STAT 418HW2 SolutionsSections 2.1-2.54,10,12,18,23,27,36,39; TE 10,11,124.a. The sample space consists of all the possible sequences of flips until a Head. The 0’sare Tails; the 1 is the first Head.b. i Because A flips first, A = {1, 0001, 0000001, . . .}ii Because B flips second, B = {01, 00001, 00000001, . . .}iii (A ∪ B) = C ∪ (000 . . .) = {000 . . . , 001, 000001, 000000001, . . .}10. Let R = ’Student wears a ring’ and N = ’Student wears a necklace’. Given:P (R) = .2; P (N) = .3; P (R ∩ N) = .6 = PR ∪ N.a.P (R ∪ N) = 1 − PR ∪ N= 1 − .6 = .4b.P (R ∪ N) = P(R) + P(N) − P (R ∩ N).ThereforeP (R ∩ N) = P (R) + P (N) − P (R ∪ N) = .2 + .3 − .4 = .112. Let S = ’Student takes Spanish’, F = ’Student takes French’, and G = ’Student takesGerman’. Given:P (S) = .28; P (F ) = .26; P (G) = .16;P (S ∩ F ) = .12; P (S ∩ G) = .04; P (F ∩ G) = .06; P (S ∩ F ∩ G) = .02.a.P (S ∪ F ∪ G) = 1 − P (S ∪ F ∪ G)= 1 − P (S) − P (F ) − P (G) + P (S ∩ F ) + P (S ∩ G) + P (F ∩ G) − P (S ∩ F ∩ G)= 1 − .28 − .26 − .16 + .12 + .04 + .06 − .02 = .5.b. By Axiom 3, P(Student taking exactly 1 language class) =P (S ∩ (F ∪ G)) + P (F ∩ (S ∪ G)) + P (G ∩ (F ∪ S)).There are 28 students taking Spanish. 12 of them are also taking French and 4 of themare also taking German. Of those 12 and 4, 2 are taking all three languages. By Axiom3,P (S ∩ F ∩ G) = 10/100 = .1,andP (S ∩ G ∩ F ) = 2/100 = .02.1BecauseS ∩ (F ∪ G), S ∩ F ∩ G, S ∩ F ∩ G,andS ∩ G ∩ Fare mutually exclusive,P (S∩(F ∪ G)) = P (S)−P (S∩F ∩G)−P (S∩G∩F)−P(S∩F ∩G) = .28−.1−.02−.02 = .14.Similarly,P (F ∩ (S ∪ G)) = .26 − .1 − .04 − .02 = .10andP (G ∩ (F ∪ S)) = .16 − .02 − .04 − .02 = .08.Therefore P(Student taking exactly 1 language class) = .14 + .10 + .08 = .32. A Venndiagram is very helpful in this problem.c. Because there are 100 students in the class and half of them are not taking a languageclass, the probability that neither one is taking a course is5021002=50∗492100∗992=49198.Because the complement of ’at least 1’ is that neither student takes a language class,P (at least 1 of the 2 students takes a language class) = 1 − P (neither takes a language class)= 1 −49198=149198.18. In order to get a Blackjack, you need one of the 4 aces and 1 of the 16 cards worth 10points, the 10, Jack, Queen, or King. The probability of Blackjack is41161522=4 ∗ 1652 ∗ 51/2= .048.23. There are6 ∗ 6 = 36outcomes in the sample space. There are5 + 4 + 3 + 2 + 1 = 15outcomes where the second die is higher than the first. The probability that the second dieis higher than the first is1536=512.227. A draws first and there are 3 red balls. Because the events that A wins on ball 1, ball3, ball 5, and ball 7 are disjoint, (A cannot win on ball 9 because there are 3 red balls), P(AWins) = P(A wins on ball 1) + P(A wins on ball 3) + P(A wins on ball 5) + P(A wins onball 7) =310+7 ∗ 6 ∗ 310 ∗ 9 ∗ 8+7 ∗ 6 ∗ 5 ∗ 4 ∗ 310 ∗ 9 ∗ 8 ∗ 7 ∗ 6+7 ∗ 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 310 ∗ 9 ∗ 8 ∗ 7 ∗ 6 ∗ 5 ∗ 4.36.a.42522=4!2!2!52∗512=626 ∗ 51= .0045.b. There are 13 ways to choose the card that will be drawn twice.13142522=134!2!2!52∗512=13 ∗ 626 ∗ 51= .05639. Assume each hotel is equally likely. The first may go to any of the 5 hotels, the secondonly has 4 choices, and the third 3. If there were no restrictions, each would have 5 choices.The probability that each of 3 people check into a different hotel is5 ∗ 4 ∗ 353=1225.TE 10. Write(E ∪ F ) ∪ Gas the union of two disjoint sets. By Axiom 3,P (E ∪ F ) ∪ G) = P ((E ∪ F ) ∪ ((E ∪ F ) ∩ G))= P (E ∪ F ) + P ((E ∪ F ) ∩ G)= P (E ∪ (¯E ∩ F )) + P ((E ∪ F ) ∩ G)= P (E) + P (¯E ∩ F ) + P (¯E ∩¯F ∩ G), by Axiom 3 and DeMorganF = (E ∩ F ) ∪ (¯E ∩ F ). By Axiom 3,P (F ) = P (E ∩ F ) + P (¯E ∩ F ).G = (¯E ∩¯F ∩ G) ∪ (E ∩¯F ∩ G) ∪ (¯E ∩ F ∩ G) ∪ (E ∩ F ∩ G). By Axiom 3,P (G) = P (¯E ∩¯F ∩ G) + P (E ∩¯F ∩ G) + P (¯E ∩ F ∩ G) + P (E ∩ F ∩ G),and thereforeP (E ∪F ∪G) = P (E)+P (F )+P (G)−P(E∩¯F ∩G)−P(¯E ∩F ∩G)−P (E ∩F ∩G)−P (E ∩F).3(E ∩ F ) = (E ∩ F ∩ G) ∪ (¯G ∩ E ∩ F ). By Axiom 3,P (E ∩ F ) = P (E ∩ F ∩ G) + P (¯G ∩ E ∩ F ).TE 11. By Axiom 1,1 ≥ P (E ∪ F ) = P (E) + P (F ) − P (E ∩ F ).ThereforeP (E ∩ F ) ≥ P (E) + P (F ) − 1.TE 12. ’Exactly E or F’ = (E ∩¯F ) ∪ (¯E ∩ F ). By Axiom 3, P(Exactly E or F)= P (E ∩¯F ) + P (¯E ∩ F ).E = (E ∩ F ) ∪ (E ∩¯F .By Axiom 3,P (E) = P (E ∩ F ) + P (E ∩¯F ),and thereforeP (E ∩¯F ) = P (E) − P (E ∩ F ).Similarly,P (¯E ∩ F ) = P (F ) − P (E ∩ F ).Therefore,P (E ∩¯F ) + P (¯E ∩ F ) = P (E) + P (F ) − 2P (E ∩ F