Example 1:

        A pair of dice is thrown. Find the probability of obtaining a sum of 8 or getting an even number on both the dice.

Solution:

        Let the events be defined as:

                A : obtaining a sum of 8

                B : getting an even number on both dice

        We are required to find out the total Probability A and B, i.e.

                P(AUB) = P(A) + P(B) - P(A∩B)

        Now cases favourable to A are

                (3, 5) (5, 3) (2, 6) (6, 2) (4, 4)

        So,    P(A) = 5/36

        Cases favourable to B : (2, 2), (2, 4), (2, 6),

                                         (4, 2), (4, 4), (4, 6),

                                         (6, 2), (6, 4), (6, 6).

                P(B) = 9/36

        Now, (2, 6) (6, 2) and (4, 4) are common to both events A and B

        So     P(A∩B) = 3/36

                => P(AUB) = = 5/36+9/36-3/36=11/36                       (Ans.)

Example 2:

A number x is selected from first 100 natural numbers. Find the probability that x satisfy the condition x + 100/x > 50.

Solution:

        Total number of ways of selecting x is 100.

Now the given condition is x + 100/x > 50, on analyzing this equation, carefully, we see that this equation is satisfied for all the numbers x such that x > 48 and also for x = 1 and 2

So, favourable number cases is 55

Therefore, probability = 55/100 = 11/20                                      (Ans.)

Example 3:

'A' has three share in a lottery in which there are 3 prizes and 6 blanks' 'B' has one share in a lottery in which there is 1 prize and 2 blanks. Show that A's chance of winning a prize of B's chance of winning a prizes in ratio 16:7

Solution:

 Method 1:

        'A' may win one, two or all the three prizes, the total probability is

      =(³C₁⁶C₂)/(⁹C₃)+(³C₂⁶C₁)/(⁹C₃)+(³C₃⁶C₀)/(⁹C₃)=(45+18+1)/84=64/84=16/21

                P(B) = ( ¹C₁)/( ³C₁ )=1/3

     (P(A))/(P(B))=(16×3)/21=16/7=> P(A) : P(B) = 16 : 7              (Proved)

Method 2:

        A can win one, two or all the prizes in the following manner

        1 prize, 2 blanks          2 prizes, 1 blank             3 prizes

                I                               II                              III

All the three cases where 'A' can win is independent and mutually exclusive. Hence probability that A can win prize is given as

                P(A) = P(I) + P(II) + P(III)

        Let's calculate P(I): it is something like this:

A bag contains 3 white balls (prizes and 6 black balls (blanks) then what is the probability that out of 3 draws random, there is 1 white ball (prize) and 2 black balls (blanks).

Which one can calculate very easily.

Favourable ways for this event = ³C₁ × ⁶C₂

        Total number of ways = ⁹C₃

        P(I) = ( ³C₁× ⁶C₂)/(⁹C₃),, similarly we can calculate P(II) and P(III)

  Hence, P(A) = (³C₁× ⁶ C₂)/(⁹C₃)+(³C₂× ⁶C₁)/(⁹C₃)+( ³C₃× ⁶C₀)/(⁹C₃) = 16/21

B has only one share in a lottery, which consists 1 prize and 2 blanks' therefore B can at most win only one prize. Hence

        P(B) = ¹C₁/³C₁ = 1/3

       (P(A))/(P(B))=(16×3)/21=16/7

        P(A) : P(B) = 16:7                                               (Ans.)

Example 4:

There are three events A, B and C one of which must, and only one can happen, the odd are 8 to 3 against and 2 to 5 for B. Find the odd, against C.

Solution:

                P(A) = 3/11, P(B) = 2/7, P(C) = x (say)

Since one most and only one can happen therefore A, B, C are mutually exclusive and exhaustive events.

So,    P(A) + P(B) + P(C) = 1

=> 3/11+2/7+x=1

x = (77-21-22)/77=34/77

Odd against C = (77 - 34) : 34 or 43 : 34                       (Ans.)