Probability: Series of Events Without Repetition
A car rental service facility has 10 foreign cars and 15 domestic cars waiting to
be serviced on a
particular Saturday morning. Because there is only one mechanic, only 6 cars can be serviced. If the 6 cars are
chosen at random, what is the probability that the first 3 cars serviced are domestic and the
last 3 cars serviced are foreign?
Correct.
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There are 15 domestic cars in 25 cars total, so the probability that the first car chosen is domestic is $$\frac{15}{25}$$, the probability for the second is $$\frac{14}{24}$$, and the probability for the third is $$\frac{13}{23}$$. The probability that the fourth car is foreign (there are 10 foreign cars in the 22 remaining cars) is $$\frac{10}{22}$$, the probability for the fifth is $$\frac{9}{21}$$, and the probability for the sixth is $$\frac{8}{20}$$.
Multiply, because multiplication is fun! (and also because there's an "and" relationship between car picks):
>$$\text{Probability} = \frac{15}{25}\times\frac{14}{24}\times\frac{13}{23}\times
\frac{10}{22}\times\frac{9}{21}\times\frac{8}{20}$$.
Incorrect.
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Note that there are only 10 foreign cars, not 12!
Incorrect.
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Incorrect.
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Incorrect.
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$$\frac{15\times14\times13\times10\times9\times8
}{25!}$$
$$\frac{15}{25}\times\frac{14}{24}\times\frac{13}{23}\times
\frac{12}{22}\times\frac{11}{21}\times\frac{10}{20}$$
$$\frac{15}{25}\times\frac{14}{24}\times\frac{13}{23}\times
\frac{10}{22}\times\frac{9}{21}\times\frac{8}{20}$$
$$\frac{15\times14\times13\times12\times11\times10
}{25!}$$
$$1-\frac{15}{25}\times\frac{14}{24}\times\frac{13}{23}\times
\frac{10}{22}\times\frac{9}{21}\times\frac{8}{20}$$
