Question

1. Two fair dice are tossed, and the face on each die is observed.

   1. Use a tree diagram to find the 36 sample points contained in the sample space.

   2. Assign probabilities to the sample points in part a.

   3. Find the probability of each of the following events:

      A = {3 showing on each die}
      B = {sum of two numbers showing is 7}
      C = {sum of two numbers showing is even}

   4. Also, for events A, B, and C, which one is simple event, which one is compound event?

   5. Event D= ?" , Event D and event C independent or not? Event D and event B mutually exclusive or not?

   6. Consider event D and event C, Find the probability of each of the following events: ?(D ∩ ?), p(D ∪ ?), ?(?|?), ?(?/|?/), ?(?/ ∩ ?/)

Answer 1

Solution

Back-up Theory

For 2 events, A and B,

P(A or B) = P(A ∪ B) = P(A) + P(B) - P(A ∩ B), in general and ……………………………….…………...............………………(1)

P(A ∪ B) = P(A) + P(B), when A and B are mutually exclusive …………………………………………..................……..………(2)

A and B are mutually exclusive if A and B have nothing common, i.e., ), P(A ∩ B) = 0……………...............................…...…(3)

If A and B are independent, P(A ∩ B) = P(A) x P(B) ..………………………………………………….................…………………(4)

If A and B are such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)……………………………....................................….(5)

Now to work out the solution,

Part (a)

1. The sample space and assigned probabilities to the sample points:

Face on Die 1	Face on Die 2
	1	2	3	4	5	6
1	1/36	1/36	1/36	1/36	1/36	1/36
2	1/36	1/36	1/36	1/36	1/36	1/36
3	1/36	1/36	1/36	1/36	1/36	1/36
4	1/36	1/36	1/36	1/36	1/36	1/36
5	1/36	1/36	1/36	1/36	1/36	1/36
6	1/36	1/36	1/36	1/36	1/36	1/36

Answer 1

Part (b)

P(A) = P{3 showing on each die} = P(3, 3) = 1/36 Answer 2

This is a simple event since it is composed of only one outcome. Answer 3

P(B) = P{sum of two numbers showing is 7} = P[(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)]
= 6/36 = 1/6 Answer 4

This is a compound event since it is composed of 6, i.e., more than one, outcomes. Answer 5

P(C) = P{sum of two numbers showing is even}

= P[(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3),

       (5, 5), (6, 2), (6, 4), (6, 6)]

= 18/36 = ½ Answer 6

This is a compound event as it is composed of 18, i.e., more than one, outcomes. Answer 7

Part (c)

2. Given Event D = ?", P(D) = 1 – P(A) = 1 – (1/36) = 35/36 Answer 8
3. To check if Event D and Event C are independent,

P(D ∩ C) = P[(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1),

                  (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)] = 34/36 = 17/18 ..…………..........................................................……………… (6)

P(D) x P(C) = (35/36) x ½ = 35/72 ≠ 17/18.

So, vide (4), D and C are NOT independent. Answer 9

To check if Event D and Event B are mutually exclusive,

D has 35 points of the sample space, i.e., excepting (3, 3).

B is composed of [(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

Clearly, D and B have a number of points, 6 to be exact, common. So, vide (3),

Event D and event B are NOT mutually exclusive. Answer 10

Part (d)

P(D ∩ ?) = 17/18 [vide (6)] Answer 10

P(D ∪ ?) = P(D) + P(C) - P(D ∩ ?) [vide (1)]

= (35/36) + ½ - 17/18 [vide Answers (8) and (6) and (6)]

= 1 Answer 11

[Note: D ∪ ? is the sample space and hence the probability is 1]

P(C|D) = P(D ∩ ?)/P(D) [vide (5)]

= (17/18)/(35/36) [vide (6) and Answer (8)]

= 34/35 Answer 12

DONE