Let X be a continuous random variable with PDF

fx(x)={4x30<x10otherwise

The value of P(X23|X>13)is

  1. 19
  2. 527
  3. 516
  4. 316

Answer (Detailed Solution Below)

Option 4 : 316
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Detailed Solution

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P(X23|X>13)=P(X23X>13)P(X>13)

=P(13<X23)P(X>13)

P(13<X23)=1/32/34x3dx=[x4]1323=1581=527

P(X>13)=1/314x3dx

=[x4]131

=1181=8081

P(X23|X>13)=5278081=527×8180=316

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