[shahkb4]

shahkb4

3 June, 2026 at 8:28 pm #1127

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) $\frac{14}{15}$

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[shahkb4]

shahkb4

3 June, 2026 at 8:29 pm #1128

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)

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