Question

Out of all the two digit numbers from 1 to 60, a two digit number is drawn at random.What is the probability that the number is not divisible by 6 ?

Answer 1

First we have to find how many two digit numbers are there from 1 to 60.

The lowest two digit number will be 10 and the highest will be 60. Thus there are  60 - 10 + 1 = 51  two digit numbers.

Hence the total no. of outcomes is 51.

To get the probability of drawing a two digit number from these 51 ones which is not divisible by 6, deducting the probability of getting a multiple of 6 from the whole probability will be easier.

Let's find the no. of two digit multiples of 6 among these 51 ones.

Consider them in an AP to get it easier.

Lowest two digit multiple of 6 will be 12. So  $\mathsf{a=12}$

Highest two digit multiple of 6 will be 60. Thus  $\mathsf{a_n=60}$

As these are multiples of 6,  $\mathsf{d=6}$

Now,

$\begin{aligned}&\mathsf{n=\frac{a_n-a}{d}+1}\\ \\ \Longrightarrow\ \ &\mathsf{n=\frac{60-12}{6}+1}\\ \\ \Longrightarrow\ \ &\mathsf{n=\frac{48}{6}+1}\\ \\ \Longrightarrow\ \ &\mathsf{n=8+1}\\ \\ \Longrightarrow\ \ &\mathsf{n=9}\end{aligned}$

Hence there are 9 two digit multiples of 6 from 1 to 60.

Thus the probability of getting a two digit multiple of 6 is 9/51 = 3/17.

∴ Probability of getting a two digit number which is not divisible by 6,

$\Longrightarrow\ \mathsf{1-\dfrac{3}{17}}\\ \\ \\ \Longrightarrow\ \mathsf{\dfrac{17-3}{17}}\\ \\ \\ \Longrightarrow\ \large \text{ \mathsf{\dfrac{14}{17}} }$

Hence the answer is 14/17.