5. Random variable and its probability distribution

Random Variables and its Probability Distributions

Random Variables:

Definition :  A random variable is a real valued function whose domain is the sample space of a random experiment.

Example:  let us consider the experiment of tossing a coin two times in succession.

The sample space of the experiment is S = {HH, HT, TH, TT}.

If X denotes the number of heads obtained, then X is a random variable and for each outcome, its value is as given below :

X(HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0.

That means  X=  0    1     2  

Example : A person plays a game of tossing a coin thrice. For each head, he is given Rs 2 by the organizer of the game and for each tail, he has to give Rs 1.50 to the organiser. Let X denote the amount gained or lost by the person. Show that X is a random variable and exhibit it as a function on the sample space of the experiment.

Solution:  X is a number whose values are defined on the outcomes of a random experiment. Therefore, X is a random variable.

Now, sample space of the experiment is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Then X(HHH) = Rs (2 × 3) = Rs 6

X(HHT) = X(HTH) = X(THH) = Rs (2 × 2 −1 × 1.50) = Rs 2.50

X(HTT) = X(THT) = (TTH) = Rs (1 × 2) – (2 × 1.50) = – Re 1

and X(TTT) = Rs (3 × 1.50) =-Rs 4.50

minus means loss and plus means gain.

X= {– 1, 2.50, – 4.50, 6}

Probability Distribution of the random variable X

Definition : The probability distribution of a random variable X is the system of numbers

                    X :     x1          x2           x3 ...    xn

               P(X) :    p(x1)    p(x2)   p(x3)... p(xn)

Example : Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of aces.

Solution: The number of aces is a random variable. Let it be denoted by X. Clearly, X can take the values 0, 1, or 2.

Now, since the draws are done with replacement, therefore, the two draws form independent experiments.

P(ace)= 4/52 = 1 / 13

P(non ace)= 48/52 = 12 / 13

Therefore,

Example : Let X denote the number of hours you study during a randomly selected school day. The probability that X can take the values x, has the following form, where k is some unknown constant.

(a) Find the value of k.

(b) What is the probability that you study at least two hours ?

      Exactly two hours?

      At most two hours?

Solution: The probability distribution of X is

   X:    0      1        2        3         4

P(X) 0.1      k        2k      2k       k

(a) We know that

 =>   0.1 + k + 2k + 2k + k = 1

=>  k = 0.15

(b) P(you study at least two hours) = P(X ³2)

= P(X = 2) + P (X = 3) + P (X = 4)

= 2k + 2k + k = 5k = 5 × 0.15 = 0.75

P(you study exactly two hours) = P(X = 2)

= 2k = 2 × 0.15 = 0.3

P(you study at most two hours) = P(X £2)

= P (X = 0) + P(X = 1) + P(X = 2)

= 0.1 + k + 2k = 0.1 + 3k = 0.1 + 3 × 0.15

= 0.55