Random Variable

A Random variable is a real valued function defined over the sample space of a random experiment.  The values of a random variables correspond to the outcomes of the random experiment.

Discrete and  Continuous random variable

A random variable is  discrete if it takes only specific values in an interval.  When X takes values 1, 2, 3, 4, 5, 6  it is a discrete variable.

A random variable is  continuous if it takes any value in a given interval.  When  X takes values  in the given interval (a,b), it is continuous variable in that interval.

Probability distribution

Let  X be a random variable which takes values x.
The probability that the random variable X takes the value x is defined as the probability distribution of X.  It is denoted by f(x)

Properties of probability distribution
Let  f(x) be a probability distribution.  Then
1.  f(x) ≥ 0, for all values of x.
2. Σ f(x) = 1.

The important discrete probability distributions are Binomial probability distribution, Poisson probability distribution, Geometric probability distribution.

Binomial Probability distribution formula

The Binomial probability formula is given by nCrprqn-r where p represents the probability of success and q represents the probability of failure.

Example : -
Probability that a batsman scores a century in a cricket match is 1/3.  Find the probability that out of 4 matches, he may score century
(i) in exactly 3 matches
(ii) in one of the matches
Solution: -
Here "success" is denoted by "scoring century"
Given probability that a batsman scores a century in a cricket match is 1/3. That is p = 1/3.
"Failure" is denoted by "not scoring century". We know that
q = 1 - p = 1 -  1/3 = 2/3.
Total number if matches n = 4.
Binomial probability formula is given by nCrprqn-r
(i) We have to find the probability that he scores century in exactly three matches.
That is r = 3.
P(scoring century in exactly 3 matches)
= 5C3(1/3)3(2/3)5-3
= 5C3(1/3)3(2/3)2
= 10 * (1/27)*(4/9)
= 40/243
P(scoring century in exactly 3 matches) = 40/243
(ii) We have to find the probability that he scores century in one of the matches.
That is r = 1
P(scoring century in one of the matches)
= 5C1(1/3)1(2/3)5-1
=5C1(1/3)1(2/3)4
= 5 * 1/3 *(16/81)
= 80/243
P(scoring century in one of the matches) = 80/243

Poisson Probability distribution formula

The Poisson probability formula is given by mx e-m/x! where m is the mean of the Poisson distribution.

Poisson distribution may be expected in cases where the chances of happening of any individual event is small.  So the distribution is used to describe the behavior of rare events.  Therefore it is also called law of improbable events.

Example: -
It is known that in a certain plant there are on an average 4 industrial accidents per year.  Find the probability that in a given year there will be less than 3 accidents.
Solution: -
Given that in a certain plant there are on an average 4 industrial accidents per year.  That is m = 4 .
We have to find that probability that in a given year there will be less than 3 accidents. So x can take values 0, 1, or 2.
We know that the Poisson probability formula is given by mx e-m/x! where m is the mean of the Poisson distribution.
So required probability is
P(x < 3) = P(x = 0) + P(x = 1) + P(x = 2)
              = 40 e-4/0! + 41 e-4/1! + 42 e-4/2!
              = 0.2382
Therefore probability that in a given year there will be less than
3 accidents =0.2382

Geometric Probability distribution formula

The Geometric probability formula is given by qk-1p where p represents the probability of success on each trial.

Example: -
To start his old mower, Tom has to pull a cord and hope for some luck.  On any particular pull, the mower has a 20% chance of starting.
Find the probability that it takes him exactly 3 pulls to start the mower.
Solution: -
We have to find the probability that it takes him exactly 3 pulls to start the mower.  That is P(k = 3)
We know that the Geometric probability formula is given by qk-1p where p represents the probability of success on each trial.
Given on any particular pull, the mower has a 20% chance of starting.  That is p = 0.20
q = 1 - p = 1- 0.20 = 0.80
So required probability is
P(k = 3) = 0.803-1 x 0.20
              =0.802 x 0.20
              = 0.128
Therefore probability that it takes him exactly 3 pulls to start the mower = 0.128