Anyone knows the solution to this problem?

You have been offered a job in Oman but would like to be in Abu Dhabi as much as possible, and so plan to go back every weekend.. You are a bit scared of flying, so you are thinking that you might instead drive every week rather than fly. However, being a statistical type, you decide to run the numbers.
a. First estimate how many total trips you would make if you planned to travel there and back twice a week every week for 20 years.
b. On the safest airlines, there is about a 1 in 19.8 million chance of dying on a single flight. If you flew to work, what is your chance of dying given the number of trips you calculated?
c. Now find the driving distance between Abu Dhabi and Muscat. In many Western European countries, there is typically around 5 traffic fatalities for every billion kilometers driven. What is the chance of a fatality if you drive there and back twice a week for 20 years?

1 Answer
Feb 4, 2018

Just a partial answer, but I've been there, so I can't help but be amused by some assumptions.

Explanation:

a. OK, not too hard. Number of weeks in a year = 52. Twice a week (good luck!!) = 104 trips in a year. #104 xx 20 = 2080# trips!

b. Fine for the exercise, but you wouldn't survive the first couple of months with that schedule. Also, your "chance of dying" is not limited to the air segments! #2080 xx 2 "ways" = 4160 "Air segments"#
#4160 xx 1/19.8xx10^6 = 0.000210 = 0.0210%#

c. Given the distance, YOUR safe driving would be considerably LESS than the "average perr km" of a general population. Plus, the driving in the Middle East is quite different - and more dangerous on average than in Western European countries.

Again, #4160 "one-way trips" xx km xx 5/(1xx10^9) = "Risk"#