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Question: A ticket is drawn from a bag containing \(100\) tickets. The tickets are numbered from one to hundre...

A ticket is drawn from a bag containing 100100 tickets. The tickets are numbered from one to hundred. What is the probability of getting a ticket with a number divisible by 1010?
A. 110\dfrac{1}{{10}}
B. 15\dfrac{1}{5}
C. 120\dfrac{1}{{20}}
D. None

Explanation

Solution

To find the probability, first we will write the sample space SS for tickets. Then, we will consider the event EE that gets a ticket with a number divisible by 1010. We will find the required probability by using the definition. That is, required probability =n(E)n(S) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}} where n(E)n\left( E \right) is the number of favourable (desired) outcomes and n(S)n\left( S \right) is the number of total outcomes.

Complete step-by-step solution:
In this problem, it is given that a bag containing 100100 tickets and the tickets are numbered from one to hundred. So, one can get a ticket with a number between 11 and 100100. The sample space is the set of all possible outcomes. Therefore, in this problem sample space for tickets is S = \left\\{ {1,2,3,...,100} \right\\}. Therefore, n(S)=100n\left( S \right) = 100.
Now we will consider the event EE that gets a ticket with a number divisible by 1010. That is, our desired outcome is a number divisible by 1010. We know that between 11 and 100100, the numbers 10,20,30,40,50,60,70,80,90,10010,20,30,40,50,60,70,80,90,100 are divisible by 1010. We can see that these are 1010 numbers. Therefore, n(E)=10n\left( E \right) = 10.
Now we are going to find the probability of an event EE by using the definition. That is, P(E)=n(E)n(S)P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}.
P(E)=10100\Rightarrow P\left( E \right) = \dfrac{{10}}{{100}}
P(E)=110\Rightarrow P\left( E \right) = \dfrac{1}{{10}}
Hence, the probability that getting a ticket with a number divisible by 1010 is 110\dfrac{1}{{10}}.

Hence, option A is the correct answer.

Note: In this problem, if we have to find the probability that getting a ticket with a number whose unit digit 00 then the answer will be the same. That is, probability of getting a ticket with a number whose unit digit 00 is also 110\dfrac{1}{{10}}. In this problem, if we have to find the probability of getting a ticket with a multiple of 1010 then the answer will be the same. That is, the probability of getting a ticket with a number multiple of 1010 is also 110\dfrac{1}{{10}}.