Question

A pair of dice are rolled. Let A be the event as “the first dice roll is 3” and event B as “the second dice roll is 4”. Let event C be as “the sum of the dice rolls are 7.”

a) Show that A and B are independent, that A and C are independent, and that B and C are independent.

b) Show that A, B, and C are not mutually independent.

Answer 1

The exhaustative cases of two times throwing of a die is (Sample Space)

S = { (1,1),(1,2), (1,3), (1,4), (1,5), (1,6),

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

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

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

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

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

n = 36

Let A be the event as “the first dice roll is 3”

i.e. A = {   (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) }

Event B as “the second dice roll is 4”.

B = { (1,4), (2,4) , (3,4), (4,4) , (5,4), (6,4) }

Let event C be as “the sum of the dice rolls are 7.”

C = ( (1,6), (2,5), (3,4) , (4,3), (5,2), (6,1) }

a) P(A) = number of elements in A / exhaustative cases

= 6/36 = 1/6

P(B) = number of elements in B / exhaustative cases

= 6/36 = 1/6

P(A and B) = 1/36

Now

P(A) P(B) = 1/6 * 1/6 = 1/36 = P(A and B)

Therefore, A and B are independent

P(C) = 6/36 = 1/6

P(B) P(C) = 1/6 * 1/6 = 1/36 = P(B and C)

Therefore, B and C are independent

P(A) P(C) = 1/6 * 1/6 = 1/36 = P(a and C)

Therefore, A and C are independent

b) P(A) P(B) P(C) = 1/6 * 1/6 * 1/6 = 1/216

P(A and B and C) = 1/36

Here P(A and B and C) not =  P(A) P(B) P(C)

Therefore, A and B and C are not independent