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Unformatted text preview: 65 a. P(H or G) = P(H) + P(G) P(H and G) = 0.26 + 0.43 0.14 = 0.55 b. P( NOT (H and G) ) = 1 P(H and G) = 1 0.14 = 0.86 c. P( NOT (H or G) ) = 1 P(H or G) = 1 0.55 = 0.45 Solution to Exercise 3.34 (p. 58) a. P(J or K) = P(J) + P(K) P(J and K); 0.45 = 0.18 + 0.37 P(J and K) ; solve to find P(J and K) = 0.10 b. P( NOT (J and K) ) = 1 P(J and K) = 1 0.10 = 0.90 c. P( NOT (J or K) ) = 1 P(J or K) = 1 0.45 = 0.55 Solution to Exercise 3.35 (p. 59) a. P(Type O or Rh ) = P(Type O) + P(Rh ) P(Type O and Rh ) a. 0.52 = 0.43 + 0.15 P(Type O and Rh ); solve to find P(Type O and Rh ) = 0.06 a. 6% of people have type O Rh blood b. P( NOT (Type O and Rh ) ) = 1 P(Type O and Rh ) = 1 0.06 = 0.94 b. 94% of people do not have type O Rh blood Solution to Exercise 3.36 (p. 59) a. P(R or F) = P(R) + P(F) P(R and F) = 0.72 + 0.46 0.32 = 0.86 b. P( Neither R nor F ) = 1 P(R or F) = 1 0.86 = 0.14 Solution to Exercise 3.37 (p. 59) . Let C be the event that the cookie contains chocolate. Let N be the event that the cookie containsLet C be the event that the cookie contains chocolate....
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This document was uploaded on 10/31/2011 for the course STA 2023 at University of Florida.
 Fall '08
 Ripol
 Statistics

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