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An airline is given permission to fly $4$ new routes of its choice. The airline is considering $12$ new routes: $4$ routes in Florida, $5$ routes in California, and $3$ routes in Texas. If the airline selects the $4$ new routes at random from the $12$ possibilities, determine the probability that:

a) $2$ are in Florida and $2$ are in Texas

b) $3$ are in California and $1$ is in Florida

c) $1$ is in Florida, $1$ is in California, and $2$ are in Texas

d) at least one is in Texas.

Please can someone help me with this question?

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2 Answers 2

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There are $\dbinom{12}{4}$ ways to choose $4$ routes from the $12$. Under our assumptions, they are equally likely.

(a) There are $\dbinom{4}{2}$ ways to choose $2$ Florida routes from the $4$ available. For each of these ways, there are $\dbinom{3}{2}$ ways to choose $2$ Texas routes, for a total of $\dbinom{4}{2}\dbinom{3}{2}$. Divide by $\dbinom{12}{4}$ for the probability.

Parts (b) and (c) are done in a way very similar to (a), so I will leave them to you.

For (d), it is easier to find first the probability that none of the routes are to Texas. There are $\dbinom{9}{4}$ ways to choose $4$ routes that don't include Texas. To obtain the probability we entirely avoid Texas, divide by $\dbinom{12}{4}$. To find the probability that at least $1$ route is to Texas, subtract the number just obtained from $1$.

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a.) $(\dbinom{4}{2}+\dbinom{3}{1})/\dbinom{12}{4}$,

b.) $(\dbinom{5}{3}+\dbinom{4}{1})/\dbinom{12}{4}$,

c.) $(\dbinom{4}{1}+\dbinom{5}{1}+\dbinom{3}{2})/\dbinom{12}{4}$,

b.) $1-(\dbinom{9}{4}/\dbinom{12}{4}$.

EDIT: added the TeX tags.