72) Suppose that if a person visits Disnelyland, theprobanbility that he will go

Question

72) Suppose that if a person visits Disnelyland, theprobanbility that he will go on the jungle cruise is 0.74, thisprobability that he will ride the Momorail is 0.70, theprobability that he will go on the Matterhorn ride is 0.62,the probability that he will go on the jungle cruise and rideMonorail is 0.52, the probabaility that he will go onthe jungle Cruise as well as the Matterhorn ride is 0.46, theprobability that he will ride the Monorail and go on the Matterhornride id 0.44 and th eprobability that he will go on all three ofthese rides in 0.34. What is the probability that aperson visiting Disnelyland will go at least one of thesethree rides? a) not go on the jungle Cruise give that he will ridethe Monorail and / or go on the Matterhorn ride? b) go on the Matterhorn and the jungle Cruise given that hewill not ride the Monorail.? CAN YOU PLEASE EXPALIN BREIFLY WITH THE FOLLWING STEPS. 72) Suppose that if a person visits Disnelyland, theprobanbility that he will go on the jungle cruise is 0.74, thisprobability that he will ride the Momorail is 0.70, theprobability that he will go on the Matterhorn ride is 0.62,the probability that he will go on the jungle cruise and rideMonorail is 0.52, the probabaility that he will go onthe jungle Cruise as well as the Matterhorn ride is 0.46, theprobability that he will ride the Monorail and go on the Matterhornride id 0.44 and th eprobability that he will go on all three ofthese rides in 0.34. What is the probability that aperson visiting Disnelyland will go at least one of thesethree rides? a) not go on the jungle Cruise give that he will ridethe Monorail and / or go on the Matterhorn ride? b) go on the Matterhorn and the jungle Cruise given that hewill not ride the Monorail.? CAN YOU PLEASE EXPALIN BREIFLY WITH THE FOLLWING STEPS.

Explanation / Answer

let the events be as follows: the person goes on jungle cruise be J the person goes on momorail ride be M and the person goes on Matterhorn ride be Ma given: P(J)= 0.74 P(M)=0.7 P(Ma)=0.62 P(J M) =0.52   P(J Ma)0.46 P(M Ma)=0.44 P(J M Ma)=0.34 probability that person goes atleast one ride is P(J U M UMa) P(J U M U Ma)= P(J) + P(M)+P(Ma) -P(J M)-P(JMa)-P(M Ma)+P(J M Ma)                         =0.74+0.7+0.62-0.52-0.46-0.44+0.34                          =0.98 P(J'/M UMa) = P(J' (M U Ma))                            P(M U Ma) now we first find P(M U Ma ) P(M U Ma) = P(M)+P(Ma)- P(M Ma)=0.7+0.62-0.44=0.88 now we find P(J' (M U Ma)) let (M U Ma) be A say for ease of calculation P(J' A) +P(J A)=P(A) =0.88....................(I) now we find P(J A) P(J (M U Ma)) = P((J M) U (J Ma)) P((J M) U (J Ma))= P(J M) +P(J Ma) - P(J M Ma) note (J M Ma)= (J M)(J Ma) hence the above statement substituting the given values we get P((J M) U (J Ma))= 0.52+0.46-0.34=0.64 i.e P(J A)=0.64 substitute the value in eqn I we get P(J' A) + 0.64 =0.88 P(J' A)=0.88-0.64=0.24 this is the answer for part a now similar way solve part b hope this helps

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