# Probabilty of choosing 5 defective buses

Printable View

• June 19th 2010, 04:46 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 $\left({25\atop 5}\right)=\frac{25!}{5!(25-5)!}=53130.$
If the order does matter the answer is $\left(25\right)_{5}=\frac{25!}{(25-5)!}=6,375,600.$

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 $\left({8\atop 4}\right)$ways to choose the 4 buses with the visible cracks
$17*\frac{8!}{4!(8-4)!}=\frac{17*8*7*6*5}{4*3*2*1}=1190.$

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 $\left(\frac{1190}{53130}\right)=.0224$

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 $\left({8\atop 5}\right)=\frac{8!}{5!(8-5)!}=56.$ Probability of choosing exactly 5 buses with visible cracks is $\left(\frac{56}{53130}\right)=.0011.$
Probability of choosing 5 buses with at least 4 with visible cracks .0224 + .0011 = .0235.

Please citique the above calculations
• June 20th 2010, 12:39 AM
undefined
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 $\left({25\atop 5}\right)=\frac{25!}{5!(25-5)!}=53130.$
If the order does matter the answer is $\left(25\right)_{5}=\frac{25!}{(25-5)!}=6,375,600.$

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 $\left({8\atop 4}\right)$ways to choose the 4 buses with the visible cracks
$17*\frac{8!}{4!(8-4)!}=\frac{17*8*7*6*5}{4*3*2*1}=1190.$

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 $\left(\frac{1190}{53130}\right)=.0224$

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 $\left({8\atop 5}\right)=\frac{8!}{5!(8-5)!}=56.$ Probability of choosing exactly 5 buses with visible cracks is $\left(\frac{56}{53130}\right)=.0011.$
Probability of choosing 5 buses with at least 4 with visible cracks .0224 + .0011 = .0235.

Please citique the above calculations

Everything looks good to me. Only things I can think to criticise are:

(a) usually I am the one who says "the question is ambiguous" but in this case I think it's pretty clear order is not considered, since we are selecting a sample. But what you wrote doesn't hurt anything.

(d) this methodology could induce rounding error. In this case it worked out the same, but under different circumstances you might have been off by 0.0001.
• June 20th 2010, 06:38 AM
oldguynewstudent
Thanks you always give great advice!