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# Of the students at a certain business school, 60% are

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03 Oct 2005, 06:04
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Of the students at a certain business school, 60% are first-year students and 40% are second-year students. Of the first-year students, 50% are receiving some form of financial aid. Of the second-year students, 25% are receiving some form of financial aid. If a student is chosen at random, what is the probability that he or she is receiving some form of financial aid?

A. .50
B. .33
C. .10
D. .30
E. .40
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03 Oct 2005, 06:53
I got E. I usually do these sort of problems by picking a number and deducing the answer. For instance, let's say the b-school has 100 students. So there are 60 (60%) first year students and 40 (40%) second year students. Of the first year students, 30 (50%) get FA, while the other 30 does not. Of the second year students, 10 (25%) get FA while 30 do not. This adds up to 40 total students that receive FA out of 100. This equals 0.40, answer E. What is the OA?
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03 Oct 2005, 07:58
Picking 100, when possible, is always a good strategy. OA anyone?
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03 Oct 2005, 08:52
Also don't forget that drawing a table can clear up this problem considerably. I'll try this here at the forumn.

___1yr_2yr____
FA l 30 l 10 l 40 total
------------------
NFA l 30 l 30 l 60 total
------------------
-------60 -40---100 total students
FA- Financial Aid
NFA-No Financial Aid

I hope this makes sense. It assumes 100 students total, this method can also be used for various overlapping sets.

Last edited by jrabenho on 03 Oct 2005, 08:58, edited 8 times in total.
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03 Oct 2005, 08:53
probability of mutually exclusive (either 1st year or 2nd year) - dependent events:

p (aid) = p(1st year) * p (aid/first year) + p(2nd year) * p(aid/second year)
= 0.6 * 0.5 + 0.4 * 0.25
= 0.4
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03 Oct 2005, 10:53
Since we're dealing w/ %ages let's assume a total of 100 students
1st year 2nd year Total #
#: 60 40 100
Aid #: 30 (50% of 60) 10 (25% of 40) 40

40/100 = .4 or 40%
Answer is E what's the OA?
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03 Oct 2005, 23:08
OA is E. Picking the number 100 made it all come together magically.
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04 Oct 2005, 00:38
Assume
Total number of students in school: 100
1st year = 60
2nd year = 40

1st year with aid = 50/100 * 60 = 30
2nd year with aid = 25/100 * 40 = 10

Total receiving aid = 40
P(receiving aid) = 40/100 = .40

Ans: E
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05 Oct 2005, 01:15
GMATT73 wrote:
Of the students at a certain business school, 60% are first-year students and 40% are second-year students. Of the first-year students, 50% are receiving some form of financial aid. Of the second-year students, 25% are receiving some form of financial aid. If a student is chosen at random, what is the probability that he or she is receiving some form of financial aid?

A. .50
B. .33
C. .10
D. .30
E. .40

P(Ist yr) = .6
P(II yr) = .4
P(Ist yr and fin aid) = .5
P(II yr and fin aid) = .25
P(rec fin aid) = (.6 * .5) + (.4 * .25) = .40

I should say, I also "enjoyed" the solution of silentell
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05 Oct 2005, 12:49
I got E.

Since we were dealing with ood percentage numbers, I chose 100 student as the total. 60=first year and 40=second year. Then I got 30=1st w/fin aid and 10=2nd w/fin aid for a total of 40.

So, 40/100 = .4
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05 Oct 2005, 14:09
GMATT73 wrote:
Of the students at a certain business school, 60% are first-year students and 40% are second-year students. Of the first-year students, 50% are receiving some form of financial aid. Of the second-year students, 25% are receiving some form of financial aid. If a student is chosen at random, what is the probability that he or she is receiving some form of financial aid?

A. .50
B. .33
C. .10
D. .30
E. .40

another E
pick number 100 HERE

60 first year half 30 FINANCIAL AID
40 second year 10 FINANCIAL
TOTAL FINANCIAL AID 40 over total number of student first and second year
40%
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Re: PS Statistics   [#permalink] 05 Oct 2005, 14:09
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