# 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

*A*,*B*, 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

*A*,*B*,*C*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 or

*P*(*A*∩*B*) =*P*(*B*) ⋅*P*(*A*/*B*)If

*A*and*B*are independent events, thenTwo events

*A*and*B*are independent if and only if*P*(*A*∩*B*) =*P*(*A*) ⋅*P*(*B*).If

*A*,*B*are independent events, then*A*′ and*B*′ are independent events. Hence*P*(*A*′ ∩*B*′) =*P*(*A*′) ⋅*P*(*B*′)

*Notes:**P*(*A*∪*B*) = 1 –*P*(*A*′) .*P*(*B*′) [valid only when*A*and*B*are independent].- If
*A*_{1},*A*_{2}, …,*A*are independent events, then_{n}*P*(*A*_{1}∪*A*_{2}∪ … ∪*A*) = 1 –_{n}*P*(*A*′_{1}) ⋅*P*(*A*′_{2}) …*P*(*A*′)._{n} - 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*= (*A*_{1}∩*A*) ∪ (*A*_{2}∩*A*) … ∪ (*A*∩_{n}*A*)Let

*A*_{1},*A*_{2}, ...,*A*are mutually exclusive, then (_{n}*A*_{1 }∩*A*), (*A*_{2}∩*A*), …, (*A*∩_{n}*A*) would also be mutually exclusive.⇒

*P*(*A*) =*P*(*A*_{1}∩*A*) +*P*(*A*_{2}∩*A*) + ... +*P*(*A*∩_{n}*A*)=

*P*(*A*_{1}) ⋅*P*(*A*/*A*_{1}) +*P*(*A*_{2}) ⋅*P*(*A*/*A*_{2}) + ...+

*P*(*A*) ⋅_{n}*P*(*A*/*A*)_{n}⇒

*P*(*A*) =This is known as the total probability of event

*A*. Here*P*(*A*/*A*) gives up the contribution of_{i}*A*_{i}_{ }in the occurrence of*A*.# Bayes’ theorem

If

*A*_{1},*A*_{2},*A*_{3}, …,*A*be_{n}*n*mutually exclusive and exhaustive events and*A*is an event which occurs together (in conjunction) with either of*A*i.e., if_{i}*A*_{1},*A*_{2}, …,*A*form a partition of the sample space_{n}*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 | x_{1} |
x_{2} |
x_{3 }… |
x_{n} |

P(x) |
p_{1} |
p_{2} |
p_{3 }… |
p_{n} |

# 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 =*. Probability of “*^{n}C_{r}*r*” success and (*n*-*r*) failures is*p*⋅^{r}*q*^{n}^{–r}. Thus probability of having exactly*r*successes =*⋅*^{n}C_{r}*p*⋅^{r}*q*^{n}^{–r}.Let “

*X*” be random variable representing the number of successes, then*P*(*X*=*r*) =*⋅*^{n}C_{r}*p*⋅^{r}*q*^{n}^{–r}(*r*= 0, 1, 2, …,*n*).

*Notes:*- Probability of atmost “
*r*” successes in*n*trials - Probability of atleast “
*r*’ successes in*n*trials - Probability of having first success at the
*r*th trial*p*⋅*q*^{r}^{–1}*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*) =^{n}C_{r}p^{r}q^{n}^{–r},*r*= 0, 1, 2, …,*n*;*p*,*q*> 0 and*p*+*q*= 1.

# Poisson distribution

It is the limiting case of BD under the following conditions:

- Number of trials are very-very large, i.e.
*n*→ ∞ *p*→ 0*nq*→*λ*, a finite quantity (*λ*is called parameter)*r*success for Poisson distribution is given by*P*(*X*=*r*) = ,*r*= 0, 1, 2, …*P*(*r*+ 1) =

*Notes:*- For Poisson distribution, mean = variance =
*λ*=*np.* - If
*X*and*Y*are independent Poisson variates with parameters*λ*_{1}and*λ*_{2}then*X*+*Y*also has Poisson distribution with parameter*λ*_{1}+*λ*_{2}.

# Mean of BD

Mean of BD of the variable

*X*is given by =

*E*(*X*) = = =

*E*(*X*) = =*np*# Variance of BD

Var(

*X*) =

*Notes:*Mean =

*np*, variance =*npq*- Mean – variance =
*np*(1 –*q*) > 0 as 0 ≤*q*≤ 1 Thus mean > variance - SD =

# 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) = ,

*r*∈ {0, 1, 2, …,

*n*– 1}