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Probability distribution with k and k^2 terms: find k, then a probability

This type of question, asked 6 times: 4 May 2023 S1, 15 May 2023 S1, 16 May 2023 S1, 25 Apr 2025 S1, 25 Apr 2025 S2, 15 May 2026 S1. All 7 questions 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

    4 May 2023 S1

    115
    A random variable has the following probability distribution XX: 0, 1, 2, 3, 4, 5, 6, 7 P(x)P(x): 0,2p,2p,3p,p2,2p2,7p2,2p0, 2p, 2p, 3p, p^2, 2p^2, 7p^2, 2p The then value of pp is

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  3. Question 3

    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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  4. Question 4

    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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  5. Question 5

    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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  6. Question 6

    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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  7. Question 7

    15 May 2026 S1

    99
    The probability mass function (p.m.f.) of a random variable X is as follows: P(X=0)=3k3, P(X=1)=4k−10k2, P(X=2)=5k−1, P(X=x)=0P(X = 0) = 3k^3,\ P(X = 1) = 4k - 10k^2,\ P(X = 2) = 5k - 1,\ P(X = x) = 0 for any value of xx, then the value of kk is:

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