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A random variable x has following probability distribution:

Values of x: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

P(x): | a | 3a | 5a | 7a | 9a | 11a | 13a | 15a | 17a |

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

Answer
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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.

$ \Rightarrow \sum\limits_{x = 0}^n {P(x)} = 1$

Here, n=8. Therefore,

$

Â Â Â \Rightarrow \sum\limits_{x = 0}^8 {P(x)} = a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1 \\

Â Â Â \Rightarrow 81a = 1 \\

Â Â Â \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.

Complete step-by-step answer:

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,

$

Â Â Â \Rightarrow \sum\limits_{x = 0}^8 {P(x)} = a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1 \\

Â Â Â \Rightarrow 81a = 1 \\

Â Â Â \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.

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