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Question 20 of 41
Q.

The probability function of a random variable is defined as :

X=xX = x−1-1−2-2001122
P(x)P(x)kk2k2k3k3k4k4k5k5k

Then k is equal to :

  1. 115\frac{1}{15}
  2. zero
  3. one
  4. 14\frac{1}{4}
Puducherry TnboardTamil Nadu HSC (DGE) Commerce Board 2022MCQ· 1mImportance★★★★★
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k=115k = \dfrac{1}{15}.

For a discrete random variable, the total probability must equal 11:

∑P(x)=1.\sum P(x) = 1.

Adding the given probabilities:

k+2k+3k+4k+5k=15k.k + 2k + 3k + 4k + 5k = 15k.

Set the sum equal to 11:

15k=1  ⇒  k=115.15k = 1 \;\Rightarrow\; k = \frac{1}{15}. …

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