Probability Flashcards
Permutations without repetition
(n!)/(n-r)!
Permutation with repetition
n^r
Combination with repetition
nCr = n!/[r!(n-r)!]
Combinations with repetition
(n+r-1)!/[r!(n-1)!]
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?
3x4x5x2 = 120 (by generalized version of the basic principle)
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?
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.
How many different letter arrangements can be formed from the letters PEPPER?
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.
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?
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.
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?
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.
Useful combinatorial identity: nCr =?
nCr = (n-1)C(r-1) + (n-1)C(r) 1≤ r ≤ n
Binomial Theorem
(x+y)ⁿ = ∑{i=0 to n}(nCi)x^i y^(n-i)
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?
n!/(n₁! n₂!…n_r!) = nC(n₁, n₂,…,n_r)
Ten children are to be divided into an A team and a B team of 5 each. How many different divisions are possible?
10!/(5!5!) = 252 possible divisions
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?
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!
What are multinomial coefficients?
nC(n₁, n₂,…,n_r)
What are the three axioms of probability?
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)
Express the probability of the union of two events in terms of the individual probabilities
P(EUF) = P(E) + P(F) - P(EF)
What is the inclusion-exclusion identity?
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)
What is P(E) if we assume that all outcomes of an experiment are equally likely to occur?
P(E) = (# of outcomes in E)/(# of outcomes in S)
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?
P(E) = [(3C1)(6x5x4)]/(11x10x9) = 4/11
Or
P(E) = (6C1)(5C2)/(11C3) = 4/11
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?
P(E) = [(1C1)((n-1)C(k-1))]/(nCk) = k/n
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?
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)
