SOLUTION: "If something can go wrong, it will go wrong." This funny saying is called Murphy's law. Let's interpret this to mean "If something can go wrong, there is a very high probability t

Question 1104144: "If something can go wrong, it will go wrong." This funny saying is called Murphy's law. Let's interpret this to mean "If something can go wrong, there is a very high probability that it will eventually go wrong."
Suppose we look at the event of having an automobile accident at some time during a day's commute. Let's assume that the probability of having an accident on a given day is 1 in a thousand or 0.001. That is, in your town, one of every thousand cars on a given day is involved in an accident (including little fender-benders). We also assume that having (or not having) an accident on a given day is independent of having (or not having) an accident on any other given day. Suppose you commute 45 weeks per year, 5 days a week, for a total of 225 days each year. In the following parts, write each probability in decimal form rounded to three places.
(a) What is the probability that you have no accident over a year's time?

(b) What is the probability that you have at least one accident over a one-year period?

(c) Repeat part (a) for a 10-year period and for a 30-year period.
10-year period��

30-year period��

Repeat part (b) for a 10-year period and for a 30-year period.
10-year period��

30-year period��

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

(a) What is the probability that you have no accident over a year's time?

p = probability of having an accident on any given day = 0.001
n = number of days = 225
probability of having no accidents for given day = q = 1-p = 1-0.001 = 0.999

We need to compute q^n to get the probability of having no accidents over the entire year
q^n = 0.999^225 = 0.79842633081077
which rounds to 0.798 when rounded to 3 decimal places

Answer: 0.798
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(b) What is the probability that you have at least one accident over a one-year period?

We'll use the result from part (a). Subtract it from 1
1-0.798 = 0.202

This works because we can break it up into two cases: either there are no accidents at all during the entire year OR there is at least one accident. The two events are mutually exclusive (no overlap) and they span to form the entire set of possibilities.

If A is the probability of no accidents and B is the probability of at least one accident, then A+B = 1 leading to B = 1-A
This is the idea of probability complements.

Answer: 0.202
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(c) Repeat part (a) for a 10-year period and for a 30-year period.

We'll follow the same basic steps as in part (a). Instead of n = 225, we'll use n = 225*10 = 2250. This is assuming that the schedule remains the same throughout the 10 years.

q^n = (1-p)^n = (1-0.001)^2250 = 0.999^2250 = 0.1052806380874
this rounds to 0.105 which represents the probability of no accidents for the 10-year period.

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Similarly, the 30-year period is worked the same way. Now n = 225*30 = 6750
q^n = (1-p)^n = (1-0.001)^6750 = 0.999^6750 = 0.00116693193552
which rounds to 0.001

Answers:
For the 10-year period, the answer is 0.105
For the 30-year period, the answer is 0.001
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(d) Repeat part (b) for a 10-year period and for a 30-year period.

The steps and logic for this part is very similar to part (b). We'll use the answers in part (c). Subtract them from 1 to get:
1-0.105 = 0.895 which is the answer for the 10-year period
1-0.001 = 0.999 which is the answer for the 30-year period

Answers:
For the 10-year period, the answer is 0.895
For the 30-year period, the answer is 0.999

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