Question

Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is (i)8 , (ii)13, (iii)less than or equal to 12.

Answer 1

Answer:

hilu ❤️

Step-by-step explanation:

ANSWER

(i) 8

Total number of outcomes =36

Probability that sum of two numbers is 8 =

Total no. of outcomes

No. of outcomes where sum is 8

=

36

5

(ii) Less than or equal to 12

Total number of outcomes =36

Probability that sum of two numbers is less than or equal to 12 =

Total no. of outcomes

No. of outcomes where sum is less than or equal to 12

=

36

36

=1

solution

Answer 2

Answers:

(i) 5/36

(ii) 10

(iii) 1

Solution:

Given:

Two dice are thrown at the same time.

So,  

We get total possible outcomes = 6 × 6 = 36

(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)

(i) The sum of the two numbers appearing on the top of the dice is 8.

Favorable outcomes are - (2,6), (6,2) (3,5) (5,3) and (4,4)

Therefore, the number of favorable outcomes = 5

So,

P(getting sum of number 8) = $\frac{no.\ favorable\ outcome}{no.\ of\ possible\ outcome}$ = $\frac{5}{36}$

Now,

(ii) the sum of the two numbers appearing on the top of the dice is 13.

Here, there are no favorable outcomes as the maximum number can be 12.

Therefore,

P(getting sum of numbers 12) = $\frac{no.\ favorable\ outcome}{no.\ of\ possible\ outcome}$ = $\frac{0}{36}$ = 0

And,

(iii) The sum of two numbers appearing on the top of the dice is less than or equal to 12.

Here,

The favorable outcomes are all 36 outcomes

Hence,

P(getting sum of numbers less than or equal to 12) = $\frac{no.\ favorable\ outcome}{no.\ of\ possible\ outcome}$ = $\frac{36}{36}$ = 1