It is a known fact that accidents happen more often driving the car rather than flying. Jim drives 180 days and flies 120 times per year. If for every 110 days driving you get into one accident and for every 10000 flying you get into 3 accidents, what is the probability of getting into two accidents, driving and flying on the same day.
i think that it will be
p=(1.5*0.009)+0.003=0.0165
To calculate the probability of getting into two accidents, we need to find the probability of getting into an accident while driving and an accident while flying on the same day.
First, let's calculate the probability of getting into an accident while driving on any given day. We are told that for every 110 days driving, there is one accident. Therefore, the probability of getting into an accident while driving is 1/110.
Next, let's calculate the probability of getting into an accident while flying on any given day. We are told that for every 10,000 flights, there are three accidents. Therefore, the probability of getting into an accident while flying is 3/10,000.
Now, since accidents while driving and accidents while flying are independent events, the probability of both events happening on the same day is the product of their individual probabilities. Thus, the probability of getting into an accident while both driving and flying on the same day is (1/110) * (3/10,000).
Finally, Jim drives for 180 days and flies 120 times per year. Assuming every day is equally likely, we can multiply the probability by the number of opportunities to get into two accidents in a year, which is 180 * 120.
Therefore, the probability of getting into two accidents, driving and flying on the same day, for Jim in one year is:
(1/110) * (3/10,000) * (180 * 120) = (1/110) * (3/10,000) * 21,600 = 0.00054655
So, the probability of getting into two accidents, driving and flying on the same day, is approximately 0.00054655, or 0.05%.