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Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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21 Apr 2015, 06:07
Question Stats:
68% (02:34) correct 32% (02:20) wrong based on 168 sessions
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Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen? A. 1/10 B. 3/28 C. 2/9 D. 1/4 E. 1/2 Kudos for a correct solution.
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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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21 Apr 2015, 21:00
C 2/9
have to chose 2 members and 3 more out of 10
so 2c2*8c3/10c5
simple..which gives 2/9
hope it helps




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Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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Updated on: 21 Apr 2015, 11:20
update: wrong explanation Bunuel wrote: Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?
A. 1/10 B. 3/28 C. 2/9 D. 1/4 E. 1/2
Kudos for a correct solution. Let's imagine that Ashley and Vinnie this is one person and in total we have 9 salespersons. We have 4C9 = 126 possible combinations and in half of this combinations will be present Ashley+Vinnie. So we have 63 possible combinations when Ashley and Vinnie chosen. And in real situtaion we have 5C10 = 252 possible combinations. And as we already know in 63 combinations will be present Ashley and Vinnie. So probability equal: \(\frac{63}{252} = \frac{1}{4}\) Answer is D
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Originally posted by Harley1980 on 21 Apr 2015, 09:18.
Last edited by Harley1980 on 21 Apr 2015, 11:20, edited 1 time in total.



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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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21 Apr 2015, 10:22
Hi Harley1980. I did slightly different and hence obviously ended up with the answer 2/9. I assumed that because Ashley and Winni have to be chosen, the remaining selection is 3 people from 8. 8C3/10C5= 2/9. Option C. Please could you explain your answer. Harley1980 wrote: Bunuel wrote: Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?
A. 1/10 B. 3/28 C. 2/9 D. 1/4 E. 1/2
Kudos for a correct solution. Let's imagine that Ashley and Vinnie this is one person and in total we have 9 salespersons. We have 5C9 = 126 possible combinations and in half of this combinations will be present Ashley+Vinnie. So we have 63 possible combinations when Ashley and Vinnie chosen. And in real situtaion we have 5C10 = 252 possible combinations. And as we already know in 63 combinations will be present Ashley and Vinnie. So probability equal: \(\frac{63}{252} = \frac{1}{4}\)
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Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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21 Apr 2015, 11:19
update: wrong explanation shriramvelamuri wrote: Hi Harley1980. I did slightly different and hence obviously ended up with the answer 2/9. I assumed that because Ashley and Winni have to be chosen, the remaining selection is 3 people from 8. 8C3/10C5= 2/9. Option C. Please could you explain your answer. Harley1980 wrote: Bunuel wrote: Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?
A. 1/10 B. 3/28 C. 2/9 D. 1/4 E. 1/2
Kudos for a correct solution. Let's imagine that Ashley and Vinnie this is one person and in total we have 9 salespersons. We have 5C9 = 126 possible combinations and in half of this combinations will be present Ashley+Vinnie. So we have 63 possible combinations when Ashley and Vinnie chosen. And in real situtaion we have 5C10 = 252 possible combinations. And as we already know in 63 combinations will be present Ashley and Vinnie. So probability equal: \(\frac{63}{252} = \frac{1}{4}\) I'll be honest: I'm not quite sure in my answer But I think that I see mistake in your calculations You wrote: "I assumed that because Ashley and Winni have to be chosen, the remaining selection is 3 people from 8. " But I think we can't do this, for example we need to calculate 4C6 = 15 variants and if we presuppose that first position is taken and calculate remained 3 position from 5 variants: 3C5 we receive 10 variants + 1 number, but this is wrong approach because we lost 5 variants.
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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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21 Apr 2015, 19:53
Hi shriramvelamuri, The 'math' behind this question can be done in a couple of different ways; your approach IS one of them. Hi Harley1980, I understand the logic that you used, but by 'combining' Ashley and Vinnie into '1 person', you've changed the parameters of the question (and the 'inputs' into the equations). Since there are 10 people in total (not 9), you have to account for the possible outcomes that would include only 1 of the 2 people (just Ashley or just Vinnie) and that doesn't happen in your calculations. Those options decrease the probability that BOTH Ashley and Vinnie will be chosen, which is why the correct answer is LESS than 1/4. GMAT assassins aren't born, they're made Rich
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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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21 Apr 2015, 20:19
Hi Empower, Can we then safely assume the answer is 2/9. EMPOWERgmatRichC wrote: Hi shriramvelamuri,
The 'math' behind this question can be done in a couple of different ways; your approach IS one of them.
Hi Harley1980,
I understand the logic that you used, but by 'combining' Ashley and Vinnie into '1 person', you've changed the parameters of the question (and the 'inputs' into the equations). Since there are 10 people in total (not 9), you have to account for the possible outcomes that would include only 1 of the 2 people (just Ashley or just Vinnie) and that doesn't happen in your calculations. Those options decrease the probability that BOTH Ashley and Vinnie will be chosen, which is why the correct answer is LESS than 1/4.
GMAT assassins aren't born, they're made Rich
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Kudos to you, for helping me with some KUDOS.



