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From a group of 8 volunteers, including Andrew and Karen
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02 Oct 2013, 05:28
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From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not? A. 3/7 B. 5/12 C. 27/70 D. 2/7 E. 9/35 Here, I approached the problem as below:
Considering that Andrew will be there , we need to select 3 other people from 6 remaining volunteers excluding Karen.
=> 6C3 / 7C3 = 4/7.
Although finally I randomly guessed and selected correct answer i.e. 2/7, I was not able to get the answer with my approach mentioned here.
Could someone tell me what am I missing here?
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Re: From a group of 8 volunteers, including Andrew and Karen
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04 Oct 2013, 03:39
paskorntt wrote: Quote: Hi
The problem is a mix of combinations & probability. 8C4 means we can choose 4 people out of a group of 8 in 8C4 ways. 8C4 expands to 8x7x6x5/1x2x3x4 Similarly choosing a group of 4 with Andrew and no karein is 6C3 ways. Here the group is 6 because, we don't need Karein and Andrew is already in the group we want. Hence we need to choose 3 people from the remaining 6. Which is 6C3 ways.
Probability is 6C3 / 8C4 = (6x5x4/1x2x3)/(8x7x6x5/1x2x3x4) = 2/7 (answer) Still don't understand how you got the 6C3, please kindly explain. From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?A. 3/7 B. 5/12 C. 27/70 D. 2/7 E. 9/35 We need Andrew and 3 others but not Karen in the group. The # of ways to choose 3 members out of 82=6 (all but Andrew and Karen) is \(C^3_6\), thus \(P=\frac{C^3_6}{C^4_8}=\frac{2}{7}\). Answer: D.
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Re: From a group of 8 volunteers, including Andrew and Karen
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24 Oct 2013, 00:07
You can also just find the probability of selecting the desired individuals and then multiply that by the number of ways those selections can be ordered:
P(Andrew)*P(Not Karen)*P(Not Karen)*P(Not Karen)*(ways to order selections) = \((\frac{1}{8})*(\frac{6}{7})*(\frac{5}{6})*(\frac{4}{5})*(\frac{4!}{3!}) = (\frac{1}{14})*4 = \frac{2}{7}\)
The above is just the sum of the individual probabilities of each possible 4person team:
Andrew, Not Karen, Not Karen, Not Karen > \((\frac{1}{8})(\frac{6}{7})(\frac{5}{6})(\frac{4}{5}) = \frac{1}{14}\)
Not Karen or Andrew, Andrew, Not Karen, Not Karen > \((\frac{6}{8})(\frac{1}{7})(\frac{5}{6})(\frac{4}{5})= \frac{1}{14}\)
Not Karen or Andrew, Not Karen or Andrew, Andrew, Not Karen > \((\frac{6}{8})(\frac{5}{7})(\frac{1}{6})(\frac{4}{5})= \frac{1}{14}\)
Not Karen or Andrew, Not Karen or Andrew, Not Karen or Andrew, Andrew > \((\frac{6}{8})(\frac{5}{7})(\frac{4}{6})(\frac{1}{5})= \frac{1}{14}\)
Thus, \((\frac{1}{14})*4=\frac{2}{7}\)




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Re: From a group of 8 volunteers, including Andrew and Karen
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02 Oct 2013, 06:28
mba1382 wrote: From a group of 8 volunteers,including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?
A. 3/7 B. 5/12 C. 27/70 D. 2/7 E. 9/35
Here, I approached the problem as below:
Considering that Andrew will be there , we need to select 3 other people from 6 remaining volunteers excluding Karen.
=> 6C3 / 7C3 = 4/7.
Although finally I randomly guessed and selected correct answer i.e. 2/7, I was not able to get the answer with my approach mentioned here.
Could someone tell me what am I missing here? =No.of ways in selecting 4 members group with Andrew and without Karein(6C3 Ways) / Total No of ways of selecting 4 member group from 8 volunteers (8C4 ways) =>6C3/8C4 =>2/7 Ans D hope it helps



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Re: From a group of 8 volunteers, including Andrew and Karen
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03 Oct 2013, 08:54
kusena wrote: =No.of ways in selecting 4 members group with Andrew and without Karein(6C3 Ways) / Total No of ways of selecting 4 member group from 8 volunteers (8C4 ways) =>6C3/8C4 =>2/7
Hi kusena, could you please elaborate how you approach this kind of questions? what do 6C3 & 8C4 mean? thanks



