# Probabilty of choosing 5 defective buses

• Jun 19th 2010, 03:46 PM
oldguynewstudent
Probabilty of choosing 5 defective buses
Shortly after being put into service, ssome buses manufactured by a certain company have developed cracks on the underside of the main frame. Suppose a particular city has 25 of these buses, and cracks have actually appeared in 8 of them.

a) How many ways are there to select a sample of 5 buses from the 25 for a thorough inspection?
If the order in which the buses are selected does not matter the answer is $\displaystyle \left({25\atop 5}\right)=\frac{25!}{5!(25-5)!}=53130.$
If the order does matter the answer is $\displaystyle \left(25\right)_{5}=\frac{25!}{(25-5)!}=6,375,600.$

b) In how many ways can a sample of 5 buses contain exactly 4 with visible cracks?
There are 17 ways to choose the bus without the crack times $\displaystyle \left({8\atop 4}\right)$ways to choose the 4 buses with the visible cracks
$\displaystyle 17*\frac{8!}{4!(8-4)!}=\frac{17*8*7*6*5}{4*3*2*1}=1190.$

c) If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5 will have visible cracks?
The probablitiy of choosing exactly 4 buses with visible cracks out of 5 radomly selected buses is $\displaystyle \left(\frac{1190}{53130}\right)=.0224$

d) If buses are selected as in part (c), what is the probability that at least 4 of those selected will have visible cracks.
This would be the probability of exactly 4 added to exactly 5 buses chosen will have visible cracks.
The number of ways to choose exactly 5 buses with visible cracks is $\displaystyle \left({8\atop 5}\right)=\frac{8!}{5!(8-5)!}=56.$ Probability of choosing exactly 5 buses with visible cracks is $\displaystyle \left(\frac{56}{53130}\right)=.0011.$
Probability of choosing 5 buses with at least 4 with visible cracks .0224 + .0011 = .0235.

• Jun 19th 2010, 11:39 PM
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Quote:

Originally Posted by oldguynewstudent
Shortly after being put into service, ssome buses manufactured by a certain company have developed cracks on the underside of the main frame. Suppose a particular city has 25 of these buses, and cracks have actually appeared in 8 of them.

a) How many ways are there to select a sample of 5 buses from the 25 for a thorough inspection?
If the order in which the buses are selected does not matter the answer is $\displaystyle \left({25\atop 5}\right)=\frac{25!}{5!(25-5)!}=53130.$
If the order does matter the answer is $\displaystyle \left(25\right)_{5}=\frac{25!}{(25-5)!}=6,375,600.$

b) In how many ways can a sample of 5 buses contain exactly 4 with visible cracks?
There are 17 ways to choose the bus without the crack times $\displaystyle \left({8\atop 4}\right)$ways to choose the 4 buses with the visible cracks
$\displaystyle 17*\frac{8!}{4!(8-4)!}=\frac{17*8*7*6*5}{4*3*2*1}=1190.$

c) If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5 will have visible cracks?
The probablitiy of choosing exactly 4 buses with visible cracks out of 5 radomly selected buses is $\displaystyle \left(\frac{1190}{53130}\right)=.0224$

d) If buses are selected as in part (c), what is the probability that at least 4 of those selected will have visible cracks.
This would be the probability of exactly 4 added to exactly 5 buses chosen will have visible cracks.
The number of ways to choose exactly 5 buses with visible cracks is $\displaystyle \left({8\atop 5}\right)=\frac{8!}{5!(8-5)!}=56.$ Probability of choosing exactly 5 buses with visible cracks is $\displaystyle \left(\frac{56}{53130}\right)=.0011.$
Probability of choosing 5 buses with at least 4 with visible cracks .0224 + .0011 = .0235.