Coupon Accepted Successfully!


Addition Theorem of Probability

If A and B be any two events in a sample space S, then the probability of occurrence of at least one of the events A and B is given by P(A  B) = P(A) + P(B) – P(A  B).
If AB, and C are any three events in a sample space S, then
P(A  B  C) = P(A) + P(B) + P(C) – P(A  B)
– P(B  C) – P(A  C) + P(A  B  C)
If A and B are mutually exclusive, events then A  B = φ and hence P(A  B) = 0.
∴ P(A  B) = P(A) + P(B)
If ABC are mutually exclusive events, then
P(A  B  C) = P(A) + P(B) + P(C)

Conditional probability

Let A and B be any two events, B ≠ φ, then P(A/B) denotes the conditional probability of occurrence of event A when B has already occurred which is given by
64834.png or P(A  B) = P(B) P(A/B)
If A and B are independent events, then
Two events A and B are independent if and only if P(A  B) = P(A) P(B).
If AB are independent events, then A′ and B′ are independent events. Hence P(A′  B′) = P(A′) P(B′)


  • P(A ∪ B) = 1 – P(A′) . P(B′) [valid only when A and B are independent].
  • If A1A2, …, An are independent events, then P(A1 ∪ A2 ∪ … ∪ An) = 1 – P(A1) ⋅P(A2) … P(An).
  • If A and B are independent events, then
    • A and B′ are independent events.
    • A′ and B are independent events.
    • A′ and B′ are independent events.
  • If A and B are two events such that B ≠ φ, then P(A/B) + P(A′/B) = 1.
  • If A and B be two events such that A ≠ φ, then P(B) = P(A) ⋅ P(B/A) + P(A′) ⋅ P(B/A′).

Total probability theorem

Let “A” be any event of S. We can write A = (A1  A (A2  A) …  (An  A)
Let A1A2, ..., An are mutually exclusive, then (A1  A), (A2  A), …, (An  A) would also be mutually exclusive.
P(A) = P(A1  A) + P(A2  A) + ... + P(An  A)
P(A1) P(A/A1) + P(A2) P(A/A2) + ...
P(An) P(A/An)
P(A) = 64822.png
This is known as the total probability of event A. Here P(A/Ai) gives up the contribution of Ai in the occurrence of A.

Bayes’ theorem

If A1A2A3, …, An be n mutually exclusive and exhaustive events and A is an event which occurs together (in conjunction) with either of Ai i.e., if A1A2, …, An form a partition of the sample space S and A be any event, then

Probability distribution

Let S be the sample space associated with a given random experiment. Then a random variable is a real-valued function whose domain is subset of sample space of the experiment. If the experiment random variable assumes (takes) the values 0, 1, 2, …, n
x x1 x2 x3 xn
P(x) p1 p2 p3 pn

Binomial distribution

A probability distribution representing the binomial trials is said to binomial distribution. Let us consider a binomial experiment which has been repeated “n” times. Let the probability of success and failure in any trial be p and q respectively in these n trials. Now number of ways of choosing “r” success in “n” trials = nCr. Probability of “r” success and (n-r) failures is pr qnr. Thus probability of having exactly r successes = nCr pr qnr.
Let “X” be random variable representing the number of successes, then P(X = r) = nCr  pr  qnr (r = 0, 1, 2, …, n).
  • Probability of atmost “r” successes in n trials
  • Probability of atleast “r’ successes in n trials
  • Probability of having first success at the rth trial
    p ⋅ qr–1
    Thus if X be a random variable, X takes the values 0, 1, 2, …, n said to be BD. Its probability distribution or probability function is given by P(X = r) = nCrprqnrr = 0, 1, 2, …, npq > 0 and p + q = 1.

Poisson distribution

It is the limiting case of BD under the following conditions:
  1. Number of trials are very-very large, i.e. n  ∞
  2. p  0
  3. nq  λ, a finite quantity (λ is called parameter)
    Probability of r success for Poisson distribution is given by P(X = r) = 64788.pngr = 0, 1, 2, …
    For Poisson distribution recurrence formula is given by P(r + 1) = 64782.png
  • For Poisson distribution, mean = variance = λ = np.
  • If X and Y are independent Poisson variates with parameters λ1 and λ2 then XY also has Poisson distribution with parameter λ1 + λ2.

Mean of BD

Mean of BD of the variable X is given by
64776.png = E(X) = 64770.png = 64764.png
64758.png = E(X) = 64752.pngnp

Variance of BD

Var(X) = 64746.png


Mean = np, variance = npq
  • Mean – variance = np(1 – q) > 0 as 0 ≤ q ≤ 1 Thus mean > variance
  • SD = 65832.png

Mode of BD

The value of the variable (X) which occurs with the maximum (largest) probability.
Recurrence relation (formula) for BD is denoted by P(r +1) and defined as
P(r +1) = 64734.pngr  {0, 1, 2, …, n – 1}

Test Your Skills Now!
Take a Quiz now
Reviewer Name