 choose 3) Product of: Ways of choosing 2 republicans from 5 Ways of choosing 2 democrats from 6 Ways of choosing 3 independents from 4

A person has 8 friends, of whom 5 will be invited to a party. How many choices if 2 of the friends will only attend together?

(6 choose 3) + (6 choose 5) = 26

What is the probability of being dealt one pair (5 card hand)?

((13 choose 1)*(4 choose 2)*(12 choose 3)*(4 choose 1)^3) / (52 choose 5)

What is the probability of being dealt a full house (5 card hand)?

(13 choose 1)(4 choose 3)(12 choose 1)(4 choose 2)/(52 choose 5)

What is the probability of being dealt four of a kind (5 card hand)?

(13 choose 1)*(12 choose 1)*(4 choose 1)/(52 choose 5)

What is the probability of being dealt three of a kind (5 card hand)?

(13 choose 1)*(4 choose 3)*(12 choose 2)*((4 choose 1)^2)/(52 choose 5)

What is the probability of being dealt two pairs (5 card hand)?

(13 choose 2)*((4 choose 2)^2)*(11 choose 1)*(4 choose 1) / (52 choose 5)

A group of 2N men and 2N women is divided into two equal groups. Find the probability p that each group will be equally divided into men and women.

(2N choose N)^2/(4N choose 2N)

In how many ways can 3 boys and 3 girls sit in a row?

(3 + 3)! = 6!

Two cards are randomly selected from an ordinary playing deck. What is the probability that they form a blackjack?

(4 choose 1)*(4*4 choose 1) / (52 choose 2)

A person has 8 friends, of whom 5 will be invited to a party. How many choices are there if 2 of the friends, in a relationship, will only attend together?

(6 choose 3) + (6 choose 5) = 26. Break the problem up into two cases. Case i: You choose to invite the couple. Then you have 8-2 friends remaining, and you still need to choose 3, so (6 choose 3). Case ii: You choose not to invite the couple. Then you have 8-2 friends remaining, and you still need to choose 5, so (6 choose 5). The OUTCOMES in this model are disjoint, and FORM A PARTITION of the output space, so take the sum to obtain (6 choose 3) + (6 choose 5) = 26.

A person has 8 friends, of whom 5 will be invited to a party. How many choices are there if 2 of the friends are feuding and will not attend together?

(7 choose 5) + (7 choose 5) - (6 choose 5) = 36. Let F={a,...,h} be the set of eight friends. Assume WLOG that a and b are the feuding friends. Now create two sets A = F\{a} and B=F\{b}, both with seven friends. You can choose five friends from A, or five friends from B, and in either case, you have (7 choose 5) options. So take the sum of both. Finally, because you have over counted ( since A intersect B = {c,...,h} ), you need to subtract the choice of 5 friends from the intersection, which has 8-2=6 elements. Putting everything together, you get (7 choose 5) + (7 choose 5) - (6 choose 5) = 21 + 21 - 6 = 42-6 = 36. Break up the friends into two groups ----------------------------------- Also (8 choose 5) - (6 choose 3) = 36.

How many (north and east only) paths are there from (0,0) to (a,b) in a (typical) lattice?

(a+b choose a) = (a+b choose b).

Binomial(n,p)[k]

(n choose k)(p)^{k}(1-p)^{n-k}

A pair of dice are rolled, what is the probability that the second die lands on a higher value than the first?

(⚀, ⚀), (⚀, ⚁), (⚀, ⚂), (⚀, ⚃), (⚀, ⚄), (⚀, ⚅) (⚁, ⚀), (⚁, ⚁), (⚁, ⚂), (⚁, ⚃), (⚁, ⚄), (⚁, ⚅) (⚂, ⚀), (⚂, ⚁), (⚂, ⚂), (⚂, ⚃), (⚂, ⚄), (⚂, ⚅) (⚃, ⚀), (⚃, ⚁), (⚃, ⚂), (⚃, ⚃), (⚃, ⚄), (⚃, ⚅) (⚄, ⚀), (⚄, ⚁), (⚄, ⚂), (⚄, ⚃), (⚄, ⚄), (⚄, ⚅) (⚅, ⚀), (⚅, ⚁), (⚅, ⚂), (⚅, ⚃), (⚅, ⚄), (⚅, ⚅) Every pair above the main diagonal has the desired property, so count and divide by the total (36). i.e. 15/36 = 5/12

If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is i? Compute for all values of i between 2 and 12.

0 if i≤6 1/(13-i) if i > 6

A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the the coins at random; when he flips it, it shows heads. What is the probability that it is the fair coin?

1/3

A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the the coins at random; when he flips it, it shows heads. He flips the same coin a second time, and it shows heads again. Now what is the probability that it is the fair coin?

1/5

What is the probability of being dealt a straight (5 card hand)?

10*(4 choose 1)^5/(52 choose 5)

How many different letter arrangements can be made from the letters: MISSISSIPPI

11! / (4!*4!*2!) Perm(MISSISSIPPI) divided by the product of Perm(SSSS), Perm(IIII), and Perm(PP).

In how many ways can 3 boys and 3 girls sit in a row if the boys and girls are each to sit together?

2*3!3! = 72 You can have either of the patterns bbbggg, gggbbb. For each of these patterns, you can independently permute the set of boys, and the set of girls. # boys = 3, so 3! # girls = 3, so 3! and the factor of 2 comes from the fact that there are two patterns.

Consider a group of 20 people. If everyone shakes hands with everyone else, how many handshakes are possible?

