Let X be the discrete random variable with PMF (x) = 12x; x = 1,2,3,.... The value of P(X > 4) is

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SSC CGL Tier-II ( JSO ) 2019 Official Paper ( Held On : 17 Nov 2020 )
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  1. 1516
  2. 916
  3. 516
  4. 116

Answer (Detailed Solution Below)

Option 4 : 116
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Detailed Solution

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Given

PMF(x) = 1/2x, x = 1, 2, 3 -----

Formula

P(X > 4) = 1 – P(X ≤ 4)

Calculation

P(X > 4) = 1 – P(X ≤ 4)

⇒ 1 – [P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]

⇒ 1 – [1/21 + ½2  + ½3 + ½4]

⇒ 1 – [1/2 + ¼ + 1/8 + 1/16]

⇒ 1 – [(8 + 4 + 2 + 1)/16]

⇒ 1 – [15/16]

⇒ (16 – 15)/16

∴ The value of P(X > 4) is 1/16

 

Probability mass function or  PMF is defined as the set of ordered pair [x, f(x)] if for each possible outcome x, f(x) satisfies the following condition

F(x) ≥ 0

∑ f(x) = 1

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