Question

From 10,000 lottery tickets numbered from 1 to 10,000 one ticket is drawn at random. What is the probability that the number marked on the drawn ticket is divisible by 20.
A) \[\dfrac{1}{{100}}\]
B) \[\dfrac{1}{{50}}\]
C) \[\dfrac{1}{{20}}\]
D) \[\dfrac{1}{{10}}\]

Answer 1

Hint: Here the given question is based on the concept of probability. We have to find the probability of choosing a lottery ticket on which the number is divisible by 20. For this, first we need to find the total number of numbers which is divisible by 20 between 1 - 10000 then by using the definition of probability and on further simplification we get the required probability of choosing a ticket.

Complete step by step answer:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
\[\text{Probability of event to happen } P\left( E \right) = \dfrac{{\text{Number of favourable outcomes}}}{{\text{Total Number of outcomes}}}\]
Consider the given question:
One ticket drawn randomly in 10,000 lottery tickets.
Now, we have to find the probability that the number marked on the drawn ticket is divisible by 20.
Total number of lottery tickets \[ = {\,^{10000}}{C_1} = 10,000\]
Favourable cases \[ = \left\{ {20,40,60,80,....,10000} \right\}\]
The sequence, \[20,40,60,80,....,10000\] is in A.P.
To find the total number of favourable cases by using a formula \[{a_n} = a + \left( {n - 1} \right)d\]
Here, \[a = 20\], \[{a_n} = 10000\], \[d = 20\], then we have
\[ \Rightarrow \,\,\,10000 = 20 + \left( {n - 1} \right)20\]
\[ \Rightarrow \,\,\,10000 = 20 + 20n - 20\]
On simplification, we get
\[ \Rightarrow \,\,\,10000 = 20n\]
Divide both side by 20
\[ \Rightarrow \,\,\,n = \dfrac{{10000}}{{20}}\]
\[ \Rightarrow \,\,\,n = 500\]
Therefore, the number of favourable outcomes \[ = \left\{ {20,40,60,80,....,10000} \right\} = 500\]
By the definition of probability
\[ \Rightarrow \,\,P\left( {Ticket\,divisible\,by\,20} \right) = \dfrac{{Number\,of\,favourable\,outcomes}}{{Total{\text{ }}numbere{\text{ }}of\,lottery\,tickets}}\]
\[ \Rightarrow \,\,P\left( {Ticket\,divisible\,by\,20} \right) = \dfrac{{500}}{{10000}}\]
On simplification, we get
\[\therefore \,\,P\left( {Ticket\,divisible\,by\,20} \right) = \dfrac{1}{{20}}\]
This is the required solution.
Therefore, option (C) is the correct option.

Note:
The probability is a number of possible values. If the sequence in A.P. we have to use a A.P formula \[{a_n} = a + \left( {n - 1} \right)d\] to find total number \[n\], where \[a\] and \[{a_n}\] is the first and last term of sequence and common difference \[d = {a_2} - {a_1}\] and must student know to use permutation combination concept to solve the given problem because it is the first and main thing to solve the problem.