Introduction to probability distribution sample problems:

Probability Distribution is a way of expressing knowledge or belief that an event will occur or has occurred. In mathematics the concept has been given an exact meaning in probability theory, that is used extensively in such areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems. (source: wikipedia)

Here let us learn probability distribution concepts, apart from studies probability distribution  is useful through out our real life situations.Let us see probability distribution sample problems.


probability distribution sample problems:


Example 1:

A box contains brown and black dolls. We are choosing two dolls without replacement. Probability of choosing brown and black dolls is 0.45 and choosing the brown  on the first draw is 0.57. Find the probability distribution for problem if the second dolls is black if the first one is brown?


             Probability of choosing brown and black dolls is 0.45

             Probability of choosing brown in the first draw is 0.57

             So probability of choosing second dolls ids black then the probability is

             P (black | brown) = P (brown and Black) / P (brown)

             P (B | R) = 0.45 /0.57

             P (B | R) = 0.79 = 79 %

Example 2:

When a dice is thrown once. Find the probability of getting a number lying between 2 and 6.


            Total no of possible outcomes associated with random experiment of throwing a dice is 6  ( i.e., 1, 2, 3, 4, 5, 6).

            Let E be the event getting a number lying between 2 and 6.

            Favorable number of elementary events (outcomes) = 3(i.e., 3, 4, 5)

            P (E) = 3 / 6

                     = 1/ 2


probability distribution sample problems:


Example 3:

Three dice are rolled once. In this problem what is the chance that the sum of the numbers on the three dice is greater than 16?


         When three dice are rolled, the sample space S = {(1,1,1), (1,1,2), (1,1,3) ...(6,6,6)}.

         S contains 6 × 6 × 6 = 216 outcomes.

         Let A be the event of getting the sum of face numbers greater than 16.

         A = {(5,6,6), (6,5,6), (6,6,5), (6,6,6)}.

         n(S) = 216, n(A) = 4.

         Now let us use the algebraic formula to calculate probability,

         P (A) = n(A)/n(S)  = `(4/216)` =` (2/108) ` = `1/54` .