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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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21 Apr 2015, 21:30
Hi shriramvelamuri, Yes, your approach and explanation are correct. GMAT assassins aren't born, they're made, Rich
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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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22 Apr 2015, 00:24
Favourable cases : Selecting Ashley and Vinnie in 1 way, and then selecting 3 more people from remaining 8. thus , Fav = 1 * 8C3. Total cases = 10C5 Thus prob : 8C3 / 10C5 = 2/9 Ans C.
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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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27 Apr 2015, 03:05
Bunuel wrote: Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?
A. 1/10 B. 3/28 C. 2/9 D. 1/4 E. 1/2
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:This problem includes a mix of combinatorics and probability. The probability that both Ashley and Vinnie will be chosen is: the number of teams on which they're both chosen divided by the number of total possible teams. In these situations it's usually easier to start with the total. With 10 total people and 5 chosen, the number of combinations (order doesn't matter!) can be calculated as: 10!/(5!5!), which factors out to (10)(9)(8)(7)(6)/[(5)(4)(3)(2)]. That ultimately reduces to 252, but remember that you're doing all of this to find a denominator, the number of total teams. So you may want to leave the math undone; as you factor the numerator 6 with the denominator 2 and 3, and then divide the 10 by 5 and the 8 by 4, you're really left with: 2 * 9 * 2 * 7, which you'll find makes an easier denominator. Then for the number of teams that both Vinnie and Ashley are on. The logic here: if those two are on the team, then there are 8 people left to be chosen for 3 remaining spots. So the calculation is: 8!/(3!5!), which reduces to (8)(7)(6)/[(3)(2)], which is 56 (or you can just leave the 8 * 7 undone). That leaves you with a "teams they're on, divided by teams total" calculation of: (8)(7)/[(2)(2)(9)(7)], in which the 7s cancel and the 8 divided by (2 * 2) leaves a 2 in the numerator. That simplifies to 2/9.
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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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01 Jul 2015, 14:01
[2C2*8C3]/[10C5]=2/9



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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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19 Feb 2018, 23:49
Hi. Can any one suggest how to do this by probability method?
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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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18 Nov 2018, 01:08
Total Outcome = 10C5 (selecting 5 out of 10) favourable outcome = 8C3 (selecting 3 out of 8 people as 2 are already selected Ashley & Vinnie) Probability = favourable outcome / total outcome = 8C3 / 10C5 = 2/9



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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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18 Mar 2019, 18:50
Bunuel wrote: Ashley and Vinnie work on a sales staff with 8 other salespeople. If five members of the staff will be chosen at random to attend a sales workshop, what is the probability that both Ashley and Vinnie will be chosen?
A. 1/10 B. 3/28 C. 2/9 D. 1/4 E. 1/2
Kudos for a correct solution. There are 10C5 = 10!/(5! x 5!) = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2) = 2 x 9 x 2 x 7 = 252 ways to choose 5 people from 10. Assuming Ashley and Vinnie are already chosen, there are only 3 spots left for the other 8 people, and the number of ways to choose 3 people from 8 is 8C3 = 8!/(3! x 5!) = (8 x 7 x 6)/(3 x 2) = 56. Therefore, the probability is 56/252 = 8/36 = 2/9. Answer: C
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Re: Ashley and Vinnie work on a sales staff with 8 other salespeople. If f
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