Mathematics Random Variables , Probability distribution, Mean and Variance of a random variable
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Class 12 Chapter 13 - PROBABILITY

Random Variables , Probability distribution, Mean and Variance of a random variable

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`color{red} ♦` Random Variables and its Probability Distributions
`color{red} ♦` Probability distribution of a random variable
`color{red} ♦` Mean of a random variable
`color{red} ♦` Variance of a random variable

Random Variables and its Probability Distributions

`=>` We have already learnt about random experiments and formation of sample spaces. In most of these experiments, we were not only interested in the particular outcome that occurs but rather in some number associated with that outcomes as shown in following examples/experiments.

(i) In tossing two dice, we may be interested in the sum of the numbers on the two dice.

(ii) In tossing a coin 50 times, we may want the number of heads obtained.

(iii) In the experiment of taking out four articles (one after the other) at random from a lot of 20 articles in which 6 are defective, we want to know the number of defectives in the sample of four and not in the particular sequence of defective and nondefective articles.

`=>` In all of the above experiments, we have a rule which assigns to each outcome of the experiment a single real number. This single real number may vary with different outcomes of the experiment. Hence, it is a variable. Also its value depends upon the outcome of a random experiment and, hence, is called random variable. A random variable is usually denoted by X.

`=>` If you recall the definition of a function, you will realise that the random variable X is really speaking a function whose domain is the set of outcomes (or sample space) of a random experiment. A random variable can take any real value, therefore, its co-domain is the set of real numbers. Hence, a random variable can be defined as follows :

`color {blue} "Definition 4"`

`=>` A random variable is a real valued function whose domain is the sample space of a random experiment.

`=>` For 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"}`.

I`=>` f 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`.

`=>` More than one random variables can be defined on the same sample space. For example, let Y denote the number of heads minus the number of tails for each outcome of the above sample space S.

`=>` Then `color{orange} {Y(HH) = 2, Y (HT) = 0, Y (TH) = 0, Y ("TT") = – 2}`.

`=>` Thus, X and Y are two different random variables defined on the same sample space S.
Q 3187691587

A person plays a game of tossing a coin thrice. For each head, he is
given Rs 2 by the organiser 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.
Class 12 Chapter 13 Example 22
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
where, minus sign shows the loss to the player. Thus, for each element of the sample
space, X takes a unique value, hence, X is a function on the sample space whose range
is
{–1, 2.50, – 4.50, 6}
Q 3107691588

A bag contains 2 white and 1 red balls. One ball is drawn at random and
then put back in the box after noting its colour. The process is repeated again. If X
denotes the number of red balls recorded in the two draws, describe X.
Class 12 Chapter 13 Example 23
Solution:

Let the balls in the bag be denoted by `w_1, w_2, r`. Then the sample space is

`S = {w_1 w_1, w_1 w_2, w_2 w_2, w_2 w_1, w_1 r, w_2 r, r w_1, r w_2, r r}`

Now, for ω ∈ S

X(ω) = number of red balls

Therefore

`X({w_1 w_1}) = X({w_1 w_2}) = X ({w_2 w_2}) = X({w_2 w_1}) = 0`

`X({w_1 r}) = X({w_2 r}) = X ({r w_1}) = X ({r w_2}) = 1` and `X ({r r}) = 2`

Thus, X is a random variable which can take values 0, 1 or 2.

Probability distribution of a random variable

`=>` Let us look at the experiment of selecting one family out of ten families `f_1, f_2 ,..., f_10` in such a manner that each family is equally likely to be selected. Let the families `f_1, f_2, ... , f_10` have 3, 4, 3, 2, 5, 4, 3, 6, 4, 5 members, respectively.

`=>` Let us select a family and note down the number of members in the family denoting X. Clearly, X is a random variable defined as below :

`X(f_1) = 3, X(f_2) = 4, X(f_3) = 3, X (f_4) = 2, X (f_5) = 5`,

`X(f_6) = 4, X(f_7) = 3, X (f_8) = 6, X (f_9) = 4, X(f_10) = 5`

`=>` Thus, X can take any value 2,3,4,5 or 6 depending upon which family is selected.

`=>` Now, X will take the value 2 when the family `f_4` is selected. X can take the value 3 when any one of the families `f_1, f_3, f_7` is selected.

`=>` Similarly, X = 4, when family `f_2, f_6` or `f_9` is selected, X = 5, when family `f_5` or `f_10` is selected and X = 6, when family `f_8` is selected.

