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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, 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?

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

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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

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

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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? thanks

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

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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? 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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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 \(P=\frac{C^3_6}{C^4_8}=\frac{2}{7}\).

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

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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 --> \((\frac{6}{8})(\frac{5}{7})(\frac{4}{6})(\frac{1}{5})= \frac{1}{14}\)

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

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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.

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

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26 Feb 2014, 16:59

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

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

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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.

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
_________________

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

Merging similar topics. Please refer to the discussion above.

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

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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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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?

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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07 May 2016, 22:23

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.

Attachments

Screen Shot 2016-05-08 at 11.29.16 AM.png [ 125.66 KiB | Viewed 17041 times ]

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

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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).

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