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Question 50 — JEE Main 28 Jan 2026 S Ii
Question
The probability distribution of a random variable X is given below
| X | 4 k | $\frac{30}{7} \mathrm{k}$ | $\frac{32}{7} \mathrm{k}$ | $\frac{34}{7} \mathrm{k}$ | $\frac{36}{7} \mathrm{k}$ | $\frac{38}{7} \mathrm{k}$ | $\frac{40}{7} \mathrm{k}$ | 6 k |
|---|---|---|---|---|---|---|---|---|
| $\mathrm{P}(\mathrm{X})$ | $\frac{2}{15}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{1}{5}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{1}{5}$ | $\frac{1}{15}$ |
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) $\frac{14}{15}$
Solution
Expectation $\mathrm{E}(\mathrm{x})=\sum \mathrm{x}_{\mathrm{i}} \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)=\frac{263}{15}$
$\Rightarrow \frac{28 \times 2 \mathrm{k}+30 \mathrm{k}+64 \mathrm{k}+102 \mathrm{k}+36 \mathrm{k}+76 \mathrm{k}+120 \mathrm{k}+42 \mathrm{k}}{7 \times 15}=\frac{263}{15}$
$\Rightarrow 526 \mathrm{k}=263 \times 7 \Rightarrow \mathrm{k}=\frac{7}{2}$
| X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
|---|---|---|---|---|---|---|---|---|
| $\mathrm{P}(\mathrm{x})$ | $\frac{2}{15}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{3}{15}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{3}{15}$ | $\frac{1}{15}$ |
$\mathrm{P}(\mathrm{x}<20)=\frac{2+1+2+3+1+2}{15}=\frac{11}{15}$ Answer(3)