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Distribution with unknown entries: use the mean (or E(X^2)) to find them, then the variance or a probability

This type of question, asked 5 times: 2021 Paper 24, 4 Apr 2024, 7 Apr 2025, 21 Jan 2026 S2, 28 Jan 2026 S2. All 5 questions below, oldest first.

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  1. Question 1

    2021 Paper 24

    81
    Let XX be a random variable with distribution.
    xx−2-2−1-1346
    P(X=x)P(X=x)15\frac{1}{5}aa13\frac{1}{3}15\frac{1}{5}bb
    If the mean of XX is 2.3 and variance of XX is σ2\sigma^{2}, then 100σ2100\sigma^{2} is equal to:

  2. Question 2

    4 Apr 2024

    165
    If the mean of the following probability distribution of a random variable XX;
    XX02468
    P(X)P(X)aa2a2aa+ba+b2b2b3b3b
    is 469\frac{46}{9}, then the variance of the distribution is

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  3. Question 3

    7 Apr 2025

    131
    Let a random variable X take values 0,1,2,30,1,2,3 with P(X=0)=P(X=1)=P\left( X= 0 \right)=P\left( X= 1 \right)= p,P(X=2)=P(X=3)p,P(X= 2) =P(X= 3) and E(X2)=2E(X)E\left( X^{2} \right)= 2E(X). Then the value of 8p−18p- 1 is :

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  4. Question 4

    21 Jan 2026 S2

    60
    A random variable XX takes values 0,1,2,30,1,2,3 with probabilities 2a+130,8a−130,4a+130\frac{2a + 1}{30},\frac{8a - 1}{30},\frac{4a + 1}{30}, b respectively, where a,b∈Ra,b \in R. Let μ\mu and σ\sigma respectively be the mean and standard deviation of XX such that σ2+μ2=2\sigma^{2}+\mu^{2}= 2. Then ab\frac{a}{b} is equal to :

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  5. Question 5

    28 Jan 2026 S2

    54
    The probability distribution of a random variable X is given below :
    XX4k4k307k\frac{30}{7}k327k\frac{32}{7}k347k\frac{34}{7}k367k\frac{36}{7}k387k\frac{38}{7}k407k\frac{40}{7}k6k6k
    P(X)P(X)215\frac{2}{15}115\frac{1}{15}215\frac{2}{15}15\frac{1}{5}115\frac{1}{15}215\frac{2}{15}15\frac{1}{5}115\frac{1}{15}
    If E(X)=26315E(X) =\frac{263}{15}, then P(X<20)P(X< 20) is equal to :

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