Since we had assumed that each family is equally likely to be selected, the probability that family `f_4` is selected is `1/10`

`=>` Thus, the probability that X can take the value 2 is `1/10` . We write `P(X = 2) = 1/10`

Also, the probability that any one of the families `f_1, f_3` or `f_7` is selected is

`P ( { f_1, f_3, f_7 } ) = 3/10`

Thus, the probability that X can take the value` 3 = 3/10`


We write `P(X = 3) =3/10`


Similarly, we obtain

`P(X = 4) = P({f_2, f_6, f_9}) =3/10`

`P(X = 5) = P({f_5, f_10}) = 2/10`

and ` P(X = 6) = P({f_8}) = 1/10`

`=>` Such a description giving the values of the random variable along with the corresponding probabilities is called the probability distribution of the random variable X.
In general, the probability distribution of a random variable X is defined as follows:

`color{blue} "Definition "`

`=>` The probability distribution of a random variable X is the system of numbers

`tt((X :, x_1 , x_2 ,....., x_n),(P(x) :, p_1,p_2, ....., p_n))`

where, `p_i >0 sum_(i=1)^n p_i = 1 , =1,2, ...., n`

`=>` The real numbers `x_1, x_2,..., x_n` are the possible values of the random variable X and `p_i (i = 1,2,..., n)` is the probability of the random variable X taking the value `x_i` i.e., `P(X = x_i) = p_i`



`color{blue}{"Key Point"} Note` :- If `x_i` is one of the possible values of a random variable X, the statement `X = x_i` is true only at some point (s) of the sample space. Hence, the probability that X takes value `x_i` is always nonzero, i.e. `P(X = x_i) ≠ 0`.

`=>` Also for all possible values of the random variable X, all elements of the sample
space are covered. Hence, the sum of all the probabilities in a probability distribution
must be one.
Q 3117691589

Two cards are drawn successively with replacement from a well-shuffled
deck of 52 cards. Find the probability distribution of the number of aces.
Class 12 Chapter 13 Example 24
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.
Therefore, P(X = 0) = P(non-ace and non-ace)
= P(non-ace) × P(non-ace)

`= 48/52 xx48/52 = 144/169`

P(X = 1) = P(ace and non-ace or non-ace and ace)
= P(ace and non-ace) + P(non-ace and ace)
= P(ace). P(non-ace) + P (non-ace) . P(ace)

`= 4/52 xx 48/52 +48/52 xx 4/52 =24/169

and P(X = 2) = P (ace and ace)

`= 4/52 xx 4/52 = 1/169`

Thus, the required probability distribution is
Q 3117791680

Find the probability distribution of number of doublets in three throws of
a pair of dice.
Class 12 Chapter 13 Example 25
Solution:

Let X denote the number of doublets. Possible doublets are
(1,1) , (2,2), (3,3), (4,4), (5,5), (6,6)
Clearly, X can take the value 0, 1, 2, or 3.

Probability of getting a doublet = 6/36 = 1/6

Probability of not getting a doublet =1 -1/6 = 5/6

Now P(X = 0) = P (no doublet) = 5/6 xx 5/6 xx 5/6 =125/216

P(X = 1) = P (one doublet and two non-doublets)

`=1/6 xx 5/6 xx 5/6 + 5/6 xx 1/6 xx 5/6 + 5/6 xx 5/6 xx 1/6`


` = 3 (1/6 xx 5^2/6^2) = 75/216`



P(X = 2) = P (two doublets and one non-doublet)

`=1/6 xx 1/6 xx 5/6 + 1/6 xx 5/6 xx 1/6 + 5/6 xx 1/6 xx 1/6 = 3 ( 1/6^2 xx 5/6 ) = 15/216`

and P(X = 3) = P (three doublets)

`= 1/6 xx 1/6 xx 1/6 = 1/216`
Thus, the required probability distribution is



Verification Sum of the probabilities

`sum_(i=1)^n P_i = 125/216 + 75/216 + 15/216 +1/216`

`= (125 + 75 + 15 +1 )/216 = 216/216 = 1`
Q 3127791681

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.

`P(X=x ) = { tt ( (0.1 text (if) x= 0), ( kx , text(if) x=1 or 2),`
` ( k(5-x) , text (if) x= 3 text (or) 4), ( 0, text (otherwise) ) )`

(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?


