SOLUTION: Break-downs occur on an old car with a rate of 6 break-downs every four weeks. The owner of the car is planning to have a trip on his car for 4 days. What is the probability that h

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Question 1131428: Break-downs occur on an old car with a rate of 6 break-downs every four weeks. The owner of the car is planning to have a trip on his car for 4 days. What is the probability that he will return home without experiencing a car break-down?

(Please round your answer to four decimal places if necessary. Thank you!)
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
from what i've read, this is a poisson probbility type problem.

the formula is p(x,m) = e^-m * m^x / x!

x is the number of occurrences
m is the mean number of occurrences
e is the scientific constant of 2.718281818.....

e is usually stored in your scientific calculator as e^x.
2.718181818...... would show up as e^1.

your problem states:

Break-downs occur on an old car with a rate of 6 break-downs every four weeks. The owner of the car is planning to have a trip on his car for 4 days. What is the probability that he will return home without experiencing a car break-down?

since the car breaks down at an averate rate of 6 break-downs every 4 weeks, then you can calculate the mean number of breakdowns every 4 days as follows:

6 / 4 weeks = 6 / 28 days * 4 = 24 / 28 breakdowns every 4 days.
the factor, in decimal form, is equal to .8571428571.

to find the probability of 0 breakdowns in 4 days, the formula is used as follows:

p(x,m) = e^-m * m^x / x! which becomes:

p(0,24/28) = e^-(24/28) * (24/28)^0 / 0!

since 0! is equal to 1, the formula becomes:

p(0,24/28) = e^-(24/28) * (24/28)^0 which becomes:

p(0,24/28) = .4243728457

here's a reference from answers.yahoo.com

https://answers.yahoo.com/question/index?
qid=20110623030726AAE2tre

here's a reference from statrek.

https://stattrek.com/probability-distributions/poisson.aspx

statrek has a calculator you can use.

that calculator can be found here:

https://stattrek.com/online-calculator/poisson.aspx

here's the results of my use of this calculator.

the calculator says that the probability of 0 failures in 4 days is equal to .42437.

this agrees with my calculation based on the formula.

round to 4 decimal places to get .4244.

note that the calculator has to take decimal entries.
entering 24/28 is no good.
it has to be the decimal form of .8571428571.

here's another reference with examples:

https://www.intmath.com/counting-probability/13-poisson-probability-distribution.php