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From a group of 8 volunteers, including Andrew and Karen

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From a group of 8 volunteers, including Andrew and Karen  [#permalink]

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New post 02 Oct 2013, 06:28
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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  [#permalink]

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New post 04 Oct 2013, 04:39
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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 8-2=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  [#permalink]

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New post 24 Oct 2013, 01:07
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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 4-person 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  [#permalink]

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New post 02 Oct 2013, 07:28
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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  [#permalink]

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New post 03 Oct 2013, 09:54
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? :oops:
thanks
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Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

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New post 03 Oct 2013, 10:03
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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? :oops:
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  [#permalink]

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New post 12 Nov 2013, 00:29
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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 # 4-man 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  [#permalink]

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New post 26 Feb 2014, 16:59
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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  [#permalink]

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New post 24 Jun 2014, 19:04
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  [#permalink]

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New post 24 Jun 2014, 22:26
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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  [#permalink]

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New post 01 Jan 2015, 09:14
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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  [#permalink]

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New post 10 Feb 2015, 08:17
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?
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Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

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New post 10 Feb 2015, 21:29
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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  [#permalink]

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New post 14 Nov 2015, 00:00
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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From a group of 8 volunteers, including Andrew and Karen  [#permalink]

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New post 07 May 2016, 22:23
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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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From a group of 8 volunteers, including Andrew and Karen  [#permalink]

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New post 08 May 2016, 11:38
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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, \((\frac{1}{14})*4=\frac{2}{7}\)

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Re: From a group of 8 volunteers, including Andrew and Karen  [#permalink]

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New post 01 Jun 2016, 17:42
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  [#permalink]

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New post 01 Jun 2016, 21:05
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  [#permalink]

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From a group of 8 volunteers, including Andrew and Karen  [#permalink]

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New post 13 Sep 2016, 10:27
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.

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From a group of 8 volunteers, including Andrew and Karen   [#permalink] 13 Sep 2016, 10:27

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