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MCQMediumJEE Main 2026 · 28 January, Shift 2Probability Distributions

Mathematics Question from JEE Main 2026 · 28 January, Shift 2

The probability distribution of a random variable XX is given below:

x4k30k732k734k736k738k740k76kP(X)2151152151511521515115\begin{array}{c|cccccccc} x & 4k & \frac{30k}{7} & \frac{32k}{7} & \frac{34k}{7} & \frac{36k}{7} & \frac{38k}{7} & \frac{40k}{7} & 6k\\ \hline 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} \end{array}

If E(X)=26315E(X)=\dfrac{263}{15}, then P(X<20)P(X<20) is equal to:

  • A

    35\dfrac{3}{5}

  • B

    1415\dfrac{14}{15}

  • C

    815\dfrac{8}{15}

  • D

    1115\dfrac{11}{15}

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