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Friday, 9 October

JEE Main 2020 · MathsMultiple choiceSingle correctEasyCalculation

JEE Main 9 January 2020, Shift 2, Maths Q67

Question 67 of 75 in this shift, Maths question 17 of 25, Section A.

A random variable X has the following probability distribution :
X12345
P(X)K2K^22K2KKK2K2K5K25K^2
Then P(X>2)P(X>2) is equal to :
  1. (1)16\frac{1}{6}
  2. (2)712\frac{7}{12}
  3. (3)2336\frac{23}{36}Official answer
  4. (4)136\frac{1}{36}

Official answer

Option 3

NTA final key (Jan 2020).

Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.