JEE Mathematics 2026 Question with Solution

Given: The random variable $X$ has the given probability distribution and $E(X)=\dfrac{263}{15}$.

Find: $P(X<20)$.

Use the expectation formula:

Substituting the given values:

Simplifying:

Since $E(X)=\dfrac{263}{15}$,

Now substitute $k=\dfrac{7}{2}$ into the values of $X$:

These are the values less than $20$.

Therefore,

The solution concludes with $\dfrac{11}{15}$ and marks option D, but the extracted working gives $\dfrac{8}{15}$ because $\frac{34k}{7}=17$ contributes probability $\frac{1}{5}$ only once. Among the listed options, the correct value is $\dfrac{8}{15}$, so the most defensible option is C.

After finding $k=\dfrac{7}{2}$, evaluate each support point of $X$:

Hence all values strictly less than $20$ are

So one must add the corresponding probabilities:

Therefore, the correct option is D.