JEE Main 2020 — Probability Question with Solution
From: JEE Main 2020 (Online) 9th January Evening Slot
Question
A random variable X has the following
probability distribution :
Then P(X > 2) is equal to :
| X: | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| P(X): | K2 | 2K | K | 2K | 5K2 |
Then P(X > 2) is equal to :
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
= 1
K2 + 2K + K + 2K + 5K2 = 1
6K2 + 5K – 1 = 0
(6K - 1)(k + 1) = 0
K = and K = -1(rejected)
P(X 2)
= K + 2K + 5K2
=
=
K2 + 2K + K + 2K + 5K2 = 1
6K2 + 5K – 1 = 0
(6K - 1)(k + 1) = 0
K = and K = -1(rejected)
P(X 2)
= K + 2K + 5K2
=
=
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This is a previous-year question from JEE Main 2020, covering the Probability chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.