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 2 years ago
Probability of injury from car trip is 1in 50K (.00002). average trips a person will take in liftime is 16K.
Whats the probability of being injury? can i get help to set this up por favor!
 2 years ago
Probability of injury from car trip is 1in 50K (.00002). average trips a person will take in liftime is 16K. Whats the probability of being injury? can i get help to set this up por favor!

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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1poisson distribution for this one

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1oh well if that is not clear, them maybe you are supposed to do it a different way

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0im gona do a ratio of 16 to 50 and im gona in crease the numerator from taht resuilt

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1multiply 16,000 by 0.00002 and get .32

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0i got 3.125 to 16000, hope its right. hah

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1no i don't think so

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0oh wait. your right. your way makes more sense.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1this is a set up for poisson distribution, because the probabiliy is very small and the number of "experiments' miles driven is very large multiply them together and get .32 probability you have no accidents is \[e^{.32}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1and so the probabilty you have at least one accident is \[1e^{.32}\]

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0im trying to find prob of yes getting injured tho.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1yes, so compute the probability of no accidents, which is \[e^{.32}\] so the probability of getting at least one accident is \[1e^{.32}\]

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0dang bro, thats way ahead of me. but thansk anyways

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1you could also compute \((99998)^{160000}\) and subtract that from 1

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1that is the probability you do not get in an accident on one trip, to the power of the number of trips, and that will get the probability you get no accidents. subtract from one to get the probability that there is at least one accident

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0OH!! so its 1 minus .32 or .68% .. i was thinking it was 32% which was confusing me.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1actually i think that is not right. let me check with a calculator

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1this is what i thin it is http://www.wolframalpha.com/input/?i=%28.00002%29^16000

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1almost identitcal to firt answer i wrote http://www.wolframalpha.com/input/?i=1e^%28.32%29

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0alrite cool. thanks dude.

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0The problem is an example of risk over time, which we discussed on Monday. You need to multiply the probability of one event by the total number of times it is likely to occur to calculate a life time risk. Your calculations below are for successive events not cumulative events.

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0thats what the teacher said. i have no idea what the hell she is saying. lol

mr.luna
 2 years ago
Best ResponseYou've already chosen the best response.0haha, its just 1 divided by 16000 dude. wtf were we doin? lol
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