Question

an integer is chosen at random b/w 1 to 100 find the probability that it (1) divisible by 8 (2) and 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

Given:

An integer is chosen between 1 and 100.

Numbers that are divisible by 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88 and 96.

The rest are not divisible by 8.

Calculation:

The integers are 2, 3, 4, 5, 6,..., 99.

The number of possible outcomes = 98 (Excluding 1 and 100)

Therefore,  

Sample space, n(S) = 98.  

So,  

The possibility of the number divisible by 8, n(E) = 12.

To find probability,

By formula,

P(E) = n(E)/n(S)  

=12/98  

= 6/49  

Hence, P(integer divisible by 8) = 6/49  

To find the possibility of the number not divisible by 8,

Probability of an integer it is not divisible by 8,  

By formula,

P(E') = 1 - P(E)  

1 - 6/49 = (49 - 6)/49

Hence, P(integer not divisible by 8) = 43/49

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