Probability Flashcards
(27 cards)
What is the probability of an event A occurring in terms of the total success outcomes for A and the total outcomes in general?
P(A) = number of success outcomes for A / total outcomes in general
Probability of obtaining heads when flipping a fair coin
P(H) = 0.5
Union from i = 1 to infinity for all events A
The event that at least one of the events A1, A2,…, occurs
Intersection from i = 1 to infinity for all events A
The event that all of the events A1, A2,…, occurs
The Kolmogorov’s axioms of modern probability
For a sample size S, with event A contained within this, we have:
P(A) is greater than, or equal to, 0
P(S) = 1
If a countable list of pairwise disjoint events exists, then the probability of at least one of these events occurring is equal to the sum of each individual event occurring.
Simple propositions following the Kolmogorov’s axioms
Probability that no events occur is 0
If A and B are disjoint events, then the probability that either event occurs is equal to the sum of P(A) and P(B)
Probability of an event A not occurring is equal to 1 - P(A)
P(A) is equal to, or lesser than, 1
If the event A is contained within an event B, then P(A) is equal to or lesser than P(B)
Probability of an event A, or an event B occurring is equal to the sum of their individual events occurring, minus the event of both occurring at once.
When we have equally likely events, the probability of each event occurring is simply 1/n where n denotes the total amount of events. If an event contains several others, then the probability of this event occurring is k/n where k denotes the number of events that occur when the main event does.
Probability of drawing a heart from a deck of cards is 13/52, as each of the 13 hearts has a 1/52 chance of being drawn. Here k=13, n=52.
How many ways can 9 names be rearranged on a list?
There are 9! = 362, 880 possible arrangements.
If we take permutations of n objects taken r at a time, what equation do we use?
n! / (n-r)!
Suppose we want to choose r objects from among n, without regard to their order. What equation do we use here?
n! / [(n-r)! * r!]
A set of snooker balls consists of 15 red balls and 6 coloured balls. How
many ways can they be arranged in a row?
21!/15!
What is the probability of having a bridge hand with only spades and clubs
(including at least one of each suits)?
0.0000164
What is the probability of having a bridge hand with exactly two suits?
0.0000984
What is the probability that a bridge hand is void in at least one suit?
[4(39 choose 13) - 6(26 choose 13) +4]/(52 choose 13)
Binomial Theorem
for any real x,y and positive integer n
(x+y)^n
Sum from k=0 to n of (n choose k) * x^k * y^(n-k)
Suppose I draw a card from a shuffled pack and, without replacing it, draw
a second. If the first card is an ace, what is the probability that the second is also an
ace?
Since there are now only three aces left in the pack, the probability that the second
card is an ace is 3
51 . Clearly, the probability of the second event has been modified by
the knowledge that the first has occurred
The conditional probability of an event A, given that event B has occurred,
is denoted by P (A|B), and defined by
P (A|B) = P (A ∩ B) / P (B)
provided that P(B) is non zero
Two events A and B are stochastically independent if and only if
P (A ∩ B) = P (A)P (B)
A finite set of events {A1, . . . , An} are pairwise independent if every pair of
events is independent, that is
P (Ai ∩ Aj ) = P (Ai) P (Aj ) when i̸ = j
A finite set of events {A1, . . . , An} are mutually independent if and only if for
every k ≤ n and for every k-element subset of events {B1, . . . , Bk} of {A1, . . . , An}
Probability of the intersection of k subevents of B occurring is equal to the product of the probability of individual events.
The events B1, B2, . . . Bn form a partition of the sample space S if and only if
Probability of each subevent B is non zero
Union of all subevents B add up to 1
Events are pairwise disjoint.
Suppose the events B1, B2, . . . Bn form a partition of S. Then for any event A ⊆ S
P(A) = sum up to n for the product of P(A|Bi) * P(Bi)
A dice is thrown twice. What is the probability that the second throw is
greater than the first?
15/36
Let A ⊆ S, with P (A) not equal to 0, and let B1, B2, . . . Bn be a partition of S. Then
P(Bj) given that A has occurred is equal to P(A) given that Bj has occurred, times P(Bj) occurring, divided by the partition of A into n parts of B
