A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears a two- digit number.

Hint: As, it is given that total no of discs are numbered 1 to 90 and you can calculate the 2 digit number. Hence, you can find the probability using the formula .\[\dfrac{{Total\,{\rm{ }}number\,{\rm{ }}of\,{\rm{ }}disc\,{\rm{ }}with\,{\rm{ }}two\,{\rm{ }}digit\,{\rm{ }}number\,}}{{total\,{\rm{ }}number\,{\rm{ }}of\,{\rm{ }}discs}}\].

Formula used:
Here, formula to be used is \[{a_n} = a + \left( {n - 1} \right)d\] and Probability (P) = \[\dfrac{{Favourable\,{\rm{ }}outcomes\,}}{{Total\,{\rm{ }}outcomes}}\] = \[\dfrac{{Total\,{\rm{ }}number\,{\rm{ }}of\,{\rm{ }}disc\,{\rm{ }}with\,{\rm{ }}two\,{\rm{ }}digit\,{\rm{ }}number\,}}{{total\,{\rm{ }}number\,{\rm{ }}of\,{\rm{ }}discs}}\]

Complete step-by-step answer:
It is given that the total number of discs = 90
Here, Number of two digit numbers in the discs = 10, 11,...90
As we observe, that difference between the consecutive numbers is the same.
So, these numbers will form an A.P series.
Let First number be a. So, a \[ = \] 10
Let Common difference be d. So, d = \[11 - 10\]
So, here d comes out to be d \[ = \] 1
Let Last number be \[{a_n}\] . So, \[{a_n} = 90\]
Using the formula we get,
 \[{a_n} = a + \left( {n - 1} \right)d\]
Substituting all the values of \[{a_n}\], d and a to calculate the value of n.
$\Rightarrow$ \[90 = 10 + \left( {n - 1} \right)1\]
$\Rightarrow$ \[90 = 10 + n - 1\]
On simplifying we get,
$\Rightarrow$ \[90 = 9 + n\]
Hence, n = 81
 Therefore, Total number of two digit numbers between 1 and 90 is 81.
So, Total number of discs with two digit numbers = 81
P (getting a two digit number) = \[\dfrac{{Total\,{\rm{ }}number\,{\rm{ }}of\,{\rm{ }}disc\,{\rm{ }}with\,{\rm{ }}two\,{\rm{ }}digit\,{\rm{ }}number\,}}{{total\,{\rm{ }}number\,{\rm{ }}of\,{\rm{ }}discs}}\]
As, we have calculated the values of the total number of discs with two digit numbers and the total number of discs is given.
So here we get, P (getting a two digit number) = \[\dfrac{{81}}{{90}}\]
Dividing both numerator and denominator with 9.
Hence, we get the probability of P (getting a two digit number) = \[\dfrac{9}{{10}}\] .


Note: In these types of questions you should calculate a favourable number of outcomes. Then find out the probability using the formula probability = \[\dfrac{{Favourable\,{\rm{ }}outcomes}}{{Total\,{\rm{ }}outcomes}}\] .