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

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Joined: 04 Mar 2011
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Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

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15 Sep 2016, 08:05
1
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

The probability that we’re looking for can be expressed as a follows:

# ways the group can be selected with Andrew but not Karen/total # of ways to select the group

Let’s begin with the numerator. If Andrew makes the team but Karen does not, we can mentally remove Andrew and Karen from the pool of people available for the group. This leaves 6 people available for the full 3 spots (remember Andrew must be selected). 3 people can be chosen from a group of 6 people in 6C3 ways:

(6 x 5 x 4)/3! = 20 ways

For the denominator, we’re looking for the total number of ways in which the 4-person team can be made. The team can be made in 8C4 ways or:

(8 x 7 x 6 x 5)/4! = 70 ways

Thus, the probability that Andrew is selected for the group but Karen is not 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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03 Oct 2016, 02:17
goenkashreya wrote:
Bunuel

Definitely not my strong area but need to understand conceptually why this is wrong.

for andrew to be included - 8C1
For the rest three positions left 6C3

why is the 8C1 not required?

There are not 8 ways to choose Andrew. The probability of choosing Andrew out of 8 people is 1/8 but the number of ways to choose Andrew is 1 (1 out of 1 Andrew)..
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Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

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29 Oct 2016, 12:21
For me solving with probability using 2 steps is the easiest:

P(Andrew in 1 slot) = $$\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$$

Andrew OR Andrew OR Andrew OR Andrew

P(Andrew in 1 slot & Karen in none) = $$\frac{1}{2} * \frac{6}{7} * \frac{5}{6} * \frac{4}{5} = \frac{2}{7}$$

Andrew AND not Karen AND not Karen AND not Karen

Is there a list of similar questions available other than searching for the "Probability" tag?
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Joined: 27 Mar 2016
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Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

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30 Oct 2016, 06:28
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

Ans:-

A: Number of ways to select any 4 volunteers at random= 8C4
B: Number of Ways to Select Andrew= 1C1

Now that we have selected 1 volunteer, only 3 more volunteers are to be selected from a lot of 6 volunteers. [i.e. 8-1(Andrew)-1(Karen)=6volunteers]
C: Number of ways to select 4 Volunteers including Andrew but not Karen= 1C1*6C3= 6C3

Final Ans.: 6C3/8C4= 2/7 i.e. Option D
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Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

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31 Oct 2016, 02:27
Here is the simplet way to do it.
Attachments

thumbnail_IMAG0314.jpg [ 87.1 KiB | Viewed 1177 times ]

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

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13 Jan 2017, 10:47
[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
quote][/quote]

I solved this question in exactly the same way as decadecaf did (quoted above, sorry if it went wrong, first time quoting, new to GMAT club).
Is this method incorrect and simply a result of luck that I achieved the correct answer using this method?

Theory states that NOT questions and AT LEAST questions can be solved using complementary probability, I just want to know whether or not it is vital to use the combination methods to achieve the correct answer in this question.

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

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22 Jul 2017, 18:32
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?

In this case, we can start by finding the total number of choices: 8C4 = 8*7*6*5*4*3*2/4*3*2*4*3*2 = 14*5 = 70 choices.

The number of choices of a group with Andrew and without Karen is 6C3 because Andrew is already in the group (so we're down to 7 choices) and Karen is not in the group at all (so we're now down to 6 volunteers). We also deduct one from the choice (from 4 to 3) because Andrew has already been selected. 6C3 = 6*5*4*3*2/3*2*3*2 = 20.

20/70 = 2/7
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From a group of 8 volunteers, including Andrew and Karen [#permalink]

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21 Aug 2017, 18:25
Probability of selecting Andrew = 4 / 8
Then, probability of NOT selecting Karen = 4 / 7

(4 / 8) * (4 / 7 )
==> 4 / 14
==> 2 / 7

Am I missing something?
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Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

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05 Oct 2017, 17:56
Probability of selecting Andrew = 4 / 8
Then, probability of NOT selecting Karen = 4 / 7

(4 / 8) * (4 / 7 )
==> 4 / 14
==> 2 / 7

Am I missing something?

Yes, the probability of selecting Andrew is not 4/8. There are 8 people and Andrew is one person. So probability of selecting Andrew is 1/8.
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Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

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09 Dec 2017, 09:12
Think of the selection as slots from left to right. We want andrew, not karen, not karen, not karen.

(1/8)(6/7)(5/6)(4/5)=
(1/8)(4/7)
We have to multiply by four to account for permutations. (1/8)(4/7)(4)=2/7

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

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14 Dec 2017, 01:30
total: 8C4=70
andrew within the selected group and Karen is not: 1*6C3=20
20/70=2/7
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Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

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30 Jan 2018, 14:12
Hi All,

There's another way to "do the math" behind this question, although it takes a little longer - we can treat this question as a straight probability question and keep track of ALL the possible ways to get "what we want"….

We have 8 people to choose from; we'll pick 4 - we WANT Andrew in our group, we DON'T WANT Karen in our group.

_ _ _ _

If we want Andrew to be the first person we pick, then the other 3 CAN'T be Karen. We end up with the following math (notice that after picking each person, the total number of people remaining decreases):

(1/8)(6/7)(5/6)(4/5) = 4/56 = 1/14

Andrew COULD be the second, third or fourth person chosen though. Here's the math if he's SECOND and we don't pick Karen...

(6/8)(1/7)(5/6)(4/5) = 4/56 = 1/14

Notice how the product is the SAME.

If he's third, it would still be 1/14
if he's fourth, it would still be 1/14

With ALL of those possibilities, we end up with 1/14 + 1/14 +1/14 + 1/14 = 4/14 = 2/7

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

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28 Mar 2018, 05:24
jcaine wrote:
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.

I usually employ this method for all combination/permuation problems. Way easier!

Anyone can clarify why, when finding the number of ways Andrew shows up without Karen, we divide by 3! (3 x 2) instead of 4! (4 x 3 x 2) ?

We are shuffling all 4 slots. If order did matter Andrew - Member A - Member B - Member C would have been a different arrangement than Member A - Andrew - Member B - Member C

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

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Updated on: 01 Jun 2018, 03:16
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?

Hey pushpitkc

1/8*6/7*5/6*4/5=1/14 ---> 1/14*4 = 2/7 ( i multiply by 4 because there are 4 numbers of ways for andrew to be in the team ) i mean Andrew can be first, second, third or fourth

is my reasoning correct ?

what would be other way to solve through probability ?

have a great night

Originally posted by dave13 on 31 May 2018, 14:24.
Last edited by dave13 on 01 Jun 2018, 03:16, edited 1 time in total.
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Re: From a group of 8 volunteers, including Andrew and Karen [#permalink]

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31 May 2018, 14:55
1
dave13 wrote:
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?

Hey pushpitkc

1/8*6/7*5/6*4/5=1/14 ---> 1/4*4 = 2/7 ( i multiply by 4 because there are 4 numbers of ways for andrew to be in the team ) i mean Andrew can be first, second, third or fourth

is my reasoning correct ?

what would be other way to solve through probability ?

have a great night

Hey dave13

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Re: From a group of 8 volunteers, including Andrew and Karen   [#permalink] 31 May 2018, 14:55

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