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As part of her MBA program, Karen applied for three different
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25 Feb 2015, 04:47
Question Stats:
78% (01:59) correct 22% (02:26) wrong based on 264 sessions
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As part of her MBA program, Karen applied for three different Spring Break outreach projects, each of which selects its students by a random lottery of its applicants.. If the probability of her being accepted to each individual project is 20%, what is the probability that Karen will be accepted to at least one project? A. 12/125 B. 4/15 C. 61/125 D. 8/15 E. 113/125
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Re: As part of her MBA program, Karen applied for three different
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25 Feb 2015, 04:50



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As part of her MBA program, Karen applied for three different
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25 Feb 2015, 04:54
I found this question in a Veritas prep test I did. I found it interesting because it uses percentages. So, this is what I did. Being accepted to each individual project has a probability of 20%. Since we are looking for the probability of being accepted to at least one, they way I would procede would be by finding the opposite probability (being accepted to none) and subtracting it from 1. So, if she has 20/100 chances of being accepted for each, she has 80/100 chances of not being accepted for each. \(\frac{80}{100}\) \(=\)\(\frac{8}{10}\) \(=\)\(\frac{4}{5}\) \(\frac{4}{5}\)*\(\frac{4}{5}\)*\(\frac{4}{5}\) = \(\frac{64}{125}\), and \(1\)\(\frac{64}{125}\) \(=\)\(\frac{61}{125}\). So, ANS C



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Re: As part of her MBA program, Karen applied for three different
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25 Feb 2015, 06:38
pacifist85 wrote: I found this question in a Veritas prep test I did. I found it interesting because it uses percentages. So, this is what I did. Being accepted to each individual project has a probability of 20%. Since we are looking for the probability of being accepted to at least one, they way I would procede would be by finding the opposite probability (being accepted to none) and subtracting it from 1. So, if she has 20/100 chances of being accepted for each, she has 80/100 chances of not being accepted for each. \(\frac{80}{100}\) \(=\)\(\frac{8}{10}\) \(=\)\(\frac{4}{5}\) \(\frac{4}{5}\)*\(\frac{4}{5}\)*\(\frac{4}{5}\) = \(\frac{64}{125}\), and \(1\)\(\frac{64}{125}\) \(=\)\(\frac{61}{125}\). So, ANS Chi pacifist, good.. this is the best way to do.. it is the smartest and least likely way to be wrong..
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Re: As part of her MBA program, Karen applied for three different
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28 Oct 2015, 13:38
Bunuel,
I get this method: P(at least one) = 1  P(none) = 1  (8/10)^3 = 1  (4/5)^3 = 61/125. Could you show the longer method. Thank you



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Re: As part of her MBA program, Karen applied for three different
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28 Oct 2015, 21:38
Hi tytus, GMAT questions can typically be approached in a number of different ways (some of which are shorter/easier than others). In this prompt, it's easiest/fastest to calculate what we DON'T want to have happen and subtract that from 1. If you wanted to approach this the other way (calculate ALL of the different ways to get accepted to AT LEAST ONE Program, here's how you would do it)... Probability of exactly 1 program (of the 3): First Program "yes"; second and third "no" (1/5)(4/5)(4/5) = 16/125 Second Program "yes"; first and third "no" (4/5)(1/5)(4/5) = 16/125 Third Program "yes"; first and second "no" (4/5)(4/5)(1/5) = 16/125 Probability of exactly 2 programs (of the 3): First and second Program "yes"; third "no" (1/5)(1/5)(4/5) = 4/125 First and third Program "yes"; second "no" (1/5)(4/5)(1/5) = 4/125 Second and third Program "yes"; first "no" (4/5)(1/5)(1/5) = 4/125 Probability of all 3 programs: (1/5)(1/5)(1/5) = 1/125 Total of all options: 16/125+16/125+16/125+4/125+4/125+4/125+1/125 = 61/25 Final Answer: As you can see, this is somewhat tedious, so learning the most efficient ways to deal with prompts can save you LOTS of time. GMAT assassins aren't born, they're made, Rich
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Re: As part of her MBA program, Karen applied for three different
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28 Oct 2015, 23:40
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. As part of her MBA program, Karen applied for three different Spring Break outreach projects, each of which selects its students by a random lottery of its applicants.. If the probability of her being accepted to each individual project is 20%, what is the probability that Karen will be accepted to at least one project? A. 12/125 B. 4/15 C. 61/125 D. 8/15 E. 113/125 Since the probability of Karen's being accepted to each individual project is 20%, the probability of her not being accepted to each project is 80%. The probability that she will be accepted to none is, therefore, (4/5)*(4/5)*(4/5)=64/125. Since the complementary event that Karen will be accepted to at least one project is that she will be accepted to none, the probability that Karen will be accepted to at least one project is 1(64/125)= 61/125. The answer is (C)
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29 Oct 2015, 04:10
EMPOWERgmatRichC  very helpful, thank you.



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Re: As part of her MBA program, Karen applied for three different
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21 Feb 2018, 13:23
pacifist85 wrote: As part of her MBA program, Karen applied for three different Spring Break outreach projects, each of which selects its students by a random lottery of its applicants.. If the probability of her being accepted to each individual project is 20%, what is the probability that Karen will be accepted to at least one project?
A. 12/125
B. 4/15
C. 61/125
D. 8/15
E. 113/125 We can use the equation: P(accepted to at least one project) = 1  P(accepted to no projects) P(accepted to no projects) = 4/5 x 4/5 x 4/5 = 64/125 P(accepted to at least one project) = 1  64/125 = 61/125 Answer: C
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As part of her MBA program, Karen applied for three different
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11 Aug 2018, 11:15
pacifist85 wrote: As part of her MBA program, Karen applied for three different Spring Break outreach projects, each of which selects its students by a random lottery of its applicants.. If the probability of her being accepted to each individual project is 20%, what is the probability that Karen will be accepted to at least one project?
A. 12/125
B. 4/15
C. 61/125
D. 8/15
E. 113/125 164/125 is the best way to do it. Though just in case , if some body wants to kill 4 minutes as I did . probability of her selection=20%=1/5 not her selection = 4/5 we have 3 places , and she can fill 1,2 or 3 seats. no of ways she can fill 1 out of 3 = 3c1 no of ways she can fill 2 out of 3=3c2 and for all 3= 3c3 p(one seat)=3*(1/5*4/5*4/5)=3*16/5=48/5 p(two seats)=3*(1/5*1/5*4/5)=3*4/125=12/125 p(all 3 seats)=1*1/125=1/125 summing all 3 we have 61/125




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