
The probability function of a random variable X is given by, where c > 0. Find c.
Answer
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Hint: We will use the fact that the probability of a function is the sum of probability of all its outcomes and is always equal to 1.
Complete step-by-step answer:
Here we are given with a function having three outcomes, and probability of occurrence of each outcome is also given.
So, according to the question,
$ \Rightarrow $ Probability of occurrence of ${x_0} = 3c$
$ \Rightarrow $ Probability of occurrence of ${x_1} = 4c$
$ \Rightarrow $ Probability of occurrence of ${x_2} = c - 1$
And, total probability of the function will be = 1
So, Probability of occurrence of ${x_0}$ + Probability of occurrence of \[{x_1}\] + Probability of occurrence of ${x_2}$ = 1
$ \Rightarrow $ (3c) + (4c) + (c – 1) = 1
$ \Rightarrow $ 8c = 2
Hence, c = $\dfrac{1}{4}$
Note: Whenever you come up with this type of problem then we have to remember that the sum of probability of all possible outcomes is equal to 1. Using that condition, we can find the missing quantity.
Complete step-by-step answer:
Here we are given with a function having three outcomes, and probability of occurrence of each outcome is also given.
So, according to the question,
$ \Rightarrow $ Probability of occurrence of ${x_0} = 3c$
$ \Rightarrow $ Probability of occurrence of ${x_1} = 4c$
$ \Rightarrow $ Probability of occurrence of ${x_2} = c - 1$
And, total probability of the function will be = 1
So, Probability of occurrence of ${x_0}$ + Probability of occurrence of \[{x_1}\] + Probability of occurrence of ${x_2}$ = 1
$ \Rightarrow $ (3c) + (4c) + (c – 1) = 1
$ \Rightarrow $ 8c = 2
Hence, c = $\dfrac{1}{4}$
Note: Whenever you come up with this type of problem then we have to remember that the sum of probability of all possible outcomes is equal to 1. Using that condition, we can find the missing quantity.
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