An integer is chosen at random between 1 and 100. Find the probability that it is i)Divisible by 8ii)Not divisible by 8

Let's solve this question step by step.

The integer is chosen randomly between 1 and 100, so the total number of possible outcomes is 100.

Part (i): Probability that the number is divisible by 8
First, we need to determine how many integers between 1 and 100 are divisible by 8. These numbers are the multiples of 8 in this range.

The first multiple of 8 is 8, and the last multiple of 8 within 100 is 96. The multiples of 8 between 1 and 100 are:

8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96.

These numbers form an arithmetic sequence where the first term is 8, the common difference is 8, and the last term is 96.

To find the number of terms in this sequence, we use the formula for the nth term of an arithmetic sequence: a_n = a_1 + (n-1) * d, where a_1 is the first term, d is the common difference, and a_n is the nth term.

In this case, a_1 = 8, d = 8, and a_n = 96.

So, 96 = 8 + (n-1) * 8.

Solving for n:

96 - 8 = (n-1) * 8 88 = (n-1) * 8 n - 1 = 88 / 8 n - 1 = 11 n = 12.

So, there are 12 numbers between 1 and 100 that are divisible by 8.

The probability that a randomly chosen integer is divisible by 8 is the number of favorable outcomes (12) divided by the total number of possible outcomes (100): P(divisible by 8) = 12 / 100 = 3 / 25.

Part (ii): Probability that the number is not divisible by 8
To find the probability that a number is not divisible by 8, we subtract the probability that the number is divisible by 8 from 1.

P(not divisible by 8) = 1 - P(divisible by 8) = 1 - 3 / 25 = 22 / 25.

Final Answer:
i) The probability that the number is divisible by 8 is 3/25.

ii) The probability that the number is not divisible by 8 is 22/25.