SOLUTION: eith’s Florists has 15 delivery trucks, used mainly to deliver flowers and flower arrangements in the Greenville, South Carolina, area. Of these 15 trucks, 7 have brake problems. A

Algebra ->  Probability-and-statistics -> SOLUTION: eith’s Florists has 15 delivery trucks, used mainly to deliver flowers and flower arrangements in the Greenville, South Carolina, area. Of these 15 trucks, 7 have brake problems. A      Log On


   



Question 1119808: eith’s Florists has 15 delivery trucks, used mainly to deliver flowers and flower arrangements in the Greenville, South Carolina, area. Of these 15 trucks, 7 have brake problems. A sample of 6 trucks is randomly selected. What is the probability that 2 of those tested have defective brakes? (Round your answer to 4 decimal places.)
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
This would be 6C2*9C5, the number of ways 6 can have 2 with brake problems * the number of ways the other 9 would have 5 brake problems.
Denominator is 15C7 (the sum of the 6 and 9 and the 2 and 5.
This is 1890/6435 or 0.2937
Another way is to take a sample of 6 with 2
7/15*6/14*8/13*7/12*6/11*5/10=70560/(15*14*13*12*11*10)=0.0196. There are 15 of these ways where 2 will have brake problems and where the numerator is ordered differently but with the same numbers, so all 15 is 0.2937.

Answer by ikleyn(52802) About Me  (Show Source):
You can put this solution on YOUR website!
.
highlight%28cross%28eiths%29%29 Keith's Florists has 15 delivery trucks, used mainly to deliver flowers and flower arrangements
in the Greenville, South Carolina, area. Of these 15 trucks, 7 have brake problems.
A sample of 6 trucks is randomly selected. What is the probability that 2 of those tested have defective brakes?
(Round your answer to 4 decimal places.)
~~~~~~~~~~~~~~~~~~~~~~

This probability is this ratio  %28C%5B7%5D%5E2%2AC%5B8%5D%5E4%29%2FC%5B15%5D%5E6.


We start choosing 2 defective cars among of 7 defective. It can be done in C%5B7%5D%5E2 = %287%2A6%29%2F2 = 21 ways.


We complement 2 defective cars to 6 by adding 4 regular cars from 8 = 15-7 regular cars.  It can be done by C%5B8%5D%5E4 = %288%2A7%2A6%2A5%29%2F%281%2A2%2A3%2A4%29 = 70 ways.


The total number of ways to choose 6 cars of 15 is  C%5B15%5D%5E6 = %2815%2A14%2A13%2A12%2A11%2A10%29%2F%281%2A2%2A3%2A4%2A5%2A6%29 = 5005.


Thus the probability under the question is  %2821%2A70%29%2F5005 = 0.2937 = 29.37%.