JEE Advanced

Question 50 — JEE Main 28 Jan 2026 S Ii

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  • #11688
    shahkb4
    Keymaster

    The probability distribution of a random variable X is given below

    [t]{|l|l|l|l|l|l|l|l|l|}
    X
    4k $\frac{30}{7} \mathrm{k}$ $\frac{32}{7} \mathrm{k}$ $\frac{34}{7} \mathrm{k}$ $\frac{36}{7} \mathrm{k}$ $\frac{38}{7} \mathrm{k}$ $\frac{40}{7} \mathrm{k}$ 6k
    P(X) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$

    If $\mathrm{E}(\mathrm{X})=\frac{263}{15}$, then $\mathrm{P}(\mathrm{X}<20)$ is equal to [JEE MAIN 28 Jan 2026 S II]
    (1) $\frac{3}{5}$
    (2) $\frac{8}{15}$
    (3) $\frac{11}{15}$
    (4) $\frac{14}{15}$

    #11689
    shahkb4
    Keymaster

    Solution

    \[ \begin{aligned} & \text { Expectation } \mathrm{E}(\mathrm{x})=\sum \mathrm{x}_{\mathrm{i}} \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)=\frac{263}{15} \\ & \Rightarrow \frac{28 \times 2 \mathrm{k}+30 \mathrm{k}+64 \mathrm{k}+102 \mathrm{k}+36 \mathrm{k}+76 \mathrm{k}+120 \mathrm{k}+42 \mathrm{k}}{7 \times 15}=\frac{263}{15} \\ & \Rightarrow 526 \mathrm{k}=263 \times 7 \Rightarrow \mathrm{k}=\frac{7}{2} \end{aligned} \]
    [t]{|l|l|l|l|l|l|l|l|l|}
    X
    14 15 16 17 18 19 20 21
    P(x) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{3}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{3}{15}$ $\frac{1}{15}$
    \[ P(x<20)=\frac{2+1+2+3+1+2}{15}=\frac{11}{15} \text { Answer }(3) \]
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