Intro to Stat_Part_17

Intro to Stat_Part_17 - 65 a. P(H or G) = P(H) + P(G)- P(H...

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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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Intro to Stat_Part_17 - 65 a. P(H or G) = P(H) + P(G)- P(H...

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