class IX probability solved problems

                                                 CLASS -IX

                                                  Probability 

Example 1:  The probability of happening of an event is 45%. The probability of an event is

                     (a) 45   (b) 4.5   (c) 0.45     (d) 0.045

Solution: (c) 45% means 45 out of 100.

                Therefore, probability = 45/100=0.45 

Example 2:  In a class of 40 students. There are 110% girls. Then the number of girls is

                (a) 44   (b) 22     (c) 30     (d) None of these

Solution:  (d) Maximum is 100%, So, 110% is invalid.

Example 3: To know the opinion of the students about the subject statistics, a survey of 200 students

                 was conducted. The data is recorded as

                Opinion                          Like            Dislike

                Number of students         135              65

               The probability that the students chosen as random like statistics is

               (a) 1     (b) -1      (c) 0.65    (d) 0.675

Solution:  (d) Total students = 200

                Number of students like statistics =135

                .. Probability =135 /200 =0.675

Example 4: The distance (in km) for 40 engineers from their residence to their place of work were

                   found as follows 5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7,

                   9, 7, 8, 3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12. The emperical probability that

                   an engineer lives less than 7 km from his/her place of work is

                    (a) 31/40                       (b) 25/40

                     (c) 1/2                            (d) 9/40

solution: (c) There are 9 engineers whose distance is less than 7 km from place of work.

                Probability = 9/40

Example 5: Two coins are tossed simultaneously 500 times, and we get

                   Two heads: 105 times, One head: 275 times

                   No head: 120 times

                  Find the probability of occurrence of each of these events.

Solution: Let us denote the events of getting two heads, one head and no head by E1, E2 and E3,

               respectively.

              So, P(E1)105/500 ==0.21

              PŒ2)275 /500 =0.55

             P(E3) =120/500 = 0.24

Example 6: An insurance company selected 2000 drivers at random (i.e., without any preference of

                   one driver over another) in a particular city to find a relationship between age and

                   accidents. The data obtained are given in the following table:

     Age of drivers 

       (in years )                  Accidents in one year

                                 0          1          2       3      Over 3

        18-29              440       160      110     61        35

         30-50              505       125       60      22       18

        Above 50       360        45        35      15         9

     Find the probabilities of the following events for a driver chosen at random from the city:

    (i) being 18-29 years of age and having exactly 3 accident in one year.

    (ii) being 30-50 years of age and having one or more accidents in one year.

    (iii) having no accidents in one year.

Solution: Total number of drivers = 2000

              (i) The number of drivers who are 18-29 years old and have exactly 3 accidents in one

               year is 61.

              So P (driver is 18-29 years old having exactly 3 accidents)

               =61/2000 =0.0305=0.031

            (ii) The number of drivers 30-50 years of age and having one or more accidents in one

             year = 125 + 60 + 22 + 18 = 225.

            =225/2000= 0.1125=0.113

           (iii) The number of drivers having no accidents in one year = 440 + 505 + 360 = 1305

            So, P (no accidents) =1305/2000 =0.6525