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Question: A random variable x has following probability distribution: Values of x:| 0| 1| 2| 3| 4| 5| 6|...

A random variable x has following probability distribution:

Values of x:012345678
P(x):a3a5a7a9a11a13a15a17a

Determine the value of k, if a=1ka = \dfrac{1}{k} .

Explanation

Solution

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

Complete step-by-step answer:
In a probability distribution the sum of all probabilities is equal to one.
x=0nP(x)=1\Rightarrow \sum\limits_{x = 0}^n {P(x)} = 1
Here, n=8. Therefore,
x=08P(x)=a+3a+5a+7a+9a+11a+13a+15a+17a=1 81a=1 a=181  \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=1ka = \dfrac{1}{k}
181=1k\Rightarrow \dfrac{1}{{81}} = \dfrac{1}{k}
So on comparing k=81k = 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.