20 choose 2 = 190. This is combinatorially obvious since handshakes are a 2 person activity, and we want to know how many ways we can 'choose' 2 people, of the 20, to shake hands. It's also clear that the graph of this situation is complete, and therefore has 20(20-1)/2 = 190 edges (handshakes).

Twenty workers are to be assigned to 20 different jobs, one to each job. How many different assignments are possible?

20!

Suppose that 5 percent of men and 0.25 percent of women are color blind. A color blind person is chosen at random. What is the probability of this person being male?

20/21.

How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers if no letter or number can be repeated in a single license plate.

26(25)*(10)(9)(8)(7)(6)

How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers?

26^2 * 10^5

What is the probability of being dealt a royal flush (5 card hand)?

4 ways to get a royal flush, so 4/(52 choose 5)

In how many ways can 3 boys and 3 girls sit in a row if only the boys must sit together?

4!3! (Permutations of the group of boys) multiplied by (Permutations of the group of girls with the boys as one unit).

What is the probability of being dealt a flush (5 card hand)?

4*(13 choose 5) / (52 choose 5)

Suppose that 5 percent of men and 0.25 percent of women are color blind. A color blind person is chosen at random. What is the probability of this person being male if the population consists of twice as many males as females?

40/41

How many different letter arrangements can be made from the letters: FLUKE

5! Perm(Five distinct elements)

Thrm: P(A | B) = P(A) if ?

A and B are independent.

If 8 new teachers are to be divided among 4 schools how many divisions are possible?

Assume that the teachers are choosing the school where they want to teach. Then each of the eight teachers has four choices. There is no restriction on the number of teachers which can or must work at any school, so we simply need to multiply these individual choices. Specifically, 4 choose 1 eight times, so our solution is 4^8 = 2^16 = 65,536.

Suppose that n balls are randomly distributed into N compartments. Find the probability that m balls will fall into the first compartment. Assume that all Nn arrangements are equally likely.

Binomial(P=1/N, n) P_X(m)(n choose m)*(1/N^n)(N-1)^{n-m}

If 8 identical blackboards are to be divided among 4 schools, how many divisions are possible?

Denote the four schools by s1, s2, s3, and s4. Each school can have between 0 and 8 blackboards, so we want to count the number of non-negative solutions to the following equation b_1 + b_2 + b_3 + b_4 = 8. (8 + 4 - 1) choose (4-1) = (11 choose 3) = 165.

If 8 identical blackboards are to be divided among 4 schools, how many divisions are possible if each school must receive at least 1 blackboard?

Each school can have between 1 and 8 blackboards, so we want to count the number of positive integer solutions to the following equation b_1 + b_2 + b_3 + b_4 = 8 (8-1 choose 4-1) = (7 choose 3) = 35.

Independent flips of a coin that lands on heads with probability p are made. What is the probability that the first four outcomes are H,H,H,H?

Flips are independent, so P(HHHH) = P(H)^4

Independent flips of a coin that lands on heads with probability p are made. What is the probability that the first four outcomes are T,H,H,H?

Flips are independent, so P(THHH) = P(T)P(H)^3

Bayes' Theorem

If S is a probability space and {A_1...A_n} forms a partition of S and B is some event in S, then for any i in 1..n: P(A_i) = P(A_i)*P(B | A_i) ------------------------- ∑_{k=1}^n P(B|A_k)P(A_k)

Law of Total Probability

If S is a probability space, and {A_1...A_n} forms a partition of S, and B is some event in S, then P(B) = ∑_{k=1}^n P(B|A_k)P(A_k)

P(A ∩ B) =

P(A | B)*P(B)

P(A|B)

P(A ∩ B) / P(B)

Events A1...Ak are independent

P(A_i...A_k) = P(A_i)*...*P(A_k) for all subsets {A_i...A_k} of 2 or more elements of {A1...Ak}.

Inclusion-exclusion identity

P(E_1 union E_2 union ... union E_n) = ∑{i = 1 .. n} P(E_i) - ∑{i < j} P(E_j E_i) + ∑{i < j< k} P(E_j E_i E_k) ...

How many (north and east only) paths are there from (0,0) to (a,b) in a (typical) lattice if you have to make a stop at some point (c,d) where 0<c<a and 0<d<b?

Take the product of the number of NE-paths from (0,0) to (c,d), and the number of NE-paths from (c,d) to (a,b). So (c+d choose c) * (a-c + b-d choose a-c) = etc...

Events E and F are independent

The probability of the product is equal to the product of the probabilities P(EF)=P(E)P(F)

If 8 new teachers are to be divided among 4 schools, and each school must receive 2 teachers, how many divisions are possible?

There are (8 choose 2) ways for the first school to pick two teachers. Then the second school must pick two from the remaining six, then the third school picks two from four, and the last school gets the remaining two. We can multiply these quantities together to obtain our solution (8 choose 2) *(6 choose 2) * (4 choose 2) * (2 choose 2) = 2520.

Proposition 6.2 Counting distinct nonnegative integer-valued vectors.

There are (n+r-1 choose r-1) distinct nonnegative integer-valued vectors (x_1, x_2, ... , x_r) which satisfy the equation x_1 + x_2 + ... + x_r = n

Proposition 6.1 Counting distinct positive integer-valued vectors.

There are (n-1 choose r-1) distinct positive integer-valued vectors (x_1, x_2, ... , x_r) which satisfy the equation x_1 + x_2 + ... + x_r = n

Stirling's Formula

n! ~ √{2πn}(n/e)^n

Compute the probability that a bridge hand is void in at least one suit.

∑_{k=1..3}(4 choose k)*(-1)^{k+1} * (52-13k choose 13) all divided by (52 choose 13).