Aspire Faculty ID #18114 · Topic: JEE Main 2026 (28 January Evening Shift) · Just now
JEE Main 2026 (28 January Evening Shift)

The probability distribution of a random variable $ X $ is given below :

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

Solution

$ E(X) = \sum X_i P(X_i) = \frac{526k}{15 \times 7} = \frac{263}{15} \Rightarrow k = \frac{7}{2} $

$ P(X < 20) = \sum_{X=14}^{19} P(X) = \frac{11}{15} $

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