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Re: From a group of 8 volunteers, including Andrew and Karen
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03 Oct 2013, 09:03
LinaNY wrote: kusena wrote: =No.of ways in selecting 4 members group with Andrew and without Karein(6C3 Ways) / Total No of ways of selecting 4 member group from 8 volunteers (8C4 ways) =>6C3/8C4 =>2/7
Hi kusena, could you please elaborate how you approach this kind of questions? what do 6C3 & 8C4 mean? thanks Hi The problem is a mix of combinations & probability. 8C4 means we can choose 4 people out of a group of 8 in 8C4 ways. 8C4 expands to 8x7x6x5/1x2x3x4 Similarly choosing a group of 4 with Andrew and no karein is 6C3 ways. Here the group is 6 because, we don't need Karein and Andrew is already in the group we want. Hence we need to choose 3 people from the remaining 6. Which is 6C3 ways. Probability is 6C3 / 8C4 = (6x5x4/1x2x3)/(8x7x6x5/1x2x3x4) = 2/7 (answer)
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Re: From a group of 8 volunteers, including Andrew and Karen
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11 Nov 2013, 23:29
Sorry, wanted to add this approach as an alternative. Was helping someone with this question, and given that she suffers from over thinking a problem too much that it eats in to her exam time, she found the slot method easier to 'understand'... hope it helps those who are still having difficulties: Find the # 4man organizing parties from group of 8: (8 x 7 x 6 x 5) / (4 x 3 x 2) = 70 TOP: Four "slots". Number of members remaining to fill slot.  BOTTOM: 4 slots to "shuffle". Now find # of ways Andrew shows up without Karen:(1) x (6 x 5 x 4) / (1) x (3 x 2) = 20 TOP: "1" because only "1" Andrew. "6 x 5..." (1 less than above) because we don't want Karen.  BOTTOM: 3 slots to "shuffle". Andrew represented by the "(1)" Therefore, 20/70 = 2/7< Answer choice D. I recommend looking for similar OG questions, and try using the above method a few times to get a feel for it.



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Re: From a group of 8 volunteers, including Andrew and Karen
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26 Feb 2014, 15:59
uwengdori wrote: From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random. What is the probability that Andrew will be among the 4 volunteers but not Karen?
3/7
5/12
27/70
2/7
9/35 out of the 4 members who are to be selected, there should be Andrew so 3 vacant positions remain. Karen should not be selected, so no. of remaining persons 6(excluding Andrew and Karen) 3 people are to be selected for 3 Vacant positions from 6 people => 6C3 =>6!/(3!*3!) = 20 In general 4 people are to be selected from 8 people =>8C4 => 8!/(4!*4!)=70 Therefore the answer is 20/70= 2/7



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Re: From a group of 8 volunteers, including Andrew and Karen
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24 Jun 2014, 18:04
Can someone help me to solve this question?
From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random. What is the probability that Andrew will be among the 4 selected AND Karen will not?
Answer is 2/7
I understand that there is 8!/4!4! ways to choose 4 volunteers from 8 people. I am struggling with how to then get the numerator, of the number of ways that Andrew can be chosen and Karen will not.
Thanks



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Re: From a group of 8 volunteers, including Andrew and Karen
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24 Jun 2014, 21:26
achakrav2694 wrote: Can someone help me to solve this question?
From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random. What is the probability that Andrew will be among the 4 selected AND Karen will not?
Answer is 2/7
I understand that there is 8!/4!4! ways to choose 4 volunteers from 8 people. I am struggling with how to then get the numerator, of the number of ways that Andrew can be chosen and Karen will not.
Thanks There are 8 people in the volunteers group. You need to choose 4. You have already chosen Andrew. You kick out Karen. Now you are left with 6 people. To make a group of 4, you need 3 more people (Andrew is already in). From 6 people, how do you choose 3? In 6C3 (or 6!/3!*3!) = 20 ways. So required probability = 20/(8!/4!4!) = 20/70 = 2/7
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Re: From a group of 8 volunteers, including Andrew and Karen
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01 Jan 2015, 08:14
Its Andrew (Other) (Other) (Other) = (1/8) (6/7) (5/6) (4/5) (4!/3!) = 2/7 (4!/3!) is the number of ways AOOO can arrange itself. Logically speaking Andrew can be chosen in four ways  first, second, third, fourth.
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From a group of 8 volunteers, including Andrew and Karen
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10 Feb 2015, 07:17
mba1382 wrote: From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not? A. 3/7 B. 5/12 C. 27/70 D. 2/7 E. 9/35 Here, I approached the problem as below:
Considering that Andrew will be there , we need to select 3 other people from 6 remaining volunteers excluding Karen.
=> 6C3 / 7C3 = 4/7.
Although finally I randomly guessed and selected correct answer i.e. 2/7, I was not able to get the answer with my approach mentioned here.
Could someone tell me what am I missing here? Dear Bunuel, How I can answer this question by the way ALL THE BAD where all = 8C4 = 70 the bad is when Karen and Andrew both are among the 4 volunteers selected 2C2*2 =number of the ways to selected Karen and Andrew 6C2 =number of the ways to selected other volunteers 8C4 2*2C2 * 6C2 ________________ = 8C4 70  2*15 _________ = 40/70 = 4/7 I do not know where is my mistake here? 70
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Re: From a group of 8 volunteers, including Andrew and Karen
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10 Feb 2015, 20:29
Andrew's chances of being selected: 4/8 If Andrew is selected, Karen's chances of not being selected: 4/7 (4/8) * (4/7) = 16/56 or 2/7



