1 2 3 by property 3 of pmf 5k 4k 3k 2k 1k 1

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Unformatted text preview: 5k + 4k + 3k + 2k + 1k = 1 ⇒ k = 1 15 P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3) = 4 3 2 9 15 + 15 + 15 = 15 = 0.6 P(X < 3|X = 0) = 5 4 + 15 15 10 15 P(0<X <3) P(X =0) = P(X =1)+P(X =2) 1−P(X =0) From Variables to Random Variables Probability Mass Function = = 0.9 9.12 Notes...
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This note was uploaded on 01/24/2014 for the course STAT 225 taught by Professor Martin during the Spring '08 term at Purdue.

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