Question

An urn contains four tickets with numbers 112, 121, 211, 222 and one ticket is drawn. Let ${{A}_{i}}\text{ }\left( i=1,2,3 \right)$ be the event that the ${{i}^{th}}$ digit if the tickets drawn is 1. Discuss the independence of the events ${{A}_{1}},{{A}_{2}},{{A}_{3}}$. If they are independent enter 1, else enter 0.

Answer 1

We solve this problem by going through the condition for two events to be independent. Then we find the probability of events ${{A}_{1}},{{A}_{2}},{{A}_{3}}$ using the formula for probability, the probability of selecting n objects from a set of N objects is equal to $\dfrac{n}{N}$. Then we check if the events ${{A}_{1}},{{A}_{2}},{{A}_{3}}$ are independent using the condition for the independence of two events $P\left( A\cap B \right)=P\left( A \right)P\left( B \right)$. Then we enter the value 0 or 1 based on the answer we get for their independence.

Complete step by step answer:
Let us start by discussing the concept of independence. Two events A and B are said to be independent if $P\left( A\cap B \right)=P\left( A \right)P\left( B \right)$.
Given that there are four tickets with numbers 112, 121, 211, 222 and one ticket is drawn from the urn.
So, the number of tickets are 4.
Now, let us consider the event ${{A}_{1}}$, that is the first digit of the ticket drawn is 1.
Then we get $P\left( {{A}_{1}} \right)=\dfrac{\text{Number of tickets with first digit 1}}{\text{Total number of tickets}}$
$\Rightarrow P\left( {{A}_{1}} \right)=\dfrac{2}{4}=\dfrac{1}{2}$
Now, let us consider the event ${{A}_{2}}$, that is the second digit of the ticket drawn is 1.
Then we get $P\left( {{A}_{2}} \right)=\dfrac{\text{Number of tickets with second digit 1}}{\text{Total number of tickets}}$
$\Rightarrow P\left( {{A}_{2}} \right)=\dfrac{2}{4}=\dfrac{1}{2}$
Now, let us consider the event ${{A}_{3}}$, that is the third digit of the ticket drawn is 1.
Then we get $P\left( {{A}_{3}} \right)=\dfrac{\text{Number of tickets with third digit 1}}{\text{Total number of tickets}}$
$\Rightarrow P\left( {{A}_{3}} \right)=\dfrac{2}{4}=\dfrac{1}{2}$
As we need to find the independence of the events ${{A}_{1}},{{A}_{2}},{{A}_{3}}$.
Now, let us look at the events ${{A}_{1}}$ and ${{A}_{2}}$.
$P\left( {{A}_{1}}\cap {{A}_{2}} \right)=\dfrac{\text{Number of tickets with first and second digits 1}}{\text{Total number of tickets}}$
$\Rightarrow P\left( {{A}_{1}}\cap {{A}_{2}} \right)=\dfrac{1}{4}$
While $P\left( {{A}_{1}} \right)P\left( {{A}_{2}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}$.
So, $P\left( {{A}_{1}}\cap {{A}_{2}} \right)=P\left( {{A}_{1}} \right)P\left( {{A}_{2}} \right)$, that is why they are independent.

Now, let us look at the events ${{A}_{2}}$ and ${{A}_{3}}$.
$P\left( {{A}_{2}}\cap {{A}_{3}} \right)=\dfrac{\text{Number of tickets with second and third digits 1}}{\text{Total number of tickets}}$
$\Rightarrow P\left( {{A}_{2}}\cap {{A}_{3}} \right)=\dfrac{1}{4}$
While $P\left( {{A}_{2}} \right)P\left( {{A}_{3}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}$.
So, $P\left( {{A}_{2}}\cap {{A}_{3}} \right)=P\left( {{A}_{2}} \right)P\left( {{A}_{3}} \right)$, that is why they are independent.
Now, let us look at the events ${{A}_{1}}$ and ${{A}_{3}}$.
$P\left( {{A}_{1}}\cap {{A}_{3}} \right)=\dfrac{\text{Number of tickets with first and third digits 1}}{\text{Total number of tickets}}$
$\Rightarrow P\left( {{A}_{1}}\cap {{A}_{3}} \right)=\dfrac{1}{4}$
While $P\left( {{A}_{1}} \right)P\left( {{A}_{3}} \right)=\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}$.
So, $P\left( {{A}_{1}}\cap {{A}_{3}} \right)=P\left( {{A}_{1}} \right)P\left( {{A}_{3}} \right)$, that is why they are independent.
As we are asked to enter 1 if they are independent and 0 if not, so we need to enter 1.
As they are independent, we need to enter 1.

So, the correct answer is 1.

Note: There is a possibility of making a mistake while solving this problem by confusing the event ${{A}_{i}}$ with selecting the ${{i}^{th}}$ ticket instead of that the ${{i}^{th}}$ digit of the selected ticket is 1. So, one needs to read the question carefully and understand it.