Wednesday, December 26, 2012

Construct Probability Table

Construct Probability table is defined as an equation or table with its probability happenings. Usually, probability table is for constant occurrence for much number of data produces results. For construct probability table function we learn about discrete random variable, if a random variable obtains only a finite or a countable number of values, it is known as discrete random variable. To construct Probability table it contains probability mass function and moments.

Classification of Construct Probability Table:

Probability Mass Function:

The statistical description of discrete probability table function p (x)  functions and satisfies the following properties:

(1.) The probability which x can take a particular value x is p (x)

That is P(x = x) = p (x) = px.

(2). P(x) is a non – negative for every real x.

(3). The amount of p (x) over all likely values of x is one. Which is Σpi = 1 here j denotes that x and pi is the probability at x = xi

Moments:

A probable value of a function of a chance variable x is used for calculating the moments. Let us see the two types of moments.

(i) Moments about the origin.
(ii) Moments about the mean that are known as central moments.

Example Problem for Construct Probability Table:

A box has 4 green and 3 black pens. Construct a probability table distribution of number of black pens in 3 draws one by one from the box. (i) With replacement

Solution:

(i) With replacement

Let x be the approximate variable of drawing number of black pens in three draws.

X will takes the values 0,1,2,3.

P (Black pen) = `3 / 7 ` = P (B)

P (Not Black pen) = `4 / 7` = P (G)

Hence P(X = 0) = P (GGG) = `4/ 7` * `4 / 7` * `4 / 7` = `64 / 343`

P(X = 1) = P (BGG) + P (GBG) + P (GGB)

=  ( `3 / 7` * `3 / 7` * `4 / 7` )  + ( `4 / 7` * `3 / 7`   * `4 / 7` ) +  ( `4 / 7` * `3 / 7`   * `3 / 7` )

= 3 * (`48 / 343` )

= `144 / 343 `

P(X = 2) = P (BBG) + P (BGB) + P (GBB)

= (`3 / 7` * `3 / 7` * `4 / 7` ) + (`3 / 7` * `4 / 7` * `3 / 7` ) + (`4 / 7` * `3 / 7` * `3 / 7` )

= 3 * ( `3 / 7` ) * ( `3 / 7` ) * ( `4 / 7` )

= 3 * (`36 / 343` )

= `108 / 343`

P(X = 3) = P (BBB) = ( `3 / 7` ) * ( `3 / 7` ) * ( `3 / 7` )

= `27 / 343`

Therefore the wanted probability table distribution is


x 0 1 2 3
p (x =x) `64 / 43` `144 / 343` `108 / 343` `27 / 343`



Example 2:

Find out the probability mass function, and the collective distribution function for obtaining ‘3’s while two dice are thrown.

Solution:

2 dice are thrown. Let x be the approximate variable of obtaining number of ‘3’s. Hence x can take the values 0, 1, 2.

P (no ‘3’) = P (X = 0) =` 25 / 36`

P (one ‘3’) = P (X = 1) = `10 / 36`

P (two ‘3’s) = P (X = 2) = `1 / 36`

Obtained Probability mass function is


x 0 1 2
P (x = x) `25 / 36` `10 / 36` `1 / 36`

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