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1.A fair coin with  1  marked on one face and  6  on the other and a fair die are both tossed.  Find the probability that the sum of numbers that turn up  is    i) 3           ii)  12.

The coin with   1 marked on one face and 6 on the other face.

The coin and die are tossed together.

. . .  S  =  { ( 1 , 1 )  , ( 1 , 2 ) , ( 1 , 3 ) , ( 1 , 4 ) , ( 1 , 5 ) , ( 1 , 6 ) , ( 6 , 1 ) ,  ( 6 , 2 ) ,  ( 6 , 3 ) , ( 6 , 4 ) , ( 5 , 5 ) , ( 6 , 6 ) }.

i)Let  A be the event  having sum of numbers  is  3

                 

          Thus  P (A)   = 1 /12 .

ii)Let  B be the event having sum of number  is  12.

             

              . . .  P  (B)  =  1 / 12 .

2.There are four men and six women on the city council. If one council member is selected for a committee at random , how likely is it that it is a woman?

Here total members in the council  1  =  4 + 6  = 10

one member is selected out of 10 members.

                    n (S)   =  10 C1  = 10

Let  A  be the event that the member is a woman

              . . .  n  (A)  =  6 C1   = 6 .

               

3.If   2 / 11  is the probability of an event  , what is the probability of the event ' not  A' .

Here probability of event  A  is  P ( A )  =  2 / 11   

. . . probability  of event  not A is  P (A¯)  = 1-  P(A)     = 1 -   2 / 11  =  9 / 11  .

4.A letter is chosen at random from the word ASSASSINATION  : Find the probability that letter is

        i)  a vowel                       ii) a consonant

There are  13 letters in the word ASSASSINATION   which contains  6  vowels and 7 consonants.

One letter is selected out of 1 letters in 13C1  = 13 ways.

i)One vowel is selected out of 6 vowels in 6C1  =  6 ways

Thus probability of a vowel  =  6 / 13 .

ii)One consonant is selected out of 7 consonants in 7C1  =  7 ways.

Thus probability of consonant  = 7 / 13 .

5.In a lottery , a person chosen six different natural numbers at random from  1  to 20  and if these size numbers  match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?

Number of numbers in the draw = 20

Number of numbers to be selected  = 6

Let  Δ  be the event that six numbers match with the six numbers fixed by the lottery committee.

. . .  n (A)  =  6C6  =  1

Thus probability of winning the prize  P (A)

                

                 

6.Check whether the following probabilities P (A)  ans  P (B)  are consistently defined

i)P (A)  = 0.5  ,  P (B)  = 0.7  , P (A∩B)  = 0.6

ii)P (A)  = 0.5  , P (B)  = 0.4 , P (AυB)  = 0.8

i) Here P (A)  = 0.5  , P (B)  = 0.7  and  P (A∩B)  = 0.6

           Now  P (A∩B)  >  P (A)

Thus the given probabilities are not consistently defined.

ii)Here P (A)  = 0.5  , P (B)  = 0.4  and  P (AυB)  = 0.8

We know that  P (AυB)  = P (A)  + P (B)  - P (A∩B)

                     0.8   =  0.5 + 0.4  -  P (A∩B)

                        P(A∩B)  = 0.9   -  0.8  = 0.1

. . .  P  ( A∩B) < P (A)  and  P (A∩B)  <  P  (B)

Thus the given probabilities are consistently defined.

7.Given  P (A)  = 3 / 5 and  P (B)  = 1 / 5 . Find  P (A  or B ), if A  and  B are mutually exclusive  events.

Here  P (A)   =  3 / 5  ,  P (B)  = 1 / 5.

Since  A and  B  are mutually exclusive events

                  P (AυB)  =  P (A)  +  P (B)

                . . .  P (AυB)   =  3 / 5  +  1 / 5   =  4 / 5 .

8.If  E  and  F  are events  such that  P (E)  =  1 / 4 ,  P (F)  =  1 / 2  and  P  (E and F )  = 1 / 8 . Find

i)  P  ( E and F )                       ii)  P  ( not  E  and  not  F )

Here  P ( E )  =  1 / 4 ,  P (  F ) ,  P ( F )  = 1 / 2   and  P  ( E ∩F )  = 1 / 8 .

i) We know that  P ( EυF ) = P ( E )  +  P ( F )  -  P ( E∩F )

                                   

9.Events  E  and  F  are  such  that   P  (not  E  or  not  F) =  0.25.  state whether  E and  F  are  mutually exclusive.

Here  P (not  E  or not  F)  =  0.25

Thus E and  F are  not  mutually  exclusive events.

10.A   and  B  are  events  such that  P (a) = 0.42 ,  P (b)  = 0.48 and  P ( A  and  B )  =  0.16. Determine

i) P  (not A)    ii) P ( not  B )  and  iii)  P  (A or  B).

11.In class XI  of a school , 40%  of the  students study Mathematics  and  30%  study  Biology,  10%  of the class study both Mathematics  and  Biology.  If a student is selected  at random from the class, find the probability that he will be studying Mathematics or  Biology.

Let A be  the event that the student is studying Mathematics  and  B be the event that the student  is studying biology.

12.In an entrance test that is graded on the basis of two examination , the probability of a randomly chosen student passing the first examination is  0.8  and the probability of passing the  second examination is  0.7 .  The probability of passing at least one of them is  0.95.  What is the probability of passing both?

Let  A  be the  event that the student passes the first examination and  B  be the event that the student passes the second examination.

Then   P (A)  = 0.8  , P (B)  = 0.7  and  P (AυB )   = 0.95

We know that  P (AυB)  = P (A)  +  P (B)  -  P (A∩B)

                                             0.95  =  0.8  +  0.7  - P (A∩B)

                                             P (A∩M)  = 1.5  -  0.95  =  0.55.

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