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. What is the probability that the minimum number on them is 5?
$
  {\text{A}}{\text{. }}\dfrac{1}{9} \\
  {\text{B}}{\text{. }}\dfrac{2}{9} \\
  {\text{C}}{\text{. }}\dfrac{3}{9} \\
  {\text{D}}{\text{. }}\dfrac{4}{9} \\
 $

Answer 1

Hint: In order to deal with this question we will use the concept as numbers are to be selected from 1 to 10 because maximum number on the two chosen tickets is not more than 10 so according to it we will calculate total number of outcomes further to calculate favourable number of outcomes we will use the given statement as minimum number on them must be 5 so we have to choose number from 6 to 10 and at last by getting the ratio of favourable outcome by total number of outcomes we will get the required answer.

Complete step-by-step answer:
Given that the maximum number on the two chosen tickets is not more than 10.
It means numbers are to be selected from 1 to 10.
So, total number of outcomes to choose 2 tickets $ = {}^{10}{{\text{C}}_2}$
As we know that the general formula for selecting r things out of a total of n things is given under
Number of ways of selecting r things out of n things $ = {}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Using the above formula, we can write
Total number of outcomes to choose 2 tickets \[ = {}^{10}{{\text{C}}_2} = \dfrac{{10!}}{{2!\left( {10 - 2} \right)!}} = \dfrac{{10 \times 9 \times 8!}}{{2 \times 1!8!}} = \dfrac{{10 \times 9}}{{2 \times 1!}}\]
As we know that the factorial of 1 is given by \[1! = 1\]
$ \Rightarrow $ Total number of outcomes to choose 2 tickets \[ = \dfrac{{10 \times 9}}{{2 \times 1}} = 45\]
As it is given that the minimum number on the tickets is 5
It means maximum number can be any number from 6 to 10
Therefore, number of favourable outcomes = 5
As we know that the general formula for probability of occurrence of any event is given as the ratio of favourable outcomes to the total number of outcomes for any event
Probability of occurrence of any event = $\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}$
By substituting the values in the above formula, we get
Probability of choosing two tickets that the minimum number on them is 5 = $\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes to choose 2 tickets}}}} = \dfrac{5}{{45}} = \dfrac{1}{9}$
Therefore, the required probability is $\dfrac{1}{9}$

So, the correct answer is “Option A”.

Note: -The probability of any event is about how likely that event is to happen. Whenever we are uncertain of an event's result, we should think about the odds of certain outcomes i.e., how probable they are to occur or happen. The term used for the study of probability driven events is called statistics.