Question

Two dice are rolled once find the probability of getting (1)sum of numbers obtained is 5, (2) some of numbers divisible by 5 ,(3) product of numbers obtained by 5 ,(4) product of numbers obtained is divisible by 5 ,(5) some of numbers numbers obtained is 13

Answer 1

Step-by-step explanation:

The sample space on rolling two dice is

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)}

Then n(S) = 36

1.

The set for sum 5 is

A = {(1, 4), (2, 3), (3, 2), (4, 1)}

Then n(A) = 4

Therefore P(A) = n(A) / n(S)

= 4/36

= 1/9

2.

The set for sum divisible by 5 is

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

Then n(B) = 7

Therefore P(B) = n(B) / n(S)

= 7/36

3.

The set for product 5 is

C = {(1, 5), (5, 1)}

Then n(C) = 2

Therefore P(C) = n(C) / n(S)

= 2/36

= 1/18

4.

The set for product divisible by 5 is

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

Then n(D) = 11

Therefore P(D) = n(D) / n(S)

= 11/36

5.

The set for sum 13 is

E = { }

Then n(E) = 0

Therefore P(E) = n(E) / n(S)

= 0/36

= 0