Question

The probability distribution of a random variable $X$ is given below:

$X$:	0	1	2	3
$P\left( X \right)$:	$k$	$\dfrac{k}{2}$	$\dfrac{k}{4}$	$\dfrac{k}{8}$

Determine $P\left( {X \le 2} \right)$ and $P\left( {X > 2} \right)$

Answer 1

Here, we are required to find the given probabilities when the probability distribution of a random variable $X$ is already given. We will use the fact that the sum of all the probabilities of an event is equal to 1. With this, we will find the value of the variable $k$ and substituting it in the given table, we will be able to find the required probabilities.

Complete step-by-step answer:
In an experiment, the sum of all the possible cases or the sum of all the possible outcomes is always equal to 1.
Since the probability distribution of a random variable $X$ is given.
We will add all the given events and their sum will be equal to 1.
$k + \dfrac{k}{2} + \dfrac{k}{4} + \dfrac{k}{8} = 1$
Taking the LCM on the RHS, we get
$\Rightarrow 8k + 4k + 2k + k = 8$
Adding the like terms, we get
$\Rightarrow 15k = 8$
Dividing both sides by 15, we get
$\Rightarrow k = \dfrac{8}{{15}}$
We found out the volume of $k$ and now we will substitute it in the table to find the respective probabilities.
So, when the variable $X = 0$,
$P\left( 0 \right) = k = \dfrac{8}{{15}}$…………………………$\left( 1 \right)$
When $X = 1$,
$P\left( 1 \right) = \dfrac{k}{2}$
Substituting $k = \dfrac{8}{{15}}$ in the above equation, we get
$P\left( 1 \right) = \dfrac{8}{{15 \times 2}}$
Multiplying the terms in the denominator, we get
$P\left( 1 \right) = \dfrac{4}{{15}}$…………………………….$\left( 2 \right)$
When, $X = 2$,
$P\left( 2 \right) = \dfrac{k}{4}$
Substituting $k = \dfrac{8}{{15}}$ in the above equation, we get
$P\left( 2 \right) = \dfrac{8}{{15 \times 4}}$
Multiplying the terms in the denominator, we get
$P\left( 2 \right) = \dfrac{2}{{15}}$…………………………..$\left( 3 \right)$
And, when $X = 3$,
$P\left( 3 \right) = \dfrac{k}{8}$
Substituting $k = \dfrac{8}{{15}}$ in the above equation, we get
$P\left( 3 \right) = \dfrac{8}{{15 \times 8}}$
Multiplying the terms in the denominator, we get
$P\left( 3 \right) = \dfrac{1}{{15}}$…………………….$\left( 4 \right)$
Now, we have to find $P\left( {X \le 2} \right)$.
This means that we have to find the probability of the variable $X$ being either 2 or less than 2.
Hence, this can be written as:
$P\left( {X \le 2} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right) + P\left( {X = 2} \right)$
$\Rightarrow P\left( {X \le 2} \right) = P\left( 0 \right) + P\left( 1 \right) + P\left( 2 \right)$
Substituting the values from $\left( 1 \right)$, $\left( 2 \right)$ and $\left( 3 \right)$, we get
$\Rightarrow P\left( {X \le 2} \right) = \dfrac{8}{{15}} + \dfrac{4}{{15}} + \dfrac{2}{{15}}$
Since the denominators are same, hence, adding the numerators, we get
$\Rightarrow P\left( {X \le 2} \right) = \dfrac{{8 + 4 + 2}}{{15}} = \dfrac{{14}}{{15}}$
We have to also find $P\left( {X > 2} \right)$.
This means that we have to find the probability of the variable $X$ being greater than 2.
$\Rightarrow P\left( {X > 2} \right) = P\left( {X = 3} \right) = P\left( 3 \right)$
Hence, substituting the value from $\left( 4 \right)$, we get,
$\Rightarrow P\left( {X > 2} \right) = \dfrac{1}{{15}}$
Therefore, the required values of $P\left( {X \le 2} \right)$ and $P\left( {X > 2} \right)$ are $\dfrac{{14}}{{15}}$ and $\dfrac{1}{{15}}$ respectively.
Hence, this is our required answer.

Note: In this question, the value of the variable $k$ plays the most vital role as if its value is wrong then our whole question will become incorrect. Also, we should know that if there is given a ‘greater than sign’ or $>$, then it means that we have to take the values larger than that particular number. Whereas, a ‘greater than equal to sign’ or $\ge$ means that we have to take the values which are not only greater than that particular number but also that number itself should be included.