Question

Q02) The probability mass function of a random variable X is zero except at the points
x=0, 1, 2. At these points it as values P(X=0)=c, P(X=1)=2c, P(X=2)=3c. Value of c= ---
(A) 1/3 (B) 1/6 (C) 1/2 (D) other than these values
B​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

The probability mass function of a random variable X is zero except at the points x=0, 1, 2. At these points it as values P(X=0)=c, P(X=1)=2c, P(X=2)=3c.

Value of c=

(A) 1/3

(B) 1/6

(C) 1/2

(D) other than these

EVALUATION

Here it is given that the probability mass function of a random variable X is zero except at the points x=0, 1, 2.

Now at these points it as values P(X=0)=c, P(X=1)=2c, P(X=2)=3c.

Now by the property of probability mass function of a random variable we get

P(X=0) + P(X=1) + P(X=2) = 1

$\implies \displaystyle \sf{c + 2c + 3c = 1}$

$\implies \displaystyle \sf{6c = 1}$

$\implies \displaystyle \sf{c = \frac{1}{6} }$

FINAL ANSWER

Hence the correct option is

$\displaystyle \sf{ (B) \: \: \: \frac{1}{6} }$

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Answer 2

Answer:

(A) 1/3

(B) 1/6

(C) 1/2

(B) is your write answer

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