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Re: From a group of 8 volunteers, including Andrew and Karen
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13 Nov 2015, 23:00
mba1382 wrote: From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?
A. 3/7 B. 5/12 C. 27/70 D. 2/7 E. 9/35
Let's say that there are 4 places to be occupied Andrew can take one of 4 places in 4 ways 6 out of remaining 7 (leaving Karen) can take their places in 6, 5 and 4 ways respectively so favorable cases = 4*6*5*4 Total ways to occupy the places = 8*7*6*5 Probability = 4*6*5*4 / 8*7*6*5 = 2/7 Answer: Option D
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07 May 2016, 21:23
Attached is a visual that should help. Quick tip: notice that in all 4 versions, the numerators and denominators are the same, so you only have to calculate the probability once, then multiply it by 4.
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Screen Shot 20160508 at 11.29.16 AM.png [ 125.66 KiB  Viewed 37077 times ]
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08 May 2016, 10:38
courtesy of Reinfrank2011: Andrew, Not Karen, Not Karen, Not Karen > \((\frac{1}{8})(\frac{6}{7})(\frac{5}{6})(\frac{4}{5}) = \frac{1}{14}\) Not Karen or Andrew, Andrew, Not Karen, Not Karen > \((\frac{6}{8})(\frac{1}{7})(\frac{5}{6})(\frac{4}{5})= \frac{1}{14}\) Not Karen or Andrew, Not Karen or Andrew, Andrew, Not Karen > \((\frac{6}{8})(\frac{5}{7})(\frac{1}{6})(\frac{4}{5})= \frac{1}{14}\) Not Karen or Andrew, Not Karen or Andrew, Not Karen or Andrew, Andrew > \((\frac{6}{8})(\frac{5}{7})(\frac{4}{6})(\frac{1}{5})= \frac{1}{14}\) Thus, \((\frac{1}{14})*4=\frac{2}{7}\) Happy studies!
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Re: From a group of 8 volunteers, including Andrew and Karen
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01 Jun 2016, 16:42
Suppose the question had said to find the probability to pick both or pick none, how would we do that? Would it be:
Pick both: 6C2 / 8C4
Pick None: 6C4 / 8C4 ?



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Re: From a group of 8 volunteers, including Andrew and Karen
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01 Jun 2016, 20:05
Total number of combinations possible = C(8,4) Andrew is always there. So, this leaves us with 7 volunteers to choose from. But even among these 7 volunteers, Karen should not be chosen. So, this leaves us with 6 volunteers, from whom we have to choose 3 (because Andrew has anyway been chosen).
So, number of combinations = C(6,3)
Hence, probability = C(6,3)/C(8,4) = 2/7



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Re: From a group of 8 volunteers, including Andrew and Karen
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From a group of 8 volunteers, including Andrew and Karen
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Hi. Though I get the method of 6C3/ 8C4 . I tried an alternative method , leaving karen out , P (of selecting Andrew) = 1/7. Remaining 3 can be selected in 6C3/7C3 ways. Unable to understand why 1/7 * 6c3/7c3 is incorrect.1/7 being the probability of selecting 1 out of 7 people ( leaving Karen out). Bunuel . Would really appreciate your help over here to point out what I missed and why this is incorrect. Regards Megha




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