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Michael has never taken a foreign language class, but is doing a story on them for the school newspaper. The school offers French and Spanish. Michael has a list of all 25 kids in the school enrolled in at least one foreign language class. He also knows that 18 kids are in the French class and 21 kids are in the Spanish class. If Michael chooses two kids at random off his list and interviews them, what is the probability that he will be able to write something about both the French and Spanish classes after he is finished with the interviews? Express your answer as a fraction in simplest form.

Dec 31, 2018

#1
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$$\text{Let's split the students up into three non-intersecting sets}\\ F=\{\text{students taking only french}\}\\ S=\{\text{students taking only spanish}\}\\ FS=\{\text{students taking both french and spanish}\}\\$$

$$|FS|=|F|+|S|-|F\cup S| = 18+21-25 = 14\\ |F|-18-14=4\\ |S|=21-14=7$$

$$\text{4 outcomes lead to Michael being able to write about both classes}\\ (F,S),~(FS,S),~(FS,F)~,(FS,FS)\\ \text{These outcomes are non-overlapping so the probability}\\ \text{of their union is the sum of their individual probabilities}$$

$$P[(F,S)] = \dfrac{\dbinom{4}{1}\dbinom{7}{1}}{\dbinom{25}{2}}=\dfrac{7}{75}\\ P[(FS,S)] = \dfrac{\dbinom{14}{1}\dbinom{7}{1}}{\dbinom{25}{2}}\dfrac{49}{150}\\ P[(FS,F)] = \dfrac{\dbinom{14}{1}\dbinom{4}{1}}{\dbinom{25}{2}}=\dfrac{14}{75}\\ P[(FS,FS)] = \dfrac{\dbinom{14}{2}}{\dbinom{25}{2}}=\dfrac{91}{300}$$

$$P[\text{M. can write about both schools}] = \dfrac{28+98+56+91}{300} = \dfrac{91}{100}$$

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Jan 1, 2019
edited by Rom  Jan 1, 2019
edited by Rom  Jan 1, 2019