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Table with k and k^2 entries (k^2, 2k, k, 2k, 5k^2): find P(X > 2)

The same question, asked 3 times: 4 May 2023 S1*, 15 May 2023 S1*, 25 Apr 2025 S2. Every wording below, oldest first.

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  1. Question 1

    4 May 2023 S1

    131
    A random variable X has the following probability distribution XX: 1, 2, 3, 4, 5 P(X)P(X): k2k^2, 2k2k, kk, 2k2k, 5k25k^2 Then P(X>2)P(X>2) is equal to

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  2. Question 2

    15 May 2023 S1

    106
    A random variable X has the following probability distribution: X:1,2,3,4,5X: 1,2,3,4,5; p(x):k2,2k,k,2k,5k2p(x): k^{2}, 2k, k, 2k, 5k^{2}. Then p(x≥2)p(x\geq2) is equal to

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  3. Question 3

    25 Apr 2025 S2

    114
    Consider the probability distribution
    X=xX = x12345
    P(X=x)P(X = x)KK2K2KK2K^{2}2K2K5K25K^{2}
    Then the value of P(X>2)P(X > 2) is

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