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Probability distribution with an unknown constant k: find k, then a probability

This type of question, asked 3 times: 2021 Paper 21, 14 Jun 2022, 22 Jan 2026 S1. All 3 questions below, oldest first.

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  1. Question 1

    2021 Paper 21

    83
    The probability distribution of random variable XX is given by:
    XX12345
    P(X)P(X)KK2K2K2K2K3K3KKK
    Let p=P(1<X<4∣X<3)p = P(1 < X < 4 \mid X < 3). If 5p=λK5p =\lambda K, then λ\lambda equal to

  2. Question 2

    14 Jun 2022

    162
    A random variable XX has the following probability distribution:
    XX01234
    P(X)P(X)kk2k2k4k4k6k6k8k8k
    The value of P(1<X<4∣X≤2)P(1 < X < 4 \mid X \leq 2) is equal to:

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  3. Question 3

    22 Jan 2026 S1

    57
    If a random variable x has the probability distribution
    xx01234567
    p(x)p(x)02k2kkk3k3k2k22k^{2}2k2kk2+kk^{2}+k7k27k^{2}
    then P(3<x≤6)P(3 < x \leq 6) is equal to

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