If a random variable x has the probability distribution x 0 1 2 3 4 5 6 7 --- --- --- --- --- --- --- --- --- p(x) 0 2K K 3K 2 K 2 2K K 2 +K 7 K 2 then P (3< x leq 6) is equal to

Mathematics · Probability

JEE Main 2026 — 22 January, Morning Shift — Question 7

If a random variable x has the probability distribution

x	0	1	2	3	4	5	6	7
p(x)	0	2K	K	3K	2 $K^2$	2K	$K^2$ +K	7 $K^2$

then $\mathrm{P}(3<\mathrm{x} \leq 6)$ is equal to

$\quad \sum \mathrm{P}\left(\mathrm{x}_{\mathrm{i}}\right)=1$

$\Rightarrow 9 \mathrm{k}+10 \mathrm{k}^{2}=1$

$\Rightarrow 10 \mathrm{k}^{2}+9 \mathrm{k}-1=0 \Rightarrow \mathrm{k}=\frac{1}{10}$

$\mathrm{P}(3<\mathrm{x} \leq 6)=3 \mathrm{k}+3 \mathrm{k}^{2}$

$=\frac{3}{10}+\frac{3}{100}=0.33$ $=0.33$

Answer key and solution verified before publishing.

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