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Question 44 — JEE Main 22 Jan 2026 (S I)
Question
If a random variable x has the probability distribution
| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| $\mathrm{p}(\mathrm{x})$ | 0 | 2 k | k | 3 k | $2 \mathrm{k}^{2}$ | 2 k | $\mathrm{k}^{2}+\mathrm{k}$ | $7 \mathrm{k}^{2}$ |
then $\mathrm{P}(3<\mathrm{x} \leq 6)$ is equal to [JEE MAIN 22 Jan 2026 S I] (1) 0.34 (2) 0.22 (3) 0.64 (4) 0.33
Solution
$\sum \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)=0+2 \mathrm{k}+\mathrm{k}+3 \mathrm{k}+2 \mathrm{k}^{2}+2 \mathrm{k}+\mathrm{k}^{2}+\mathrm{k}+7 \mathrm{k}^{2}=1$
$\Rightarrow 10 \mathrm{k}^{2}+9 \mathrm{k}-1=0 \Rightarrow(10 \mathrm{k}-1)(\mathrm{k}+1)=0 \Rightarrow \mathrm{k}=\frac{1}{10}, \mathrm{k} \neq-1$
$\mathrm{P}(3<\mathrm{x} \leq 6)=\mathrm{P}(\mathrm{x}=4)+\mathrm{P}(\mathrm{x}=5)+\mathrm{P}(\mathrm{x}=6)$
$=2 \mathrm{k}^{2}+2 \mathrm{k}+\mathrm{k}^{2}+\mathrm{k}=3 \mathrm{k}^{2}+3 \mathrm{k}=\frac{3}{100}+\frac{3}{10}=0.33$ Answer(4)