Question

Two different dice are thrown at the same time . Find the probability that the sum of the two number appearing on the top of the dice is 7

Answer 1

Answer:

1/6

Solution:

Here ,

The experiment is throwing of two dice at the same time .

Here,

The sample space for the experiment will be :

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) }

Here,

The number of elements in sample space S or the number of total outcomes will be ;

n(S) = 6² = 36.

Let ,

E be the event of getting a sum of 7 .

Thus ,

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

Here,

The number of elements in E or the number of favourable outcomes will be ;

n(E) = 6

We know that ,

(no. of favourable outcomes)

Probability = ———————————————

(no. of total outcomes)

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

=> P(E) = 6/36

=> P(E) = 1/6

Hence,

The probability of getting a sum of 7 is 1/6 .

Answer 2

Answer:

the answer is 1/6

please let me know for any other quires I look forward to do clear then Thanks

Step-by-step explanation:

as we know we have 1 to 6 numbers on dice

so total probability or sample space of two dice is 6*6=36

now let us check the combinations of getting sum as 7

1+6=7

2+5=7

3+4=7 and we get vice versa so total combinations of getting sum as 7 is 6

=>probability of getting sum as 7= 6/36=>1/6