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For the joint density fxy(x, y) = x2 + Cy; 0 ≤ x ≤ 1, 0 ≤ y ≤ 1,  the value of constant C is:

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

Answer (Detailed Solution Below)

Option 1 : 4/3
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Detailed Solution

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Concept used

We know that the probability of joint density fxy(xy)dxdy

Calculation

 (x2 + Cy)dxdy = 1

After integration we get

⇒ (x3/3)10 + C(y2/2)10 = 1

⇒ (1/3) + (1/2)C = 1

⇒ (1/2)C = 1 – 1/3

⇒ C(1/2) = 2/3

⇒ C = 4/3

∴ The value of C is 4/3

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