# You have the numbers 1-24 written on a slip of paper. If you chose one slip at random what is the probability that you will not select a number which is divisible by 6?

May 17, 2018

The probability is $\setminus \frac{5}{6}$

#### Explanation:

Let A be the event of selecting a number divisible by 6 and B be the event of selecting a number not divisible by 6:
$P \left(A\right) = \setminus \frac{1}{6}$
$P \left(B\right) = P \left(\neg A\right) = 1 - P \left(A\right)$
$= 1 - \setminus \frac{1}{6} = \setminus \frac{5}{6}$

In general, if you have n slips of paper numbered 1 to N (where N is a big positive integer say 100) the probability of selecting a number divisible by 6 is ~1/6 and if N is exactly divisible by 6, then the probability is exactly 1/6
i.e.

$P \left(A\right) = \setminus \frac{1}{6} \iff N \equiv 0 \mod 6$

if N is not divisible exactly by 6 then you would calculate the remainder, for example if N = 45:
$45 \equiv 3 \mod 6$
(6*7 = 42, 45-42 = 3, the remainder is 3)

The greatest number less than N that is divisible by 6 is 42,
and $\because \setminus \frac{42}{6} = 7$ there are 7 numbers divisible between 1 to 45
and they would be $6 \cdot 1 , 6 \cdot 2 , \ldots 6 \cdot 7$

if you instead chose 24 there would be 4: and they would be 61,62, 63,64 = 6,12,18,24

Thus the probability of choosing a number divisible by 6 between 1 and 45 is $\setminus \frac{7}{45}$ and for 1 to 24 this would be $\setminus \frac{4}{24} = \setminus \frac{1}{6}$

and the probability of choosing a number not divisible by 6 would be the complement of that which is given by $1 - P \left(A\right)$
For 1 to 45 it would be: $1 - \setminus \frac{7}{45} = \setminus \frac{38}{45}$
For 1 to 24 it would be: $1 - \setminus \frac{1}{6} = \setminus \frac{5}{6}$