Class 12 Chapter 13 Example 26
Solution:

The probability distribution of X is

(a) We know that `sum_(i=1)^n P_i = 1`

Therefore 0.1 + k + 2k + 2k + k = 1
i.e. k = 0.15

(b) P(you study at least two hours) = P(X ≥ 2)
= 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

Mean of a random variable

`color {blue} "Definition"`

`=>`Let X be a random variable whose possible values `x_1, x_2, x_3, ..., x_n` occur with probabilities `p_1, p_2, p_3,..., p_n`, respectively.
`=>`The mean of X, denoted by μ, is the number `sum_(i=1)^n x_i p_i` i.e. the mean of X is the weighted average of the possible values of X, each value being weighted by its probability with which it occurs.

`=>`The mean of a random variable X is also called the expectation of X, denoted by E(X).

Thus , `color{orange} {E(X) = μ = sum_(i=1)^n x_i p_i = x_1 p_1 + x_2 p_2 + ...... + x_n p_n}` .

`=>` In other words, the mean or expectation of a random variable X is the sum of the products of all possible values of X by their respective probabilities.
Q 3167191985

Let a pair of dice be thrown and the random variable X be the sum of the
numbers that appear on the two dice. Find the mean or expectation of X.
Class 12 Chapter 13 Example 27
Solution:

The sample space of the experiment consists of 36 elementary events in the
form of ordered pairs `(x_i , y_i)`, where `x_i = 1, 2, 3, 4, 5, 6` and `y_i = 1, 2, 3, 4, 5, 6`.
The random variable X i.e. the sum of the numbers on the two dice takes the
values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12.

Now P(X = 2) = P({(1,1)}) =1/36

P(X = 3) = P({(1,2), (2,1)}) = 2/36

P(X = 4) = P({(1,3), (2,2), (3,1)}) = 3/36

P(X = 5) = P({(1,4), (2,3), (3,2), (4,1)}) = 4/36

P(X = 6) = P({(1,5), (2,4), (3,3), (4,2), (5,1)}) = 5/36

P(X = 7) = P({(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}) = 6/36

P(X = 8) = P({(2,6), (3,5), (4,4), (5,3), (6,2)}) = 5/36

P(X = 9) = P({(3,6), (4,5), (5,4), (6,3)}) = 4/36

P(X = 10) = P({(4,6), (5,5), (6,4)}) = 3/36

P(X = 11) = P({(5,6), (6,5)}) = 2/36

P(X = 12) = P({(6,6)}) = 1/36

The probability distribution of X is

Therefore,

` μ = E(X) = x_i p_i = 2 xx 1/36 +3 xx 2/36 + 4 xx 3/36 + 5 xx 4/36`

`+ 6 xx 5/36 + 7 xx 6/36 + 8 xx 5/36 + 9 xx 4/36 + 10 xx 3/36 +11 xx 2/36 +12 xx 1/36`

`= ( 2+ 6 +12 +20 +30 +42 +40+ 36+ 30 +22 +12 )/36 = 7`

Thus, the mean of the sum of the numbers that appear on throwing two fair dice is 7.

Variance of a random variable

`=>` The mean of a random variable does not give us information about the variability in the values of the random variable. In fact, if the variance is small, then the values of the random variable are close to the mean. Also random variables with different probability distributions can have equal means, as shown in the following distributions of X and Y.








Clearly `E(X) = 1 xx 1/8 +2 xx 2/8 +3 xx 3/8 + 4 xx 2/8 = 22/8 = 2.75`

and `E(Y) = -1 xx 1/8 + 0 xx 2/8 + 4 xx 3/8 + 5 xx 1/8 = 6 xx 1/8 = 22/8 = 2.75`

`=>` The variables X and Y are different, however their means are same. It is also easily observable from the diagramatic representation of these distributions (Fig).



`=>` To distinguish X from Y, we require a measure of the extent to which the values of the random variables spread out. In Statistics, we have studied that the variance is a measure of the spread or scatter in data. Likewise, the variability or spread in the values of a random variable may be measured by variance.

`color{blue} "Definition 7"` Let X be a random variable whose possible values `x_1, x_2,...,x_n` occur with probabilities p(x1), p(x2),..., p(xn) respectively.

