Advertisements
Advertisements
Question
If X is a random variable with probability mass function
P(x) = kx , x=1,2,3
= 0 , otherwise
then , k=..............
(a) 1/5
(b) 1/4
(c) 1/6
(d) 2/3
Solution
(c)
| x | 1 | 2 | 3 |
| P(x) | k | 2k | 3k |
Since, the function is a p.m.f.
∴ ∑P(xi) = 1
∴ k + 2k + 3k = 1
∴ k = 1/6
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Let the p. m. f. (probability mass function) of random variable x be
`p(x)=(4/x)(5/9)^x(4/9)^(4-x), x=0, 1, 2, 3, 4`
=0 otherwise
find E(x) and var (x)
The probability mass function (p.m.f.) of X is given below:
| X=x | 1 | 2 | 3 |
| P (X= x) | 1/5 | 2/5 | 2/5 |
find E(X2)
P(x) = `{(1/"k" ((5),(x))"," (x=0,1,2,3,4,5); k > 0),(0 "," "otherwise"):},`
is p.m.f. of a r.v. X. Then `1/"k"` is equal to
The p.m.f. of an r.v. X is ______
P(x) = `{{:(x^2/"k"";", x = - 4"," - 3"," - 2"," - 1),(0";", "otherwise"):}`
Then, E(X) = ______
