SOLUTION: a) A city has four fire engines. For each fire engine the probability that it is available is 0.9. A fire accident happens somewhere in the city one Thursday afternoon. It is deter

Algebra ->  Probability-and-statistics -> SOLUTION: a) A city has four fire engines. For each fire engine the probability that it is available is 0.9. A fire accident happens somewhere in the city one Thursday afternoon. It is deter      Log On


   

Question 1183439: a) A city has four fire engines. For each fire engine the probability that it is available is 0.9. A fire accident happens somewhere in the city one Thursday afternoon. It is determined that fire engines will be needed:
(i) What is the chance that exactly 2 engines are available?
(ii) What is the probability that at least two of the engines are available?
(iii) What is the chance that none of the engines are available?
Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
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A city has four fire engines. For each fire engine the probability that it is available is 0.9.
A fire accident happens somewhere in the city one Thursday afternoon.
It is determined that fire engines will be needed:
(i) What is the chance that exactly 2 engines are available?
(ii) What is the probability that at least two of the engines are available?
(iii) What is the chance that none of the engines are available?
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All three problems are the binomial distribution type of problems. 


(i)   The number of trials n= 4; the probability of the success trial is 0.9,  and the number of success trials k = 2. 


          P = C%5B4%5D%5E2%2A0.9%5E2%2A%281-0.9%29%5E%284-2%29 = %28%284%2A3%29%2F2%29%2A0.9%5E2%2A0.1%5E2 = 0.0486  (rounded).     ANSWER



(ii)  The number of trials n= 4; the probability of the success trial is 0.9,  and the number of success trials k >= 2.


      We need calculate  P(n=4; k>=2; p=0.9).      


      To facilitate calculations, I use an appropriate online (free of charge) calculator at this web-site 

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


      It provides nice instructions  and  a convenient input and output for all relevant options/cases.


          P(n=4; k>=2; p=0.9) = 0.9963   (rounded).       ANSWER



(iii)  This probability is


          P = %281-0.9%29%5E4 = 0.1%5E4 = 0.0001.      ANSWER

Solved.     //     All questions are answered and explained.

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If you want to see many similar  (or different)  solved problems,  look into the lessons
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in this site.

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