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From a seven-member dance group, four will be chosen at random to volu [#permalink]

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17 Mar 2015, 08:45

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From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

a) 1/21 b) 1/7 c) 4/21 d) 2/7 e) 3/7

Total number of ways to choose 4 out of 7 = 7C4 = 35.

Number of ways to choose 2 (any but Kori and Jason) out of 5 (excluding Kori and Jason) = 5C2 = 10.

Re: From a seven-member dance group, four will be chosen at random to volu [#permalink]

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17 Mar 2015, 09:05

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ynaikavde wrote:

From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

a) 1/21 b) 1/7 c) 4/21 d) 2/7 e) 3/7

-> If Kori and Jason have to be in group than we have to just choose 2 others out of remaining 5. 5C2 total ways to choose 4 out of 7 . 7C4 10/35 = 2/7 Answer D.
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Re: From a seven-member dance group, four will be chosen at random to volu [#permalink]

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17 Mar 2015, 09:09

Bunuel wrote:

ynaikavde wrote:

From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

a) 1/21 b) 1/7 c) 4/21 d) 2/7 e) 3/7

Total number of ways to choose 4 out of 7 = 7C4 = 35.

Number of ways to choose 2 (any but Kori and Jason) out of 5 (excluding Kori and Jason) = 5C2 = 10.

P = favorable/total = 10/35 = 2/7.

Answer: D.

Bunuel , I tried below approach and got incorrect answer . Where am i going wrong. ( 1- both are not selected/total no of ways of selecting) = 1-5C4/7C4= 1-5/35= 6/7 .
_________________

From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

a) 1/21 b) 1/7 c) 4/21 d) 2/7 e) 3/7

Total number of ways to choose 4 out of 7 = 7C4 = 35.

Number of ways to choose 2 (any but Kori and Jason) out of 5 (excluding Kori and Jason) = 5C2 = 10.

P = favorable/total = 10/35 = 2/7.

Answer: D.

Bunuel , I tried below approach and got incorrect answer . Where am i going wrong. ( 1- both are not selected/total no of ways of selecting) = 1-5C4/7C4= 1-5/35= 6/7 .

Both are not selected is not an opposite event of both selected. The opposite event of both selected is (i) both are not selected and (ii) any of those two are selected and another is not.
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Re: From a seven-member dance group, four will be chosen at random to volu [#permalink]

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19 Apr 2015, 15:17

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This post was BOOKMARKED

Bunuel wrote:

ynaikavde wrote:

From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

a) 1/21 b) 1/7 c) 4/21 d) 2/7 e) 3/7

Total number of ways to choose 4 out of 7 = 7C4 = 35.

Number of ways to choose 2 (any but Kori and Jason) out of 5 (excluding Kori and Jason) = 5C2 = 10.

P = favorable/total = 10/35 = 2/7.

Answer: D.

Bunel is this also a legitimate way to sold this question? (or do the numbers just coincidentally work) ?

Probability of choosing K and J (2/7) * (1/6) = 2/42

Multiply by 4C2 (or 6) for the number of ways you can arrange K and J amongst the four chosen.

Your method ABSOLUTELY works and is just as viable an approach as any other. It goes to show that Permutation/Combination/Probability questions can often be attempted in a variety of ways (in much the same way that Quant questions in general can be approached in multiple ways). Staying flexible with your thinking (and learning more than one way to deal with GMAT questions) can help you immensely on Test Day.

From a seven-member dance group, four will be chosen at random to volu [#permalink]

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24 Jan 2016, 19:29

Bunuel wrote:

ynaikavde wrote:

From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

a) 1/21 b) 1/7 c) 4/21 d) 2/7 e) 3/7

Total number of ways to choose 4 out of 7 = 7C4 = 35.

Number of ways to choose 2 (any but Kori and Jason) out of 5 (excluding Kori and Jason) = 5C2 = 10.

P = favorable/total = 10/35 = 2/7.

Answer: D.

Bunuel,

Sorry i am struggling to understand the highlighted part...Can you please through some light?

From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

a) 1/21 b) 1/7 c) 4/21 d) 2/7 e) 3/7

Total number of ways to choose 4 out of 7 = 7C4 = 35.

Number of ways to choose 2 (any but Kori and Jason) out of 5 (excluding Kori and Jason) = 5C2 = 10.

P = favorable/total = 10/35 = 2/7.

Answer: D.

Bunuel,

Sorry i am struggling to understand the highlighted part...Can you please through some light?

Thanks,A

Hi, there are 7 members including Kori and Jason. We are to find probability of choosing these two if we are selecting 4 out of these 7.. way to choose 4 out of 7= 7C4=35..

Now two out of 4 are Kori and Jason. the remaining 2 are to b choosen from remaining 5.(K and J are not included as they are already choosen).. so 5C2=10.. prob=10/35.. hope it helps you
_________________

From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

a) 1/21 b) 1/7 c) 4/21 d) 2/7 e) 3/7

We are given that from 7 members, 4 people will be selected for a dance team. We need to determine the probability that both Kori and Jason will be selected.

Let’s first determine the total number of ways to select 4 people from a group of 7 people.

The number of ways to select 4 people from a group of 7 is:

7C4 = (7 x 6 x 5 x 4)/4! = (7 x 6 x 5 x 4))/(4 x 3 x 2 x 1) = 7 x 5 = 35

Next, we need to determine how many ways the group can be selected when both Kori and Jason are selected. Since they MUST BE SELECTED, there are 5 remaining people for 2 spots. We can select 2 people from 5 in the following ways:

5C2 = (5 x 4)/2! = 20/2 = 10

Thus, the probability of creating a dance team of 4 with both Kori and Jason is 10/35 = 2/7.

Answer: D
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Scott Woodbury-Stewart Founder and CEO

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Re: From a seven-member dance group, four will be chosen at random to volu [#permalink]

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25 Feb 2017, 15:44

This is an old post but I was hoping somebody could explain something for me.

I understand the approaches in this question and I solved correctly. Is there a way to solve by taking the opposite of the desired outcome. aka 1-(probability that j and k are not included)

denominator for this is 7C4=35

numerator is the number of arrangements where j and k are not included ( arrangements of j an k in the three excluded spots) arrangements where x is the third member excluded from the group: j-k-x k-j-x x-k-j x-j-k j-x-k k-x-j

6 arrangements total. since x can be 5 different members multiple 6 by 5 = 30

probability that j and k are excluded is 30/35. therefore probability that they are included is 1-30/35= 5/35

The answer is wrong but is this approach possible?

From a seven-member dance group, four will be chosen at random to volunteer at a youth dance event. If Kori and Jason are two of the seven members, what is the probability that both will be chosen to volunteer?

This problem blends combinatorics with probability. To begin, determine the total number of outcomes. With N = 7 and K = 4, the calculation is:

7!/(4!3!) = 35 outcomes.

Then consider all the favorable outcomes, those in which Jason and Kori are both chosen. If they're both chosen, that leaves 5 more people to be chosen for the 2 remaining spots, for a calculation of:

5!/(2!3!)=10 favorable outcomes.

Since the probability is then 10/35, that reduces to 2/7 for answer choice D.
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