Probability

Question 0

Permutations without repetition

Answer 0

(n!)/(n-r)!

Question 1

Permutation with repetition

Answer 1

n^r

Question 2

Combination with repetition

Answer 2

nCr = n!/[r!(n-r)!]

Question 3

Combinations with repetition

Answer 3

(n+r-1)!/[r!(n-1)!]

Question 4

A college planning committee consists of 3 freshman, 4 sophomores, 5 juniors, and 2 seniors. A subcommittee of 4, consisting of 1 person from each class, is to be chosen. How many different subcommittees are possible?

Answer 4

3x4x5x2 = 120 (by generalized version of the basic principle)

Question 5

Ms. Jones has 10 books that she is going to put on her bookshelf. Of these, 4 are mathematics books, 3 are chemistry, 2 are history, and 1 is language. Ms. Jones wants to arrange her books so that all the books dealing with the same subjects are together on the shelf. How many different arrangements are possible?

Answer 5

There are 4! possible arrangements of the SUBJECTS, and there are 4!3!2!1! possible arrangements PER SUBJECT, so it’s 4!4!3!2!1! = 6912.

Question 6

How many different letter arrangements can be formed from the letters PEPPER?

Answer 6

First, there are 6! permutations of the letters. However, since there are 2 E’s, and 3 P’s that repeat, 3!2! permutations are the same. Therefore, 6!/(2!3!) = 60 possible letter arrangements.

Question 7

From a group of 5 women and 7 men, how many different committees consisting of 2 women and 3 men can be formed? What if 2 of the men are feuding and refuse to serve on the committee together?

Answer 7

a) There are (5C2)(7C3) = 350 possible choices.
b) There is a total of (2C2)(5C1) = 5 out of the (7C3) = 35 possible groups of 3 men that consist of the two feuding men. So 35-5 = 30 possible choices for men, times (5C2) for women, so 30x10 = 300 choices.

Question 8

Consider a set of n antennas of which m are defective and n-m are functional and assume that all of the defectives and all of the functionals are indistinguishable. How many linear orderings are there in which no two defectives are consecutive?

Answer 8

In the n-m+1 possible positions → NYNY—NYNY→ we must place m antennas between n-m functional antennas. Hence, there are (n-m+1Cm) possible orderings.

Question 9

Useful combinatorial identity: nCr =?

Answer 9

nCr = (n-1)C(r-1) + (n-1)C(r) 1≤ r ≤ n

Question 10

Binomial Theorem

Answer 10

(x+y)ⁿ = ∑{i=0 to n}(nCi)x^i y^(n-i)

Question 11

A set of n distinct items is to be divided into r DISTINCT groups of respective sizes n₁, n₂,…, n_r, where ∑{i=1 to r}n_i = n. How many different versions are possible?

Answer 11

n!/(n₁! n₂!…n_r!) = nC(n₁, n₂,…,n_r)

Question 12

Ten children are to be divided into an A team and a B team of 5 each. How many different divisions are possible?

Answer 12

10!/(5!5!) = 252 possible divisions

Question 13

In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?

Answer 13

NOTE that there is no A team or B team, but just a division consisting of 2 groups of 5 each. Therefore, the answer is: [10!/(5!5!)]/2!

Question 14

What are multinomial coefficients?

Answer 14

nC(n₁, n₂,…,n_r)

Question 15

What are the three axioms of probability?

Answer 15

Axiom 1: 0≤ P(E) ≤ 1
Axiom 2: P(S) = 1
Axiom 3: for any sequence of mutually exclusive events E_1, E_2,…,
P(U{i=1 to ∞} E_i) = ∑{i=1 to ∞}P(E_i)

Question 16

Express the probability of the union of two events in terms of the individual probabilities

Answer 16

P(EUF) = P(E) + P(F) - P(EF)

Question 17

What is the inclusion-exclusion identity?

Answer 17

P(E_1 U E_2 U … U E_n) = ∑{i=1 to n}P(E-i) - ∑{i_1<i_r}P(Ei_1Ei_2…Ei_r)+…+(-1)^(n+1)P(E_1E_2…E_n)

Question 18

What is P(E) if we assume that all outcomes of an experiment are equally likely to occur?

Answer 18

P(E) = (# of outcomes in E)/(# of outcomes in S)

Question 19

3 balls are randomly drawn from a bowl containing 6 white and 5 black balls. What is the probability that one of the ball is white and the other two is black?

Answer 19

P(E) = [(3C1)(6x5x4)]/(11x10x9) = 4/11
Or
P(E) = (6C1)(5C2)/(11C3) = 4/11

Question 20

An urn contains n balls, one of which is special. If k of these balls are withdrawn one by one, what is the probability that the special ball is chosen?

Answer 20

P(E) = [(1C1)((n-1)C(k-1))]/(nCk) = k/n

Question 21

A poker hand consists of 5 cards. If the cards have distinct consecutive values and are not all the same suit, then we have a straight. What is the probability of being dealt a straight?

Answer 21

Let’s look at the probability of getting ace, 2,3,4, and 5. Since there are 4 of each, we have 4^5 possibilities. And since we want different suits, we get (4^5 - 4) possibilities. There are 10 ways to make a straight (1-5,2-6,3-7…10-1) so 10(4^5 - 4) are all the possible ways to make a straight out of 52C5 possibilities. So P(E) = 10(4^5-4)/(52C5)