Question

A box contains 100 tickets numbered 1, 2, 3,...., 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number of them is 5, with probability.
(a)$\dfrac{1}{9}$
(b)$\dfrac{2}{11}$
(c)$\dfrac{3}{19}$
(d) None

Answer 1

Firstly, we will find the total number of the ways in which numbers can be selected from the range 1 to 100 with a condition that none of the numbers is larger than 10. Then, we will find the number of possibilities of both numbers being less than 10 by using the combination formula
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$. Then, we will find the possibilities of the cases in which the minimum number is 5. Then, find the ratio of the number of the cases which has a minimum of 5 to the total possibilities of both the numbers being less than 10 to get the probability.

Complete step-by-step answer :
In this question, we are supposed to find the number of the ways in which the ticket is chosen from the range 1 to 100 so that the number on the both tickets will not be larger than 10.
So, the only possibility of choosing the number for the two tickets must lie in the range 1 to 10.
Now, to get the total number of possibilities for get for getting both the numbers less than 10 is given by the formula of the combination:
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{aligned} & 4!=4\times 3\times 2\times 1 \\\ & \Rightarrow 24 \\\ \end{aligned}$
To get both the numbers less than 10, we substitute n=10 and r=2 as we want to select 2 numbers from the range of 10 in the above formula and get the total possibilities of the cases:
$\begin{aligned} & ^{10}{{C}_{2}}=\dfrac{10!}{\left( 10-2 \right)!2!} \\\ & \Rightarrow \dfrac{10\times 9\times 8!}{8!2!} \\\ & \Rightarrow \dfrac{10\times 9}{2\times 1} \\\ & \Rightarrow \dfrac{90}{2} \\\ & \Rightarrow 45 \\\ \end{aligned}$
So, there are a total of 45 possibilities in which both the numbers are less than 10.
Now, to find the possibilities for getting the minimum number as 5 only. There cases are such as:
(5,6), (5,7), (5,8), (5,9), (5,10)
So, there are 5 such possibilities for getting the minimum number as 5 only.
Now, to get the probability that both the numbers are less than 10 and minimum number is only 5, find the ratio of the number of the cases which has a minimum of 5 to the total possibilities of both the numbers being less than 10.
$\dfrac{5}{45}=\dfrac{1}{9}$
Hence, option (a) is correct.

Note :The common mistakes done by you in this type of the question is that you will forget to take the total number of cases as 45 by combination and consider it as 100 due to the fact that the number range for selection is 1 to 10 according to the condition for two numbers. But if we follow the rule of combination, it rejects the repeating cases as (5,6) and (6,5) is the same for us and doesn’t require two times.