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.