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exitwound
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Review problem. I have the answers. Don't know how to get them. Answers in brackets.
Shortly after being put into service, some buses manufactured by a certain company have developed cracks on the underside of the main frame. Suppose a paricular city has 25 of these buses, and cracks have actually appeared on 8 of them.
a.)How many ways are there to select a sample of 5 buses from the 25 for a thorough inspection? [53130]
b.) In how many ways can a sample of 5 buses contain exactly 4 with visible cracks? [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? [.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? [.0235]
[tex]Permutations = P_{k,n} = \frac{n!}{(n-k)!}[/tex]
[tex]Combinations = \begin{pmatrix}
n\\
k
\end{pmatrix}= \frac{n!}{k!(n-k)!}[/tex]
a.) I'm looking for the total number of combinations of choosing 5 buses out of the 25 buses. Therefore
[tex]Combinations = \begin{pmatrix}
n\\
k
\end{pmatrix} = \frac{n!}{k!(n-k)!} = \frac{25!}{5!(25-5)!} = 53130[/tex]
b.) I don't know how to set it up.
c.) Didn't attempt it because I can't do (b).
d.) Didn't attempt it because I can't do (b).
Homework Statement
Shortly after being put into service, some buses manufactured by a certain company have developed cracks on the underside of the main frame. Suppose a paricular city has 25 of these buses, and cracks have actually appeared on 8 of them.
a.)How many ways are there to select a sample of 5 buses from the 25 for a thorough inspection? [53130]
b.) In how many ways can a sample of 5 buses contain exactly 4 with visible cracks? [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? [.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? [.0235]
Homework Equations
[tex]Permutations = P_{k,n} = \frac{n!}{(n-k)!}[/tex]
[tex]Combinations = \begin{pmatrix}
n\\
k
\end{pmatrix}= \frac{n!}{k!(n-k)!}[/tex]
The Attempt at a Solution
a.) I'm looking for the total number of combinations of choosing 5 buses out of the 25 buses. Therefore
[tex]Combinations = \begin{pmatrix}
n\\
k
\end{pmatrix} = \frac{n!}{k!(n-k)!} = \frac{25!}{5!(25-5)!} = 53130[/tex]
b.) I don't know how to set it up.
c.) Didn't attempt it because I can't do (b).
d.) Didn't attempt it because I can't do (b).
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