Question

Two dice are rolled together. Find the probability of getting a sum as a multiple of 3

Answer 1

The outcomes if two dice are rolled are as follows:
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

Outcomes in which the sum is a multiple of 3: 1,2; 1,5; 2,1; 2,4; 3,3; 3,6; 4,2; 4,5; 5,1; 5,4; 6,3; 6,6
Now the number of times the outcomes can be a multiple of three is 12
$probability \: of \: something \: to \: happen = \frac{number \: of \: successful \: outcomes}{number \: of \: possible \: outcomes}$
$therefore \: the \: probability \: of \: the \: outcome \: coming \: as \: a \: multiple \: of \: 3 = \frac{12}{36} = \frac{1}{3}$
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Answer 2

Answer:

ether. Find the probability of getting a sum as a multiple of 3.

Solution:-

Total number of possible outcomes

Number of favourable outcomes P(E) 

Two dice are rolled together, sample space is

(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)

The sample space of getting a sum as a multiple of 3 is (1,2) (1,5) (2,1) (2,4) (3,3) (3,6) (4,2) (4,5) (5,1) (5,4) (6,3) (6,6) . So , probability is 12 /36 =1/3