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York University (33,712)
MATH 2565 (19)
Lecture

# PPS5_Solution_.doc

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School
York University
Department
Mathematics and Statistics
Course
MATH 2565
Professor
Jagmohan Chawla
Semester
Winter

Description
PPS5 (Solution) Problem14.19: Homes. Using obvious notations, we are given that P(G) = 0.64, P(S) = 0.21, P( G and S) = 0.17. a) P( S or G) = P(S) + P(G) – P( S and G) = 0.21 + 0.64 – 0.17 = 0.68 b) P(S and G ) = P(S or G) =1− P(S or G) =1−0.68 = 0.32 c c) P(S and G ) = P(S )− P(S and G) = 0.21−0.17 = 0.04 Problem 15.7: Dice. a) (1/6)(1/6)(1/6) = 1/216 =.0046 b) (3/6)(3/6)(3/6) = 27/216 = 0.125 c) (4/6)(4/6)(4/6) = 64/216 = 0.2963 d) 1 – (5/6)(5/6)(5/6) = 1-125/216= 91/216 = 0.4213 e) 1 – (1/6)(1/6)(1/6) = 1-1/216=215/216 = 0.9954 Problem 15.17: Cards a)P(Heart|Red) =P(Heart and Red)=13/52 = 1 P(Red) 26/52 2 P(Heart and Red) 13/52 b) P(Red|Heart) = = =1 P(Heart) 13/52 P(Ace and Red) 2 /52 2 1 c) P(Ace|Red) = = = = P(Red) 26 /52 26 13 P(Queen and Face) 4/52 1 d) P(Queen|Face) = = = (Jack, King and Queen P(Face) 12/52 3 are face cards) Problem 15.21: Movies. a) 15/38 = 0.3947 b) 15/51 = 0.2941 c) (21 + 22 + 38)/(69 + 46 + 56) = 81/171 = 0.4737 d) P(Over 30 and didn't select a comedy ) P(over 30|didn't select a comedy) = P(didn't select a comedy) (9+ 7 + 6+ 39+17 +12) / 240 90 45 = = = = 0.6618 (51+85) / 240 136 68 Problem 15.33: Men’s health, again. Consider the two-way table l r Blood from Exercise 19. t Pressure l High OK Tota o C l Hig 0.11 0.21 0.32 h OK 0.16 0.52 0.68 Tota 0.27 0.73 1.00 High blood pressure and high cholesterol are not independent events. 28.8% of men with OK blood pressure have high cholesterol, while 40.7% of men with high blood pressure have high cholesterol. If having high blood pressure and high cholesterol were independent, these percentages would be the same. Problem 15.37: Luggage. Organize using a tree diagram. a) No, the flight leaving on time and the luggage making the connection are not independent events. The probability that the luggage makes the connection is dependent on whether or not the flight is on time. The probability is 0.95 if the flight is on time, and only 0.65 if it is not on time. b) P(Luggage)=P(On timeand Luggage)+ P(Noton timeandLuggage) =(0.15)(0.95(0.85)(0.65) =0.695 Problem15.39: Late luggage. Refer to the tree diagram constructed for Exercise 37. P(Notontime No Lug.)= P(Noton timeandNoLug.) = (0.85)(0.35) ≈ 0.975 P(NoLug.) (0.15)(0.05(0.85)(0.35) If you pick Leah up at the Denver airport and her lugga
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