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k(x+1)(1/5)^x for x = 0, 1, 2, ...: find k or P(X = 0)

The same question, asked 4 times: 2 May 2023 S1, 11 May 2023 S2, 16 May 2023 S2, 20 Apr 2025 S2. Every wording below, oldest first.

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  1. Question 1

    2 May 2023 S1

    134
    If a discrete random variable X takes values 0,1,2,3,…0, 1, 2, 3, \ldots with probability P(X=x)=k(x+1)5−xP(X=x) = k(x+1)5^{-x}, where kk is a constant, then P(X=0)P(X=0) is

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  2. Question 2

    11 May 2023 S2

    140
    A random variable X takes the values 0,1,2,3,…0,1,2,3,\ldots with probability P(X=x)=k(x+1)(15)xP(X=x) = k(x+1)\left(\frac{1}{5}\right)^x, where k is a constant, then P(X=0)P(X=0) is

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  3. Question 3

    16 May 2023 S2

    121
    If a discrete random variable X is defined as follows: P[X=x]=k(x+1)5xP[X=x] = k\frac{(x+1)}{5^x}, if x=0,1,2,…x = 0,1,2,\ldots, 0 otherwise, then k=k =

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  4. Question 4

    20 Apr 2025 S2

    145
    A random variable X takes the values 0, 1, 2, 3, ____\_\_\_\_ with probability P(X=x)=k(x+1)(15)xP(X =x) = k(x+ 1)\left( \frac{1}{5} \right)^{x}, where k is a constant.Then P(X=0)P(X = 0) is

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