Random Variables and Probability Distributions,Probability Mass Function,Discrete Probability Distribution

Random Variables and Probability Distributions

Random Variables

A random variable has defined as a numerical (or real) valued function defined over the sample points in a sample space S of a random experiment.

In simple terms, a variable whose numerical values are associated with or determined by the outcomes of a random experiment is called a random variable (in short r.v.).

Types of Random Variables

There are two types of random variable:

i) Discrete Random variable

A random variable which can take only a finite number or countably infinite number of values is called a discrete random variable.In simple terms, a discrete random variable is numerical valued function defined on a discrete sample space.The natural1,2,3....are the countably infinite numbers.

Examples

a)Total number or sum of the numbers shown in throwing two fair dice simultaneously

b) the number of heads turned up in tossing two coins together

c) the number of members in each family in a certain village( family size)

ii) Continuous Random variable

A random variable is said to be a continuous random variable if it assumes any (or all possible) value (s) within a certain interval (a,b) say, where a and b may be either any real number R or \((- \infty , + \infty \))

b) Income,expenditure, import, export , rainfall, temperature etc.

Distinction between discrete and continuous random variables:

Some fundamental theorems on random variables are stated without proof as follows:

1.) If X is a random variable, then any function f(x) is also random variable viz. , \(f(x)=X^2\)

2. ) If X is a random variable and C is a constant, then CX is also a random variable.

3.) If X and Y are two random variables, then X+Y,X-Y,XY and \(\frac {X}{Y}\) are also random variables.

Probability Mass Function

Let X be a discrete random variable , then a functionp(x)=P(X=x) is called probability mass function of r.v. X , if for each possible numerical value x, p(x) satisfies the following properties or conditions:

$$ i. p(x) \geq 0, \; \; \; \; \; ii. \sum_{x} p(x)=1$$

Discrete Probability Distribution

The set of ordered pairs {x,p(x)} for each possible values of discrete random variables X is called discrete probability distribution of the random variable X.Thus, a set of the different values, \(x_i\) of a discret random variables X together with their respective probabilities \(p_i (i=1,2,...)\) gives the probability distribution of the random variable X.

In a probability distribution , the total probability 1 is distributed among the various possible values assumed by the random variable.

The probability distribution of a discrete random variable X can be presented graphically also which helps us to obtain a probability curve.There are two methods of representation a discrete probability distribution : (i) line chart or probability chart and (ii) probability histogram .

Example

Three bulbs are randomly selected without replacement from a lot of 12 bulbs of which 2 bulbs are defective. Find the probability distribution of the number of defective bulbs selected.Also find the probability that the bulbs selected contains (I) exactly 1 defective bulb (ii) at least 1 defective bulbs.

Solution:

Let X denotes the number of defective bulbs in the 3 bulbs selected. Since a lot of 12 bulbs contains 2 defective bulbs, the random variable X can take the values 1, 1, and 2. Thus,

$$P(X=0)=P(No \; defective \; bulbs)$$

$$=P( All \; 3 \; non-defective \; bulbs)$$

$$=\frac {^{10} C_3}{^{12}C_3}=\frac{12}{22}$$

$$P(X=1)=P(1 \; defective \; and \; 2 \; non-defective \; bulbs)$$

$$ =\frac {^2C_1 \times ^{10}C_2}{^{12}C_3}=\frac{9}{22}$$

$$P(X=2)=P( 2 \; defective \; and \; 1\; non-defective \; bulbs)$$

$$=\frac {^2C_2 \times ^{10}C_1}{^{12}C_3}=\frac{1}{22}$$

Hence, the probability distribution of X is as follows:

The pobability histogram of the above probability distribution is given below:

Also,

(i) The probability that the 3 bulbs selected contains exactly 1 defective bulb is given by P(X=1)=\(\frac{9}{22}\).

(ii) The probability of selecting at least 1 defective bulbs is given by

$$P(at \; least \; 1 \; defective \; bulb)=1-P(No \; defective \; bulb)$$

$$ i.e. \; P(X \geq 1)=1-P(X=0)=1-\frac{12}{22}=\frac{5}{11}$$

References

Sukubhattu,Narendra Prasad. Probability and Inference-I. Asmita Books Publishers & Distributors (P)Ltd. 2013