Question 47461: 40 students enrolled in three areas of study.
23 students in music
25 students in math
27 students in comp. science
7 took mathematics and music only
5 took music and computer science only
6 took math and comp. sci. only
7 took all three subjects
What is the probability a student took exactly one of the three courses?
Found 2 solutions by venugopalramana, mszlmb:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 40 students enrolled in three areas of study.
LET US USE A ^ B TO SHOW A INTERSECTION B
23 students in music(S..SAY)
25 students in math(M..SAY)
27 students in comp. science(C...SAY)
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7 took mathematics and music DISABLED_event_only= M&S ONLY
5 took music and computer science DISABLED_event_only= S&C ONLY
6 took math and comp. sci. DISABLED_event_only= M&C ONLY
7 took all three subjects=M&C&S=7
===================================================
HENCE THOSE WHO TOOK MATHS DISABLED_event_ONLY= M-M&S ONLY-M&C ONLY - M&S&C=25-7-6-7=5
THOSE WHO TOOK MUSIC DISABLED_event_ONLY= S-S&M ONLY-S&C ONLY-S&C&M=23-7-5-7=4
THOSE WHO TOOK COMPUTRERS DISABLED_event_ONLY= C-C&M ONLY-C&S ONLY-C&S&M=27-6-5-7=9
HENCE PROBABILITY OF HAVING THOSE WHO TOOK ONLY ONE SUBJECT=(5+4+9)/40=18/40=9/20
What is the probability a student took exactly one of the three courses?
You may edit the question. Maybe convert formulae to the same formula notatio
Answer by mszlmb(115)  (Show Source):
You can put this solution on YOUR website! I'm sorry, but every time I've tried this problem, it adds up to 43 students, not 40.::
23
/ | \
7 7 5
/ / \ \
25---6---27
As you can see, this is my attempt at a triangle.
The numbers are thence explained:
the corners of the triangle, 23, 25, and 27, are respectively the number of students enrolled in
music, math, and comp. sci..
the 7 connecting just music and math is the 7 students in only both classes.
the 6 connecting the math and comp. sci. is the students in only the two classes.
the 5 connecting the music and comp. sci. is the only 5 doing those two classes.
and lastly the 7 connecting the 3 represents the 7 students taking all three classes.
To get the sum of all of the students taking only one course, one must take all three numbers
and subtract from each the numbers connecting to it.
For example, to get the total number of students ONLY enrolled in music, I
would equate 23-7-7-5.
this means 23 minus the 7 math/music minus the 7 math/music/comp minus the 5 music/comp., it would give me 4.
If we do this for all of the numbers, we'd have 23-29, 25-20, and 27-18 for (respectively)
music, math, and comp. The answers for each are 6, 5, and 9.
We'd be fine up to this point, saying that p(only one course) is (5+6+9)/40, or .5, until we
check to make sure it really is 40. if it were, then
5+6+9+5+6+7+7=40.
with key:
students enrolled in just one class
students in just two classes
students in all three
total number of students
However, this is not the case:
5+6+9+5+6+7+7=43, not 40. So the problem is, I think, wrong. Were it not for the bit of
information that the total number of students were 40, we could assume it were 43,
and our probability would be 30/43. Also, suppose it were the other way around
and we were told there were 43 students, we could assume 3 are unenrolled in any classes.
Alas, that is as much assistance as I can give. But ask for more if you need to!
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