Question

an integer is chosen at random between 1 and 100 find the probability that it is
i) divisible by 8
ii) not divisible by 8

Answer 1

In the question it is Given that the  numbers are "between 1 and 100", we should exclude both 1 and 100.So we are left with 98 numbers.

i) Total numbers BETWEEN 1 and 100 are 98
Total Number of Outcomes= 98
Number divisible by 8 are : 8,16,24,32,49,48,56,64,72,80,88,96
Number of favourable outcomes = 12
Probability = Number of favourable outcomes / Total number of outcomes
Probability (divisible by 8) = 12/98 = 6/49
P((divisible by 8) = 6/49

ii) Number not divisible by 8 are : 98 -12 = 86
Number of favourable outcomes = 86
Probability = Number of favourable outcomes / Total number of outcomes
Probability (not divisible by 8) = 86/98 = 43/49
P((not divisible by 8) = 43/49

HOPE THIS WILL HELP YOU….

Answer 2

Find the number of terms that can be divided by 8:

In numbers from 1 to 100 :

• The first number that can be divide by 8 is 8 ( ⇒  a1 = 8)
• The last number that can be divided by 8 is 96 ( ⇒ an = 96)
• It is multiples of 8 ( ⇒ d = 8)

an = a1 + ( n - 1) d

96 = 8 + (n - 1)8

96 = 8 + 8n - 8

8n = 96

n = 96 ÷ 8

n = 12

Find the probability that a number chosen can be divisible by 8:

Total number of numbers between 1 and 100 = 100 - 2 = 98

Total number of numbers that can be divided by 8 = 12

P(divisible by 8) = 12/98 = 6/49

Find the probability that a number chosen cannot be divisible by 8:

P(divisible by 8) = 12/98 = 6/49

P(not divisible by 8) = 1 - 6/49 = 43/49

Answer: (i) 6/49 (ii} 43/49