Let μ = E (X) be the mean of X. The variance of X, denoted by Var (X) or `x^2` is defined as

`sigma_x^2 = Var (X) = sum_(i =1)^n ( x _i - mu )^2 p(x_i )`

or equivalently `sigma_x^2 = E(X -mu)^2`

The non-negative number

`sigma_x = sqrt(Var (X)) = sqrt ( sum_(i =1)^n ( x _i - mu)^2 p(x_i))`

is called the standard deviation of the random variable X.

`color{orange}" Another formula to find the variance of a random variable"`. We know that,

`Var (X) = sum_(i =1 ) ^n ( x_i - mu )^2 p(x_i)`

`= sum_(i =1)^n ( x_i^2+ mu^2 - 2 mu x_i ) p (x_i)`

`= sum_(i =1)^n x_i^2 p (x_i ) + sum_(i =1) ^n mu^2 p (x _i ) - sum_(i =1 )^n 2 mu x_i p (x _i )`

`= sum_(i =1) ^n x _i^2 p ( x_i ) + mu^2 sum_(i =1 ) ^n p (x _i ) - 2 mu sum _(i =1) ^n x _i p (x _i )`

`= sum_(i =1) ^n x _i^2 p (x_i ) + mu^2 - 2 mu^2 \ \ \ \ \ \ [ "since" sum _(i =1)^n p (x_i) =1 "and" mu = sum _(i =1)^n x_i p ( x_i)]`

` = sum_(i=1)^n x_i^2 p (x_i ) - mu^2`

or `Var (X) = sum_(i=1)^n x_i^2 p (x_i ) - ( sum_(i=1) ^n x _i p (x_i ) )^2`

or `color {green} { Var (X ) = E(X^2) - [ E(X)]^2}`, where `color{green} {E(X^2) = sum_(i=1)^n x_i^2 p (x_i )}`
Q 3138101002

Find the variance of the number obtained on a throw of an unbiased die.
Class 12 Chapter 13 Example 28
Solution:

The sample space of the experiment is S = {1, 2, 3, 4, 5, 6}.
Let X denote the number obtained on the throw. Then X is a random variable
which can take values 1, 2, 3, 4, 5, or 6.

Also P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

Therefore, the Probability distribution of X is

Now `E(X) = sum_(i=1)^n x_i p(x_i)`

`= 1 xx 1/6 + 2 xx 1/6 + 3 xx 1/6 +4 xx 1/6 + 5 xx 1/6 + 6 xx 1/6 = 21/6`

Also `E(X^2) = 1^2 xx 1/6+ 2^2 + 1/6 + 3^2 xx 1/6 + 4^2 xx 1/6 + 5^2 xx 1/6 + 6^2 xx 1/6 = 91/6`



Thus,` Var (X) = E (X^2) – (E(X))^2`

`= 91/6 - (21/6)^2 = 91/6 - 441/36 = 35/12`
Q 3188101007

Two cards are drawn simultaneously (or successively without replacement)
from a well shuffled pack of 52 cards. Find the mean, variance and standard deviation
of the number of kings.
Class 12 Chapter 13 Example 29
Solution:

Let X denote the number of kings in a draw of two cards. X is a random
variable which can assume the values 0, 1 or 2.

Now P(X = 0) = P (no king) `= (text()^(48)C_2 )/( text()^(52)C_2) = ( (48!)/(2! (48 -2)!) )/( (52!)/(2! (52 -2)! )) = (48 xx 47 )/(52 xx 51) = 188/221`

P(X = 1) = P (one king and one non-king) = ` ( text()^(4)C_1 text()^(48)C_1)/(text()^(52)C_2)`

`= (4 xx 48 xx 2)/( 52 xx 51) = 32/221`

and P(X = 2) = P (two kings) ` = (text()^(4)C_2)/( text()^(52)C_2) = (4 xx 3)/( 52 xx 51) = 1/221`

Thus, the probability distribution of X is


Now Mean of X = E(X) = ` sum_(i=1)^n x_i p(x_i)`

`= 0 xx 188/221 +1 xx 32/221 +2 xx 1/221 = 34/221`

Also `E(X^2) = sum_(i=1)^n x_(i)^2 p (x_i)`

`= 0^2 xx 188/221 + 1^2 xx 32/221 + 2^2 xx 1/221 = 36/221`


Now `Var(X) = E(X^2) – [E(X)]^2`

`= 36/221 - ( 34/221)^2 = 6800/(221)^2`

Therefore `σ_x = sqrt(var (X)) = (sqrt 6800)/221 = 0.37`
 
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