Question

Two different digits are chosen at random from the set 1, 2, 3, 4, 5, 6, 7, 8. Find the

probability that sum of two digits exceeds 13.

Answer 1

Answer:

6,8

7,8

only this two pairs can make sum exceed 13

so, 2/8 is the probability

so

1/4

25%

Answer 2

Answer:

$\frac{1}{14}$ is the required probability that sum of two digit exceeds 13

Step-by-step explanation:

Explanation:

Given , numbers are , 1 ,2 ,3 ,4 ,5 ,6 ,7 ,8

Two digit are chosen randomly  from the given numbers , so that  their sum exceeds 13 .

Therefore , we have (6,8) and (7 , 8) whose sum is more than 13 .

Step 1:

Total  possible out come is 2

Total  number of  favourable out come  = $8C_{2}$ = $\frac{8!}{(8-2)! 2! }$ = 28

Probability that sum of two digits exceed 13 = $\frac{2}{28} = \frac{1}{14}$

Final answer:

Hence , the  Probability that sum of two digits exceed 13 is  $\frac{1}{14}$

#SPJ3