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From a group of 8 volunteers, including Andrew and Karen, 4 people are

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EMPOWERgmatRichC
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink] New post   21 Sep 2020, 17:15
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Hi hennaja,

If you've read through the thread of posts, you'll see that there are a number of different ways that you can answer this question. Depending on which method you prefer, you might find it easiest to treat this question as a Probability question (and calculate 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 in any of the other spots...

(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, 4 people are [#permalink] New post   11 Oct 2020, 15:26
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Total volunteers: 8
To be selected: 4

Total ways: \(^8{C_4} = 70\)

Condition: Andrew will be always there and Karen will be not.[Andrew being always there we are left with 8 - 1 = 7 volunteers and Karen will never be there, therefore, we have now 7 - 1 = 6 volunteers]

Selecting remaining '3' from '6': \(^6{C_3} = 20\)

Probability: \(\frac{20 }{ 70} = \frac{2}{7}\)

Answer D
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink] New post   11 Mar 2021, 08:55
Reinfrank2011 wrote:
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, \((\frac{1}{14})*4=\frac{2}{7}\)


Hi Bunuel,

Can you explain why it's 4!/3! and not just 4!. I do not understand why the "not karens" who are actually individuals and not identical items do not count determining the number of orders. Also, in case there is a page with theory about this aspect, determining the number of orders in a selection, would be great.

Thank you!
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From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink] New post   14 Mar 2021, 13:45
With all the explanations given above, I am maybe wrong but I did this:

Probability that Andrew is selected : 4/8 = 1/2
If we now count the probability that Karen is not selected given that Andrew is: 4/7

1/2 * 4/7 = 4/14 = 2/7

Am I wrong ?

Bunuel EMPOWERgmatRichC
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink] New post   26 Mar 2021, 03:31
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Moussaillon wrote:
With all the explanations given above, I am maybe wrong but I did this:

Probability that Andrew is selected : 4/8 = 1/2
If we now count the probability that Karen is not selected given that Andrew is: 4/7

1/2 * 4/7 = 4/14 = 2/7

Am I wrong ?

Bunuel EMPOWERgmatRichC
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It is correct.

Think of it this way - You make everyone stand randomly in a line and select first 4 and drop the rest. So probability that Andrew is in the first 4 is 4/8. Now one slot is gone.
Next, Karen should be in the last four so probability of that is 4/7.

Overall probability = (4/8)*(4/7) = 2/7
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink] New post   26 Mar 2021, 03:50
VeritasKarishma wrote:
Moussaillon wrote:
With all the explanations given above, I am maybe wrong but I did this:

Probability that Andrew is selected : 4/8 = 1/2
If we now count the probability that Karen is not selected given that Andrew is: 4/7

1/2 * 4/7 = 4/14 = 2/7

Am I wrong ?

Bunuel EMPOWERgmatRichC
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VeritasKarishma
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It is correct.

Think of it this way - You make everyone stand randomly in a line and select first 4 and drop the rest. So probability that Andrew is in the first 4 is 4/8. Now one slot is gone.
Next, Karen should be in the last four so probability of that is 4/7.

Overall probability = (4/8)*(4/7) = 2/7


Thanks a lot for the confirmation !!
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink] New post   02 Apr 2021, 09:21
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Moussaillon wrote:
With all the explanations given above, I am maybe wrong but I did this:

Probability that Andrew is selected : 4/8 = 1/2
If we now count the probability that Karen is not selected given that Andrew is: 4/7

1/2 * 4/7 = 4/14 = 2/7

Am I wrong ?

Bunuel EMPOWERgmatRichC
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VeritasKarishma
MathRevolution



Hello,

This solution is absolutely correct.

Good luck.

Punit Joshi
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From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink] New post   18 Feb 2022, 18:45
KarishmaB

I'm not very familiar with probability & combinations.
Could you please check if my logic is correct here?
Big thanks! :-)

Method 1. Probability + unarrange

1/8 (Andrew)
*
6/7*5/6*4/5 (Other 3 members excluding Karen)
*
4!/3!*1 (unarrange) (numerator: choose 4 people/ denominator: the order for the remaining members doesn't member


Method 2. Combination

1 * 6!/3!3! (Choose Andrew > Choose 3 members from the remaining 6. Their orders don't matter, so do the other 3 not selected)
_________
8!/4!4! ( Choose 4 members from the 8 in total. Their orders don't matter, so do the other 4 not selected)
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink] New post   24 Feb 2022, 06:48
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rachelphwu wrote:
KarishmaB

I'm not very familiar with probability & combinations.
Could you please check if my logic is correct here?
Big thanks! :-)

Method 1. Probability + unarrange

1/8 (Andrew)
*
6/7*5/6*4/5 (Other 3 members excluding Karen)
*
4!/3!*1 (unarrange) (numerator: choose 4 people/ denominator: the order for the remaining members doesn't member


Method 2. Combination

1 * 6!/3!3! (Choose Andrew > Choose 3 members from the remaining 6. Their orders don't matter, so do the other 3 not selected)
_________
8!/4!4! ( Choose 4 members from the 8 in total. Their orders don't matter, so do the other 4 not selected)


Yes, both methods are correct. You might want to think in terms of "selection" or "selection and arrangement" too. If you break it into two steps, it simplifies your thought process because either you only select or you select and then arrange.
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink] New post   23 Mar 2024, 02:38
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Re: From a group of 8 volunteers, including Andrew and Karen, 4 people are [#permalink]
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