Question

Let n be a random number between 1 and 100000 chosen with uniform probability. Compute

a) The probability that n can be divided by 3

b) The probability that n can be divided by 6

c) The probability that n can be divided by 9

d) The probability that n can be divided by 9 given that in can be divided by 6

e) The probability that n can be divided by 6 given that in can be divided by 9

Answer 1

I am assuming that 1 and 100000 are included in range. So total numbers = 100000

a.)

First number divisible by 3 = 3

Last number divisible by 3 = 99999

common difference between any two numbers divisible by 3 = 3

So, a_n = a + (n-1)*d

99999 = 3 + (n-1) * 3

99996 = (n-1)*3

33332 = n - 1

n = 33333

So, Probability that n can be divided by 3 = 33333/100000 = 0.33333

b.)

First number divisible by 6 = 6

Last number divisible by 6 = 99996

common difference between any two numbers divisible by 6 = 6

So, a_n = a + (n-1)*d

99996 = 6 + (n-1) * 6

99990 = (n-1) * 6

16665 = n - 1

n = 16666

So, Probability that n can be divided by 6 = 16666/100000 = 0.16666

c.)

First number divisible by 9 = 9

Last number divisible by 9 = 99999

common difference between any two numbers divisible by 9 = 9

So, a_n = a + (n-1)*d

99999 = 9 + (n-1) * 9

99990 = (n-1) * 9

11110 = n - 1

n = 11111

So, Probability that n can be divided by 9 = 11111/100000 = 0.11111

d.)

Given that number is divisible by 6, so the series is:

6, 12, 18, 24, 30, 36, 42, 48, 54, ....... 99990, 99996

So from above series numbers which are divisible by 9 are:

18, 36, 54, 72, ..... 99990

Total numbers in above series:

99990 = 18 + (n-1)*18

99972 = (n-1)*18

n-1 = 5554

n = 5555

This is the case of conditional probabaility.

The formula is: P(9|6) = P(6 9) / P(6)

So, probability that n can be divided by 9 given that it can be divided by 6 = P(9|6) = (5555 / 100000) / (16666 / 100000)

P(9|6) = 5555 / 16666 = 0.3333