Question

If the probability distribution of $ X $ is

$ X $	$ 0 $	$ 1 $	$ 2 $	$ 3 $	$ 4 $	$ 5 $	$ 6 $
$ P\left( X \right) $	$ k $	$ 3k $	$ 5k $	$ 7k $	$ 9k $	$ 11k $	$ 13k $

Then the values of $ P\left( x\ge 2 \right) $ and $ P\left( 0 < x < 4 \right) $ are
A. $ \dfrac{1}{49} $ , $ \dfrac{45}{49} $
B. $ \dfrac{45}{49} $ , $ \dfrac{15}{49} $
C. $ \dfrac{15}{49} $ , $ \dfrac{34}{49} $
D. None of these.

Answer 1

Hint: In this problem all the probabilities are given in terms of $ k $ . So, we need to calculate the value of $ k $ firstly. To calculate the value of $ k $ , we will use the probability rule i.e. the sum of the all probabilities in a given sample space is equal to one. Mathematically $ \sum{P\left( X \right)=1} $ . From this equation we will obtain the value of $ k $ . After calculating the value of $ k $ , we need to calculate the probabilities of different events they are $ P\left( x\ge 2 \right) $ and $ P\left( 0 < x < 4 \right) $ . For calculating $ P\left( x\ge 2 \right) $ , we will add the probabilities of $ x=2,3,4,5,6 $ and for calculating $ P\left( 0 < x < 4 \right) $ we will add the probabilities of $ x=1,2,3 $ .

Complete step by step answer:
Given that,
The probability distribution of $ X $ is given by

$ X $	$ 0 $	$ 1 $	$ 2 $	$ 3 $	$ 4 $	$ 5 $	$ 6 $
$ P\left( X \right) $	$ k $	$ 3k $	$ 5k $	$ 7k $	$ 9k $	$ 11k $	$ 13k $

We have the rule of probability as the sum of the probabilities in a sample space is one.
 $ \begin{align}
  & \Rightarrow \sum{P\left( X \right)=1} \\
 & \Rightarrow P\left( 0 \right)+P\left( 1 \right)+P\left( 2 \right)+P\left( 3 \right)+P\left( 4 \right)+P\left( 5 \right)+P\left( 6 \right)=1 \\
 & \Rightarrow k+3k+5k+7k+9k+11k+13k=1 \\
 & \Rightarrow 49k=1 \\
 & \Rightarrow k=\dfrac{1}{49} \\
\end{align} $
Now the value of $ P\left( x\ge 2 \right) $ can be obtained by adding the probabilities of $ x=2,3,4,5,6 $ .
 $ \begin{align}
  & \therefore P\left( x\ge 2 \right)=P\left( X=2 \right)+P\left( X=3 \right)+P\left( X=4 \right)+P\left( X=5 \right)+P\left( X=6 \right) \\
 & \Rightarrow P\left( x\ge 2 \right)=5k+7k+9k+11k+13k \\
 & \Rightarrow P\left( x\ge 2 \right)=45k \\
\end{align} $
Substituting the value of $ k=\dfrac{1}{49} $ in the above equation, then we will have
 $ \begin{align}
  & \therefore P\left( x\ge 2 \right)=45\times \dfrac{1}{49} \\
 & \Rightarrow P\left( x\ge 2 \right)=\dfrac{45}{49} \\
\end{align} $
Now the value of $ P\left( 0 < x < 4 \right) $ can be obtained by adding the probabilities of $ x=1,2,3 $ .
 $ \begin{align}
  & \Rightarrow P\left( 0 < x < 4 \right)=P\left( X=1 \right)+P\left( X=2 \right)+P\left( X=3 \right) \\
 & \Rightarrow P\left( 0 < x < 4 \right)=3k+5k+7k \\
 & \Rightarrow P\left( 0 < x < 4 \right)=15k \\
\end{align} $
Substituting the value of $ k=\dfrac{1}{49} $ in the above equation, then we will have
 $ \begin{align}
  & P\left( 0 < x < 4 \right)=15\times \dfrac{1}{49} \\
 & \Rightarrow P\left( 0 < x < 4 \right)=\dfrac{15}{49} \\
\end{align} $
Finally the values of $ P\left( x\ge 2 \right) $ and $ P\left( 0 < x < 4 \right) $ are $ \dfrac{45}{49} $ and $ \dfrac{15}{49} $ respectively.

Note:
We can also find the value of $ P\left( x\ge 2 \right) $ in another method. i.e.
 $ \begin{align}
  & P\left( x\ge 2 \right)=1-\left[ P\left( X=0 \right)+P\left( X=1 \right) \right] \\
 & \Rightarrow P\left( x\ge 2 \right)=1-\left( k+3k \right) \\
 & \Rightarrow P\left( x\ge 2 \right)=1-4k \\
\end{align} $
Substituting the value of $ k=\dfrac{1}{49} $ in the above equation, then we will have
 $ \begin{align}
  & \Rightarrow P\left( x\ge 2 \right)=1-\dfrac{4}{49} \\
 & \Rightarrow P\left( x\ge 2 \right)=\dfrac{49-4}{49} \\
 & \Rightarrow P\left( x\ge 2 \right)=\dfrac{45}{49} \\
\end{align} $
From both the methods we got the same result.