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Unit 6: Probability



                                                                                                  Notes
                   Example 2: What is the probability of obtaining at least one head in the simultaneous
            toss of two unbiased coins?
            Solution.
            The equally likely, mutually exclusive and exhaustive outcomes of the experiment are (H, H),
            (H, T), (T, H) and (T, T), where H denotes a head and T denotes a tail. Thus, n = 4.
            Let A be the event that at least one head occurs. This event corresponds the first three outcomes
            of the random experiment. Therefore, m = 3.

                                                 3
            Hence, probability that A occurs, i.e.,  ( ) =  .
                                           P A
                                                 4

                   Example 3: Find the probability of obtaining an odd number in the roll of an unbiased
            die.
            Solution.

            The number of equally likely, mutually exclusive and exhaustive outcomes, i.e., n = 6. There are
            three odd numbers out of the numbers 1, 2, 3, 4, 5 and 6. Therefore, m = 3.

                                                       3  1
            Thus, probability of occurrence of an odd number  =  =  .
                                                       6  2

                   Example 4: What is the chance of drawing a face card in a draw from a pack of 52 well-
            shuffled cards?
            Solution.
            Total possible outcomes n = 52.

            Since the pack is well-shuffled, these outcomes are equally likely. Further, since only one card is
            to be drawn, the outcomes are mutually exclusive.
            There are 12 face cards,    m = 12.

                                               12  3
            Thus, probability of drawing a face card  =  =  .
                                               52  13


                   Example 5: What is the probability that a leap year selected at random will contain 53
            Sundays?
            Solution.

            A leap year has 366 days. It contains 52 complete weeks, i.e, 52 Sundays. The remaining two days
            of the year could be anyone of the following pairs:
            (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday,
            Saturday), (Saturday, Sunday), (Sunday, Monday). Thus, there are seven possibilities out  of
            which last two are favourable to the occurrence of 53rd Sunday.


                                        2
            Hence, the required probability  =  .
                                        7





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