Question

If the random variables X takes the values ${x_1}$ , ${x_2}$ , ${x_3}$ ,………., ${x_{10}}$ with probabilities $P\left( {X = {x_i}} \right) = ki$ , then the value of $k$ is
$1){\text{ }}\dfrac{1}{{10}}$
$2){\text{ }}\dfrac{1}{4}$
$3){\text{ }}\dfrac{1}{{55}}$
$4){\text{ }}\dfrac{7}{{12}}$
$5){\text{ }}\dfrac{3}{4}$

Answer 1

Hint : First try to understand that this is a question of probability and we know that the maximum value of probability of any event is $1$ and minimum value is $0$ . Ignore those options which are $>$ $1$ and $<$ $0$. But in this question all the options satisfy these conditions. Mainly general terms contain $'i'$ in their expressions but the given range in this question is 1 to 10 .Use the given and known conditions like here we have to do the sum of the probabilities of different random variables. So use given conditions to find the answer. Like first find the probability of every value of X. Then use the condition , $\sum\nolimits_x {P\left( {X = x} \right) = 1}$ .Then put the values in the formula $P\left( {X = {x_1}} \right) + P\left( {X = {x_2}} \right) + P\left( {X = {x_3}} \right) + .............. + P\left( {X = {x_{10}}} \right) = 1$ and find the value of $k$ .

Complete step-by-step answer :
Random variable $X$ is given. Where $X = {x_i}$ and ${x_i} = {x_1},{x_2},{x_3},........,{x_{10}}$ .
It is also given that $P\left( {X = {x_i}} \right) = ki$
Therefore the probability of ${x_1}$ $= P\left( {{x_1}} \right) = k$ ,
The probability of ${x_2}$ $= P\left( {{x_2}} \right) = 2k$ ,
The probability of ${x_3}$ $= P\left( {{x_3}} \right) = 3k$ ,
The probability of ${x_4}$ $= P\left( {{x_4}} \right) = 4k$ ,
The probability of ${x_5}$ $= P\left( {{x_5}} \right) = 5k$ ,
The probability of ${x_6}$ $= P\left( {{x_6}} \right) = 6k$ ,
The probability of ${x_7}$ $= P\left( {{x_7}} \right) = 7k$ ,
The probability of ${x_8}$ $= P\left( {{x_8}} \right) = 8k$ ,
The probability of ${x_9}$ $= P\left( {{x_9}} \right) = 9k$ ,
The probability of ${x_{10}}$ $= P\left( {{x_{10}}} \right) = 10k$
As we know that the sum of probabilities distribution is $1$ . Therefore, we can write
$P\left( {X = {x_1}} \right) + P\left( {X = {x_2}} \right) + P\left( {X = {x_3}} \right) + .............. + P\left( {X = {x_{10}}} \right) = 1$
Put the probability values in the above equation ,
$k + 2k + 3k + 4k + 5k + 6k + 7k + 8k + 9k + 10k = 1$
By taking $k$ common we get
$k\left( {1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10} \right) = 1$
$\because$ $1 + 2 + 3 + ........... + n = \dfrac{{n\left( {n + 1} \right)}}{2}$ . So, by using this in the above equation we get
$\dfrac{{10\left( {10 + 1} \right)}}{2}k = 1$
$\dfrac{{10\left( {11} \right)}}{2}k = 1$
Further simplifying we get,
$\dfrac{{110}}{2}k = 1$
$k = \dfrac{2}{{110}}$
So finally the value of $k = \dfrac{1}{{55}}$
Hence the correct option is $3){\text{ }}\dfrac{1}{{55}}$
So, the correct answer is “Option 3”.

Note : A random variable X is said to be discrete if it can assume only a finite or countable infinite number of distinct values. We may also denote $P\left( {X = x} \right)$ by $p\left( x \right)$ or by ${p_X}\left( x \right)$ . The expression $P\left( {X = x} \right)$ is a function that assigns the probabilities to each value $x$ ; therefore this function is called the probability function for the random variable X. Try to remember all the basic properties of probability.