107 total foreign language students
42 take spanish
46 take french
45 take german
8 take spanish & french but not german
6 take french & german but not spanish
8 take spanish & german but not french
24 students are taking at least 2 languages
What is the probability that a student randomly selected from those taking any language does take all three languages in this semester?
What is the probability that a student randomly selected from those taking exactly one language takes French in this semester?
well, 24 are taking at least two languages, if you add the above taking 2 languages, you get 8+8+6=22, which means 2 are taking three languages.
Pr(all three)=2/106
well, 42 are taking french, but of those, 6 are taking german as well, 2 are taking spanish and german, and 8 are taking spanish with french.
Pr(only French)= (42-6-2-8)/(106-24-2)
To find the answer to these probability questions, we need to use the concept of set intersection and set union.
Let's define the following sets:
A: Set of students taking Spanish
B: Set of students taking French
C: Set of students taking German
Using the given information:
|A| = 42 (number of students taking Spanish)
|B| = 46 (number of students taking French)
|C| = 45 (number of students taking German)
|A ∩ B ∩ C| = ? (number of students taking all three languages)
Now let's find the number of students taking all three languages:
|A ∩ B ∩ C| = |A| + |B| + |C| - |A ∪ B| - |B ∪ C| - |A ∪ C| + |A ∪ B ∪ C|
= 42 + 46 + 45 - 8 - 6 - 8 + 24
= 135 - 22
= 113
So, there are 113 students taking all three languages.
Now, we can find the first probability:
P(student takes all three languages) = |A ∩ B ∩ C| / (|A| + |B| + |C| - |A ∪ B| - |B ∪ C| - |A ∪ C| + |A ∪ B ∪ C|)
= 113 / (42 + 46 + 45 - 8 - 6 - 8 + 24)
= 113 / 135
≈ 0.837
Therefore, the probability that a randomly selected student from those taking any language takes all three languages is approximately 0.837.
Now let's find the second probability:
P(student takes exactly one language and it is French) = |B| / (|A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |A ∩ C| + |A ∪ B ∪ C|)
= 46 / (42 + 46 + 45 - 8 - 6 - 8 + 24)
= 46 / 135
≈ 0.341
Therefore, the probability that a randomly selected student from those taking exactly one language takes French in this semester is approximately 0.341.