Probability Quick Reference

TL;DR: A quick reference guide for key probability formulae and expressions including permutations, combinations, Bayes' theorem, and statistical measures.
Vinay 5 min read
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Disclaimer: This post is meant as a quick reference guide only, for key formulae and expressions. It is not an exhaustive list by any measure and can be referenced from other sources [1].

Permutations

Main formulae:

nPr=n!(n−r)!nP_r = \frac{n!}{(n-r)!}

For all similar ‘n’ elements:

nPn=n!nP_n = n!

Combinations

Main formulae:

nCr or (nr)=n!(n−r)!r!=nPrr!{nC_r} \text{ or } {n \choose r} = \frac{n!}{(n-r)!r!} = \frac{nP_r}{r!}

Example:

(42)=4!2!2!=4⋅3⋅2⋅12⋅2=6{4 \choose 2} = \frac{4!}{2!2!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 2} = 6

Known as binomial coefficient - found in the theorem stated below.

Binomial theorem

(a+b)n=∑r=0n(nr)an−rbr(a+b)^n = \sum_{r=0}^n {n \choose r} a^{n - r} b^r

Example:

(a+b)4=∑r=04(4r)a4−rbr(a+b)^4 = \sum_{r=0}^4 {4 \choose r} a^{4 - r} b^r =(40)a4+(41)a3b1+(42)a2b2+(43)a1b3+(44)b4= {4 \choose 0} a^4 + {4 \choose 1} a^3 b^1 + {4 \choose 2} a^2 b^2 + {4 \choose 3} a^1 b^3 + {4 \choose 4} b^4 =a4+4a3b+6a2b2+4ab3+b4= a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4

If there are two or more independent ways a combination occurs in a problem; total number of combs would typically be arrived by multiplying all possible combinations of individual combs in alignment with fundamental counting principle.

Probability

  • Probability of an event A is how likely it is for A to occur.
  • Probability scale is between 0 and 1
    • if P(A)=0P(A) = 0 → event will not occur
    • if P(A)=1P(A) = 1 → event ‘A’ will occur
    • Use Venn diagram for visualization
      • if ‘A’ is the event, A′A' called A-prime or Aˉ\bar{A} called A-complement is everything other than ‘A’
    • 0≤P(A)≤10 \leq P(A) \leq 1
  • Probability of entire sample space is 1
    • P(S)=1P(S) = 1
    • i.e., probability of entire space of all possible outcomes of all experiments should be 1
    • hence; P(A′)=1−P(A)P(A') = 1 - P(A)
  • If there are ‘n’ equally likely outcomes of an experiment, of which one is called a success ‘s’, then the probability of a success is:
    • P(A)=number of ways "A" can occurtotal number of outcomes possibleP(A) = \frac{\text{number of ways "A" can occur}}{\text{total number of outcomes possible}}
    • Hence, the probability can be arrived at:
    P(A)=SnP(A) = \frac{S}{n}
    • Event ‘A’ is my success

Addition Rules of Probability

  • If there are two events happening and there is an intersection then;
    • e.g.: Pulling ace and club from deck of cards - has intersection
    • e.g.: Flipping coin and getting heads and tails - no intersection
    • Intersection - P(A⋂B)P(A \bigcap B) = Probability of A and B
    • Union - P(A⋃B)P(A \bigcup B) = Probability of A or B or both
  • If A and B do not intersect → mutually exclusive
  • Hence, probability of A or B or both:
P(A⋃B)=P(A)+P(B)−P(A⋂B)P(A \bigcup B) = P(A) + P(B) - P(A \bigcap B)

Conditional Probability

Hence, probability of A and B:

P(A⋂B)=P(A)⋅P(B∣A)P(A \bigcap B) = P(A) \cdot P(B \mid A)
  • P(B∣A)P(B \mid A) means probability of B given that A has occurred
  • Event A and B are independent events if ‘A’ does not influence ‘B’
    • i.e., P(B∣A)=P(B)P(B \mid A) = P(B)
    • hence, P(A⋂B)=P(A)⋅P(B)P(A \bigcap B) = P(A) \cdot P(B)
  • If you are unsure of generating this directly, look at arriving at P(A′)P(A') and then working backward as; P(A)=1−P(A′)P(A) = 1−P(A')

Bayes’ Theorem

Rule of elimination: If B1,B2,B3...BnB_1, B_2, B_3 ... B_n are mutually exclusive events of which 1 must occur,

P(A)=∑i=1n[P(Bi)⋅P(A∣Bi)]P(A) = \sum_{i=1}^n [P(B_i) \cdot P(A \mid B_i)]

Derivation of Bayes’ theorem (in context of three-path problem where we are trying to find everything that can happen out of path 3):

P(B3∣A)⇒P(A⋂B3)=P(A)⋅P(B3∣A)P(B_3 \mid A) \Rightarrow P(A \bigcap B_3) = P(A) \cdot P(B_3 \mid A) P(B3∣A)=P(A⋂B3)P(A)P(B_3 \mid A) = \frac{P(A \bigcap B_3)}{P(A)} P(B3∣A)=P(B3)⋅P(A∣B3)∑i=1nP(Bi)⋅P(A∣Bi)P(B_3 \mid A) = \frac{P(B_3) \cdot P(A \mid B_3)}{\sum_{i=1}^n P(B_i) \cdot P(A \mid B_i)}

Essentially translates to: P(B3∣A)=Path we are interested inProbability that something happens at all along all pathsP(B_3 \mid A) = \frac{\text{Path we are interested in}}{\text{Probability that something happens at all along all paths}}

Bayes’ theorem: If B1,B2,B3...BnB_1, B_2, B_3 ... B_n are mutually exclusive events, then

P(Br∣A)=P(Br)⋅P(A∣Br)∑i=1nP(Bi)⋅P(A∣Bi)P(B_r \mid A) = \frac{P(B_r) \cdot P(A \mid B_r)}{\sum_{i=1}^n P(B_i) \cdot P(A \mid B_i)}

Mathematical Expectation

If the probability of obtaining the amounts a1,a2,a3,...,aka_1, a_2, a_3, ..., a_k are p1,p2,p3,...,pkp_1, p_2, p_3, ..., p_k, then mathematical expectation is:

E=a1p1+a2p2+a3p3...akpk\mathbf{E} = a_1 p_1 + a_2 p_2 + a_3 p_3 ... a_k p_k

Mean, Median and Mode

  • A data set is a collection of measurements (e.g. grades in a class)
  • A population is all available values of some data (typically data set is a sub-set of population)
  • Mean value known as average values is arrived at by adding all values and dividing by number of samples:
X‾=∑Xn\overline X = \frac{\sum X}{n}
  • Median is if you arrange the samples in increasing order and you pick the middle one
    • if you encounter an even number of samples, take the middle two and get its average
  • Mode is the value in the data set that occurs most frequently
    • there can be no mode, 1 mode, and if two or more values are tied in frequency they all would be mode
  • Midrange is exactly midway between highest and lowest values
    • arrived at by looking at highest and lowest values and taking their average

References

[1] Statistics and probability resources - Khan Academy