Question 1104144
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(a) <font color=blue>What is the probability that you have no accident over a year's time?</font>


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: <font color=red>0.798</font>

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(b) <font color=blue>What is the probability that you have at least one accident over a one-year period? </font>


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: <font color=red>0.202</font>
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(c) <font color=blue>Repeat part (a) for a 10-year period and for a 30-year period. </font>


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 <font color=blue>10-year period</font>, the answer is <font color=red>0.105</font>
For the <font color=blue>30-year period</font>, the answer is <font color=red>0.001</font>
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(d) <font color=blue>Repeat part (b) for a 10-year period and for a 30-year period.</font>


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 <font color=blue>10-year period</font>, the answer is <font color=red>0.895</font>
For the <font color=blue>30-year period</font>, the answer is <font color=red>0.999</font>
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