Section

(iii) Number of elements in inverse images are shown in the table. Values of the Random Variable Total Number of elements in inverse image Example . Suppose a pair of unbiased dice is rolled once. If X denotes the total score of two dice, write down (i) the sample space (ii) the values taken by the random variable X , (iii) the inverse image of , and (iv) the number of elements in inverse image of X .

(i) The sample space S = , , , , , , , , , , consists of ordered pairs where α and β can take any integer value between and as shown. X is assigned to each point (α, β) the sum of the numbers on the dice . That is X Therefore X , = X = X X = X X X = X X X X = X X X X X = X X X X X TT TH HT HH Sample space Real line S X  A mapping X (.) from S to Fig. .

S X = X X X X X = X X X X = X X X = X = . (ii) Then the random variable X takes on the values , , , , , , , , , , . (iii) The inverse images of is , , (iv) The number of inverse images are given below Values of the random variable Total Number of elements in inverse image Example . An urn contains white balls and red balls.

A sample of balls are chosen at random from the urn. If X denotes the number of red balls chosen, find the values taken by the random variable X and its number of inverse images. Let us denote white and red balls as w w r r , , ,and The sample space consists of c = different samples of size . That is S w w r w w r w w r w rr w r r w rr w rr w r r , w rr rr r The random variable X takes on the values , , and .

A mapping X (.) from S to real numbers Fig. . Values of the Random Variable X Total Number of elements in inverse images Remark If X denotes the number of white balls, then X takes on the values , , and and the elements in inverse images are X= X= X= X= X= X= X= X= X= X= S w w r w r r w r r w r r w r r w r r r w r w w r w w r r r r Real line Sample space S Probability Distributions Values of the Random Variable X Total Number of elements in inverse images Illustration . A batch of students is taken in buses to an excursion.

There are students in the first bus, in second bus, in the third bus, and the remaining students in the fourth bus. When the buses arrive at the destination, one of the students is randomly chosen. Suppose that X denotes the number of students on the bus of that randomly chosen student. Then X takes on the values , , , and .

Example . Two balls are chosen randomly from an urn containing white and black balls. Suppose that we win ` for each black ball selected and we lose ` for each white ball selected. If X denotes the winning amount, find the values of X and number of points in its inverse images.

The possible events of selection are (i) both balls may be black, or (ii) one white and one black or (iii) both are white. Therefore X is a random variable that take the values, X (both are black balls) = ` ( ) = ` X (one black and one white ball) = ` − ` = ` X (both are white balls) = ` ( − ) ` Therefore X takes on the values , , and − . Note : The inverse image of is b b b b b b b b b b b b Values of the Random Variable X – Total Number of elements in inverse images Illustration . A coin is tossed until head occurs.

The sample space is S H TH TTH TTTH ,  . Suppose X denotes the number of times the coin is tossed until head occurs. Then the random variable X takes on the values , , ,  Illustration . Suppose N is the number of customers in the queue that arrive at a service desk during a time period, then the sample space should be the set of non-negative integers.

That is S = { , , , , }  and N is a random variable that takes on the values , , , ,  Illustration . If an experiment consists in observing the lifetime of an electrical bulb, then a sample space would be the life time of electrical bulb. Therefore the sample space is S [ , . Suppose X denotes the lifetime of the bulb, then X is a random variable that takes on the values in , Illustration .

Let D be a disk of radius r . Suppose a point is chosen at random in D . Let X denote the distance of the point from the centre. Then the sample space S D and X is the random variable that takes on any number from to r .

That is X S ) . = Therefore the cumulative distribution function is Fig. . (ii) P X .

(iii) P X . Example . A six sided die is marked ‘ ’ on one face, ‘ ’ on two of its faces, and ‘ ’ on remaining three faces. The die is rolled twice.

If X denotes the total score in two throws. (i) Find the probability mass function. (ii) Find the cumulative distribution function. (iii) Find P X (iv) Find P X ≥ .

O F ( x ) Cumulative distribution function F ( x ) , , , F( x ) = Fig. . Solution: Since X denotes the total score in two throws, it takes on the values , , , , and . From the Sample space S , we have Values of the Random Variable Total Number of elements in inverse images = = , = = = = , = = , and = = .

(i) Probability mass function is (ii) Cumulative distribution function By definition of the cumulative distribution function for discrete random variable we have F x ( ) = P X x i i = for X . F ( ) = P X = Probability mass function O f ( x ) Fig. . Sample space S II I Probability Distributions F ( ) = P X = P X = Therefore the cumulative distribution function is for for for for for −∞< or ≤ < ∞               (iii) P X i = (iv) P X ≥ = x i = P X = Therefore = = (iii) P X = = f ( x ) Area .

. f ( x ) Area = = ∫ ) P x ≤ O f ( x ) Area = ( P O Fig. . Check: (i) Whether F x ( ) is continuous everywhere.

(ii) From the Fig. . , triangle area = bh . .

Probability density function from Probability distribution function. Let us learn the method to determine the probability density function f x ( ) from the distribution function F x ( ) of a continuous random variable X . Suppose F x ( ) is the distribution function of a continuous random variable X . Then the probability density function f x is given by dF x F x , wherever derivative exists.

Example . If X is the random variable with distribution function F x ( ) given by, F x then find (i) the probability density function f x ( ) (ii) P X ( . . ) £ £ O .

. . . F x Distribution function F x Fig.

. (i) Differentiating F x ( ) with respect to x at continuity points of f x ( ) , we get F x The pdf f x ( ) is not continuous at x = , or at x = . We can define f ( ) and f ( ) in any manner. Choosing f ( ) = , and f ( ) Therefore the probability density function f x ( ) is otherwise (ii) P X ( .

