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Table 0, k, 2k, 2k, 3k, k^2, 2k^2, 7k^2 + k: find P(X >= 6)

The same question, asked 2 times: 16 May 2023 S1, 25 Apr 2025 S1. Every wording below, oldest first.

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  1. Question 1

    16 May 2023 S1

    106
    If a random variable X has the following probability distribution values: X: 0,1,2,3,4,5,6,7 and P(X): 0,k,2k,2k,3k,k2,2k2,7k2+k0,k,2k,2k,3k,k^2,2k^2,7k^2+k. Then P(X≥6)P(X \geq 6) has the value

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  2. Question 2

    25 Apr 2025 S1

    146
    If a random variable XX has the following probability distribution of X
    X=xX = xP(X=x)P(X = x)
    00
    1kk
    22k2k
    32k2k
    43k3k
    5k2k^{2}
    62k22k^{2}
    77k2+k7k^{2} + k
    Then P(X≥6)=P(X\geq 6) =

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