chap-3 - Chapter 3 Problems 1. P{6 different} = P{6,...

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20 Chapter 3 Chapter 3 Problems 1. P {6 different} = P {6, different}/ P {different} = 6 / 5 } 6 nd 2 , 6 s 1 { } 6 nd 2 , 6 st 1 { = + = t P P = 6 . 5 6 / 5 6 / 1 2 = 1/3 could also have been solved by using reduced sample space—for given that outcomes differ it is the same as asking for the probability that 6 is chosen when 2 of the numbers 1, 2, 3, 4, 5, 6 are randomly chosen. 2. P {6 sum of 7} = 6 / 1 )} 1 , 6 {( P = 1/6 P {6 sum of 8} = 36 / 5 )} 2 , 6 {( P = 1/5 P {6 sum of 9} = 36 / 4 )} 3 , 6 {( P = 1/4 P {6 sum of 10} = 36 / 3 )} 4 , 6 {( P = 1/3 P {6 sum of 11} = 36 / 2 )} 5 , 6 {( P = 1/2 P {6 sum of 12} = 1. 3. P { E has 3 N S has 8} = 8} has { 8} has 3, has { S N P S N E P = 26 52 18 39 8 13 13 26 26 52 10 21 3 5 18 39 8 13 = .339 4. P {at least one 6 sum of 12} = 1. Otherwise twice the probability given in Problem 2. 5. 12 8 13 9 14 5 15 6 6. In both cases the one black ball is equally likely to be in either of the 4 positions. Hence the answer is 1/2. 7. P 1 g and 1 b at least one b } = 4 / 3 2 / 1 = 2/3
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Chapter 3 21 8. 1/2 9. P { A = w 2 w } = } 2 { } 2 , { w P w w A P = = } 2 { } , , { } , , { w P w C w B w A P w C w B w A P = = + = = = 11 7 4 1 3 2 3 2 4 1 3 1 3 1 4 3 3 2 2 1 4 1 3 1 3 1 4 3 3 2 3 1 = + + + 10. 11/50 11. (a) P ( B A s ) = 13 31 () 1 52 21 52 51 2 1 7 52 s s PBA PA + == Which could have been seen by noting that, given the ace of spades is chosen, the other card is equally likely to be any of the remaining 51 cards, of which 3 are aces. (b) P ( B A ) = 43 1 52 51 48 47 3 3 1 52 51 PB 12. (a) (.9)(.8)(.7) = .504 (b) Let F i denote the event that she failed the i th exam. 496 . ) 2 )(. 9 (. 504 . 1 ) ( ) ) ( 2 1 3 2 1 2 = = F F P F F F F P c c c c c = .3629 13. P ( E 1 ) = 13 52 12 48 1 4 , P ( E 2 E 1 ) = 13 39 12 36 1 3 P ( E 3 E 1 E 2 ) = 13 26 12 24 1 2 , P ( E 4 E 1 E 2 E 3 ) = 1. Hence, p = 13 26 12 24 1 2 13 39 12 36 1 3 13 52 12 48 1 4 14. 768 35 18 9 16 7 14 7 12 5 .
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22 Chapter 3 15. Let E be the event that a randomly chosen pregnant women has an ectopic pregnancy and S the event that the chosen person is a smoker. Then the problem states that P ( E S ) = 2 P ( E S c ), P ( S ) = .32 Hence, P ( S E ) = P ( SE )/ P ( E ) = ) ( ) ( ) ( ) ( ) ( ) ( C c S P S E P S P S E P S P S E P + = ) ( ) ( 2 ) ( 2 c S P S P S P + = 32/66 .4548 16. With S being survival and C being C section of a randomly chosen delivery, we have that .98 = P ( S ) = P ( S C ).15 + P ( S C 2 ) .85 = .96(.15) + P ( S C 2 ) .85 Hence P ( S C c ) .9835.
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This note was uploaded on 03/31/2008 for the course STAT 418 taught by Professor G.jogeshbabu during the Spring '08 term at Penn State.

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chap-3 - Chapter 3 Problems 1. P{6 different} = P{6,...

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