Question

# A random variable x has following probability distribution:Values of x:012345678P(x):a3a5a7a9a11a13a15a17aDetermine the value of k, if $a = \dfrac{1}{k}$ .

Hint: Here we will apply the property of the probability i.e. Sum of all probabilities is equal to one.

In a probability distribution the sum of all probabilities is equal to one.
$\Rightarrow \sum\limits_{x = 0}^n {P(x)} = 1$
Here, n=8. Therefore,
$\Rightarrow \sum\limits_{x = 0}^8 {P(x)} = a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1 \\ \Rightarrow 81a = 1 \\ \Rightarrow a = \dfrac{1}{{81}} \\$
Now it is given that $a = \dfrac{1}{k}$
$\Rightarrow \dfrac{1}{{81}} = \dfrac{1}{k}$
So on comparing $k = 81$.

Note: In such types of questions the key concept we have to remember is that the sum of the probability distribution is always 1 so, simply add all the probabilities and equate to 1.Probability for a particular value or range of values must be between 0 and 1.