Probability

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jsbeckton

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After Calc I, II , III, Differential Equations and Linear algebra done I thought that statistics would be a walk in the park! :D The first one was simple but I can't seem to figure out how to go about the others and my book is a bit vague. Can anyone walk me through these?

Shortly after being put into service, some buses have developed cracks. Suppose a city has 25 of these busses, and cracks in 8 on them.

a) How many ways are there to select a sample of 5 buses from the 25 for inspection?

\(\displaystyle C_{5,25} = \frac{{25!}}{{20!5!}} = 53130\)


b) In how many ways can a sample of 5 buses contain exactly 4 with visable cracks?

c) If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5 buses will have cracks?

d) If buses are selected as in part (c), what is the probability that at least 4 of those selected will have visable cracks?[/tex]
 
jsbeckton said:
After Calc I, II , III, Differential Equations and Linear algebra done I thought that statistics would be a walk in the park! :D The first one was simple but I can't seem to figure out how to go about the others and my book is a bit vague. Can anyone walk me through these?

Shortly after being put into service, some buses have developed cracks. Suppose a city has 25 of these busses, and cracks in 8 on them.

a) How many ways are there to select a sample of 5 buses from the 25 for inspection?

\(\displaystyle C_{5,25} = \frac{{25!}}{{20!5!}} = 53130\) correct!


b) In how many ways can a sample of 5 buses contain exactly 4 with visable cracks?
\(\displaystyle C_{4,5}\)

c) If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5 buses will have cracks?
Divide (b) by (a).
d) If buses are selected as in part (c), what is the probability that at least 4 of those selected will have visable cracks?
4 or 5
Click to expand...
 
For part (b) you do you mean:

\(\displaystyle C_{4,5} = \frac{{5!}}{{1!4!}} = 5\)

The anwsers in the back of the book are:
a) 53130
b) 1190
c) .022
d) .023

Am I misunderstanding what you mean? Thanks for the help bye the way.
 
Sorry, but I simply misread the question!
There are 8 with defects so in a sample of 5 for exactly 4 to have defects:
\(\displaystyle \L
\left( {C_{4,8} } \right)\left( {C_{1,17} } \right)\)
 
Thanks

For part (d) you said 4 or 5, i'm not too sure what you mean by that. I'm guessing that they want me to add the probability of exactly 4 (part c) to the probability of finding exactly 5, is that right? and how would I go about showing that?

would prob of exactly 5 be (C 5,8)(C 1,17) ?
 
NO!
exactly 5 be (C 5,8)(C 0,17)

[C(4,8)C(1,17)+C(5,8)]/C(5,25)
 
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