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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

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#### Solution

(c)

x | 1 | 2 | 3 |

P(x) | k | 2k | 3k |

Since, the function is a p.m.f.

∴ ∑P(x_{i}) = 1

∴ k + 2k + 3k = 1

∴ k = 1/6

Is there an error in this question or solution?

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