Learn some topics about Probability!

   According to Ya-Lin Chou "Probability is the science of decision-making with calculated risks in the face of uncertainty.

    If a random experiment or a trail results in 'n' exhaustive, mutually exclusive and equally likely outcomes, out of which m are favorable to the occurrence of an event E, then the probability 'p' of occurrence of E, usually denoted by P(E), is given by: 

                

              

    

    There are number of events in day-to-day life about which one is not sure whether it will occur or not. But there is a curiosity to know what chance their event is to occur. For example, one may be interested to estimate profit or loss. The numerical evaluation of chance factor of an event is known as probability. An event is the collection of possible outcomes which may be favorable to an happening out of total outcomes which may be count or list out. The result of a random experiment is said to be outcome.  If in each trail of an experiment conducted under identical conditions, the outcome is not unique, but may be one possible outcome, then it is said to be random experiment. An event having only one sample point is called simple event. Two events A and B are said to be equal if A subset B and B subset A. If A, B and C are 3 events as A subset B and B subset C, it implies that A subset C. It is said to be transitivity of events. An event which is not simple is said to be compound event. If two or more events joined by the conjunction 'or' are called derived events. An event which is certain not to occur is called an impossible event. Events are said to be mutually exclusive or incompatible if the happening of any of them precludes the happening of all the others, if no two or more of them can happen simultaneously in the same trail. Outcomes of trail are said to be equally likely if taking into consideration all the relevant evidence, there is no reason to expect one in preference to the others. Several events are said to be independent if the happening of an event is not affected by the supplementary knowledge concerning the occurrence af any number of the remaining events.

1. Some important theorems on probability:

1.1 Addition theorem of probability

Statement: If A and B are any two arbitrary events in the sample space and are not disjoint, then P (A Union B) = P (A) + P(B) - P (A Intersection B) 

Proof: From the Venn diagram, we have 

1.2 Multiplication theorem of probability

Statement: For two events A and B, 

                                

    Where P (B | A) represents conditional probability of occurrence of B when the event A has already happened and P (A | B) is the conditional probability of happening of A, given that B has already happened.

Proof: In the usual notations, we have

   

 For the conditional event A | B, the favorable outcomes must be one of the sample points of B, i.e., for the event A | B, the sample space is B and out of the n(B) sample points, n(A intersection B) pertain to be occurrence of the event A. Hence

1.3 Baye's Theorem.

Statement: If E1, E2, E3, .......En are mutually disjoint events with P(Ei) is not equals to 0, (i = 1, 2, ......, n), then for any arbitrary event E which is a subset of

Ei such that P(E) > 0, 

Proof: 

Some problems:

1. Two unbiased dice are thrown.

    In a random throw of two dice, since each of the six faces of one die can be associated with each of six faces of the other die, the total number of cases is 6 * 6 = 36, as given below:

       (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3), (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6).

    Here, the expression, say (i, j) means that the first die shows the number i and the second die shows the number j. Obviously, (i, j) is not equals to (j, i) if i is not equals to j.

    Therefore, Exhaustive number of cases (n) = 36.

   

    a) The favorable cases that both the dice show the same number are:

  

                (1, 1), (2, 3), (3, 3), (4, 4), (5, 5) and (6, 6), i.e., m = 6.

        Therefore, Probability that has two dice show the same number = 6 / 36 = 1 / 6.

    b) The favorable cases that the first die shows 6 are:

               (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) and (6, 6), i.e., 6 in all.

        Therefore, Probability that the first die shows 6 = 6 / 36 = 1 / 6.

    c) The cases favorable to getting a total of 8 on the two dice are:

               (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), i.e., m = 5.

        Therefore, Probability that total of numbers on two dice is 8 = 5 / 36.

    d) The cases favorable to getting a total of more than 8 are:

               (3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6), i.e., m = 10.

        Therefore, Probability that the total of numbers on two dice is greater than 8 = 10 / 36 = 5 / 18.

    e) The total of the numbers on the dice is 13: This is an example of an impossible event, since the maximum total can be 6 + 6 = 12. Therefore, the required probability is 0.

2. What is the chance that a leap year selected at random will contain 53 Sundays?

    In a leap year, there are 52 complete weeks and 2 days over. The following are the possible combinations for these two 'over' days.

•  Sunday and Monday
• Monday and Tuesday
• Tuesday and Wednesday
• Wednesday and Thursday
• Thursday and Friday
• Friday and Saturday
• Saturday and Sunday.

    In order that a leap year selected at random should contain 53 Sundays, one of the two over days must be Sunday. Since out of the above 7 possibilities, 2, to this event.

    Required probability = 2 / 7.

3. n persons are seated on n chairs at a round table. Find the probability that two specified persons are sitting next to each other.

    n persons can be seated in n chairs at a round table in (n - 1)! ways, the exhaustive number of cases = (n -1)!

    

    Assuming the two specified persons A and B who sit together as one, we get (n -1) persons in all, who can be seated at a round table in (n - 2)! ways. Further, since A and B can interchange their positions in 2! ways, total number of favorable cases of getting A and B together is (n-2)! * 2!

    Required probability = (n - 2)! * 2! / (n - 1)! = 2 / n - 1. 

4. A and B throw with three dice; if A throws 14, find B's chance of throwing a higher number.

    To throw higher number than A, B must throw either 15 or 16 or 17 or 18. Now a throw amounting to 18 must be made up of (6, 6, 6), which can occur in one way; 17 can be made up of (6, 6, 5), which can occur in 3! / 2!1! = 3 ways; 16 may be made up of (6, 6, 4), and (6, 5, 5), each of which arrangements can occur in 3! / 2!1! = 3 ways; 15 can be made up of (6, 4, 5), or (6, 3, 6), or (5, 5, 5), which can occur in 3! 3 and 1 way, respectively.

    The number of favorable cases = 1 + 3 + 3 + 3 + 6 + 3 + 1 = 20.

In a random throw of 3 dice, the exhaustive number of cases = 6 * 6 * 6 = 216.

    Required probability = 20 / 216 = 5 / 54.

5. An urn contains 4 tickets numbered 1, 2, 3, 4 and another contains 6 tickets numbered 2, 4, 6, 7, 8, 9. If one of the two urns is chosen at random and a ticket is drawn at random from the chosen urn, find the probabilities that the ticket drawn bears the number (i) 2 or 4, (ii) 3, (iv) 1 or 9.

    (i) p = P(I) + P(II) = 1/2 * 2/4 + 1/2 * 2/6 = 5/12

    (ii) 1/2 * 1/4 + 1/2 * 0 = 1/8

    (iii) 1/2 *1/4 + 1/2 * 1/6 = 5/24.