. ) . . Remark By definition, P X F x f u du if is discrete if is continuous The expected value is in general not a typical value that the random variable can take on.

It is often helpful to interpret the expected value of a random variable as the long-run average value of the variable over many independent repetitions of an experiment. Theorem . (Without proof) Suppose X is a random variable with probability mass ( or ) density function f x . The expected value of the function g X , a new random variable is E g X g x f x g x g x f x dx g x ( ( )) ∑ if is discrete if is continuous −∞ ∞       If g X x k ) = the above theorem yield the expected value called the k-th moment about the origin of the random variable X.

Therefore the k-th moment about the origin of the random variable X is E X x f x X x f x dx X k k k if is discrete if is continuous Probability Distributions Note When k = , by definition, E X X     ∑ −∞ ∞ if is discrete if is continuous   E aX = aE X ) + Similarly, when X is a continuous random variable, we can prove it, by replacing summation by integration. Corollary : E aX ) = aE X ) ( when b = ) Corollary : E b ( ) = b (when a = ) (ii) Var X E X E X − ( Proof We know E x ( ) = μ Var X ) = E X = E X X = E X E X (Since μ is a constant) = E X E X Var X ) = E X E X An alternative formula to compute variance of a random variable X is σ = Var ( X E X E X (iii) Var(aX +b) a Var X ( ) where a and b are constants Proof Var aX = E aX E aX = E aX aE X = E aX aE X = E a X E X = a E X E X ) . Hence Var aX = a Var X Corollary : V aX ) = a V X ( ) (when b = ) Corollary : V b ( ) = (when a = ) Variance gives information about the deviation of the values of the random variable about the mean μ . A smaller σ implies that the random values are more clustered about the mean, similarly, a bigger σ implies that the random values are more scattered from the mean.

Fig. . µ µ Deviation from mean Deviation from mean Different variance with same mean Smaller Variance Bigger variance Probability Distributions The above figure shows the pdfs of two continuous random variables whose curves are bell-shaped with same mean but different variances. Example .

Suppose that f x ( ) given below represents a probability mass function, c c c c c Find (i) the value of c (ii) Mean and variance. (i) Since f x ( ) is a probability mass function, f x ( ) ≥ for all x , and . Thus, = x p x q px qx pq x The mean and the variance are respectively px qx and pq x (b) Symmetrical Case: When p = , the two point distribution becomes for for and the cumulative distribution function is F x if if if The mean and variance respectively are x and x Let p denote the probability of success on a single trial. Then, by using the binomial expansion with a and , we see that the sum of the probabilities for a binomial random variable is .

Since each trial in the experiment is classified into two outcomes, {success, failure}, the distribution is called a “bi’’-nomial. If X is a binomial random variable which follows binomial distribution with parameters and , the mean μ and variance σ are np and np The expected value is in general not a typical value that the random variable can take on. It is often helpful to interpret the expected value of a random variable as the long-run average value of the variable over many independent repetitions of an experiment. The shape of a binomial distribution is symmetrical when p = .

or when n is large. When p = , the binomial distribution becomes n x , , ,..., . Probability Distributions That is , , ,..., . The mean and variance are respectively are n and n Example .

Find the binomial distribution for each of the following. (i) Five fair coins are tossed once and X denotes the number of heads. (ii) A fair die is rolled times and X denotes the number of times appeared. (i) Given that five fair coins are tossed once.

Since the coins are fair coins the probability of getting an head in a single coin is p = and q Let X denote the number of heads that appear in five coins. X is a binomial random variable that takes on the values , , , , and , with n and .That is X B  Therefore the binomial distribution is ( ) = n x p n x , , ,..., becomes ( ) = , , ,..., . That is ( ) = , , ,..., (ii) A fair die is rolled ten times and X denotes the number of times appeared. X is binomial random variable that takes on the values , , , ,  , with n = and p = .

That is X B  Probability of getting a four in a die is p = and q . Therefore the binomial distribution is ( ) = , , ,..., Example . A multiple choice examination has ten questions, each question has four distractors with exactly one correct answer. Suppose a student answers by guessing and if X denotes the number of correct answers, find (i) binomial distribution (ii) probability that the student will get seven correct answers (iii) the probability of getting at least one correct answer.

(i) Since X denotes the number of success, X can take the values , , ,..., The probability for success is p = and for failure q , and n = . Therefore X follows the binomial distribution X B  . This gives, ( ) = , , ,..., (ii) Probability for seven correct answers is = = f ( ) Probability that the student will get seven correct answers is . (iii) Probability for at least one correct answer is ≥ = = .

Probability that the student will get for at least one correct answer is . Example . The mean and variance of a binomial variate X are respectively and . .

Find (i) P X = (ii) P X = (iii) P X ≥ To find the probabilities, the values of the parameters n and p must be known. Given that Mean = np = and variance = npq . This gives npq np = . = q = and p np = , gives n .

Therefore X B  . Therefore the binomial distribution is = f x , . ,... (i) = = f ( ) Probability Distributions (ii) = = f ( ) (iii) ≥ = Example .

On the average, % of the products manufactured by ABC Company are found to be defective. If we select of these products at random and X denotes the number of defective products find the probability that (i) two products are defective (ii) at most one product is defective (iii) at least two products are defective. Given that n = Probability for selecting a defective product is , that is p = . Since X denotes the number defective products, X can take on the values , , ,..., The probability for defective (success) is p = and for failure q , and n = Therefore X follows the binomial distribution denoted by X B  This gives ( ) = , , ,..., .

(i) Probability for two defective products is = = f ( ) (ii) Probability for at most one defective product is ≤ = P X = = Probability for at most one defective product is . (iii) Probability for at least two defective products is ≥ = Probability for at least two defective products is .