Question 422396: At a university with 1,000 business majors, there are 200 business students enrolled in an introductory statistics course. Of these 200 students, 50 are also enrolled in an introductory accounting course. There are an additional 250 business students enrolled in accounting but not enrolled in statistics.
If a business student is selected at random, what is the probability that the student is either enrolled in accounting or statistics, but not both?
ii. Given the same problem shown above, if a business student is selected at random, what is the probability that the student is enrolled in accounting?
iii.Given the same problem shown above, if a business student is selected at random ,what is the probability that the student is enrolled in both statistics and accounting?
IV. Given the same problem shown above, if a business student is selected at random and found to be enrolled in statistics, what is the probability that the student is also enrolled in accounting?
V. Given the same problem shown above, if a business student is selected at random,what is the probability that the student is enrolled in neither accounting nor statistics?
VI.Given the same problem shown above, if a business student is selected at random, what is the probability that the student is not enrolled in statistics?
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Statistics (and possibly accounting) : 200
Statistics And Accounting: 50
Statistics only: 200 minus 50 = 150.
Accounting only: 250
Total Business students: 1000.
The probability of anything is the number of ways it can happen that you would consider a success divided by the number of ways that it can happen that are either a success or a failure.
For your first problem, the number of ways that it can happen is the total number of business students, namely 1000. That's your denominator. The number of ways that it can happen and be a success is the number of statistics only students, i.e. 150 plus the number of accounting only students, namely 250, or a total of 400. That's your numerator.
So the answer to the first one is 400 divided by 1000 or 40%.
You should be able to handle the rest of them. Be careful about how you calculate your denominator in part 4.
John
My calculator said it, I believe it, that settles it
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! At a university with 1,000 business majors, there are 200 business students enrolled in an introductory statistics course. Of these 200 students, 50 are also enrolled in an introductory accounting course. There are an additional 250 business students enrolled in accounting but not enrolled in statistics.
Draw the Venn Diagram:
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If a business student is selected at random, what is the probability that the student is either enrolled in accounting or statistics, but not both?
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Ans: (200 + 250)/1000 = 450/1000 = 0.45
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ii. Given the same problem shown above, if a business student is selected at random, what is the probability that the student is enrolled in accounting?
250/1000 = 1/4
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iii.Given the same problem shown above, if a business student is selected at random ,what is the probability that the student is enrolled in both statistics and accounting?
50/1000 = 1/20
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IV. Given the same problem shown above, if a business student is selected at random and found to be enrolled in statistics, what is the probability that the student is also enrolled in accounting?
50/1000 = 1/20
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V. Given the same problem shown above, if a business student is selected at random,what is the probability that the student is enrolled in neither accounting nor statistics?
500/1000 = 1/2
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VI.Given the same problem shown above, if a business student is selected at random, what is the probability that the student is not enrolled in statistics?
(1000-250)/1000 = 3/4
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Cheers,
Stan H.
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