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

Question

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

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

Bunuel · Accepted Answer

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.

Lucky2783 · Answer

-> 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.

adityadon · Answer

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 .

Bunuel · Answer

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.

adityadon · Answer

Bunuel , Thanks a lot .. you clarified an important Mis-concept of mine:)

Sneakysam · Answer

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.

EMPOWERgmatRichC · Answer

Hi Sneakysam,

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.

GMAT assassins aren't born, they're made,
Rich

roro555 · Answer

Thank you Rich for pointing this out. Really helpful!

ArunpriyanJ · Answer

Bunuel,

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

Thanks,A

chetan2u · Answer

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

ScottTargetTestPrep · Answer

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

Joejohn101 · Answer

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?

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

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.

KanishkM · Answer

IF both will be selected this will mean that they have already occupied 1 spot each in the 4 volunteer team

Remaining 2 will be selected in 5C2 ways

Total number of ways of selecting will be 7C4

Probability will be 5C2 / 7C4

= 2/7

D

Balkrishna · Answer

Hi,  EMPOWERgmatRichC

Can you help me understand why there is 2/7 in the solution posted by  Sneakysam ?

My understanding:
Probability of selecting Kori 1/7.
Probability of selecting Jason 1/6.
No. of ways in which Kori and Jason can be included to volunteer = 4C2 (KJXX, KXJX, KXXJ, JKXX, JXKX, JXXK)

1/7 * 1/6 * 6 = 1/7

Thank you.

EMPOWERgmatRichC · Answer

Hi Balkrishna,

In this prompt, we're forming a 'group' of people - meaning that it does not matter who is chosen 'first', 'second', etc. Thus, either Kori OR Jason could be selected from the initial 7 (that's the 2/7) and whoever is left can be selected from the remaining 6 (that's the 1/6). The 4c2 means that there are 6 different ways to form that specific group of four, thus...

(2/7)(1/6)(6) = 2/7

GMAT assassins aren't born, they're made,
Rich

Kashu18 · Answer

Can someone please help where is my approach going wrong:

Let's say we select K and J as first 2 people: 1/7 * 1/6
Let's say we select K and J as second 2 people: 1/6 * 1/5
Let's say we select K and J as third 2 people: 1/5 * 1/4
Now there's a way to select them in opposite way as well (J first and then K) so,

Probability = 2* (1/7 * 1/6 + 1/6 * 1/5 + 1/5 * 1/4) = 3/14.
 
This answer seems way off, can someone please explain why?

AbhishekP220108 · Answer

Kashu18

Think of this way, there are seven people and assume you are at the place of Kori and there is one of yoyr fried named J. There is a selection committee who needs to select 4 people. Including J and You.

For finding the probability 1st we need to find total no. Of outcomes. To select 4 people out of 7 there 7C4 ways=35

Now since J and You are already in 4 person team there is need to select 2 more out of remaining 5 which can be done in 5C2 ways I.e 10 so probability is 10/35=2/7

Now looking at your way ordering doesn't matter and neither it matters you are selecting K and J at which no. questions just asks to find the probability of selecting 4 people including K and J out of 7 people.

Kashu18 · Answer

Thanks this helps. We are applying the strategy of selecting the other 2 people here whilst keeping K and J stagnant. I was wondering how would we approach the problem if we thought the other way around and chose K and J.

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