Probability, seven-member dance group, help from veritas

[kornerua]

I need help understanding this:

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?

I have looked at the explained solution, bur really dont understand it.

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Hi kornerua,

For future reference, when posting a GMAT question, you should make sure to include the 5 answer choices. In certain cases, the answers provide a hint as to how to go about solving the problem. Without having the answers, we're forced to just 'do math.' This question can be solved in a couple of different ways, but is essentially a 'Combination' question wrapped in a Probability question.

We have 7 dancers to choose from and we're going to select a group of 4. We can use the Combination Formula to figure out the total number of possible foursomes...

7!/(4!)(3!) = (7)(6)(5)/(3)(2)(1) = 35 possible foursomes

Of those foursomes, we want the ones that include both Kori and Jason. If we 'lock in' those two people, then we'll have 2 open spots left and 5 dancers to choose from. Again, we can use the Combination Formula:

5!/(2!)(3!) = (5)(4)/(2)(1) = 10 possible foursomes that include Kori and Jason

10/35 = 2/7 = probability that Kori and Jason are in the foursome that is selected.

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

[OptimusPrep]

I need help understanding this:

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?

I have looked at the explained solution, bur really dont understand it.

Hi kornerua,

Total members = 7
# of people to be chosen = 4

Kori and Jason are to be chosen.
This means two slots are already filled.

Hence we are left with 2 more slots.

We need to chose 2 people from 5 people (Kori and Jason have already been chosen)

Number of ways = 5C2 = 10

Total ways to choosing 4 people out of 7 = 7C4 = 35

Probability = 10/35 = 2/7

[GMATGuruNY]

I need help understanding this:

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?

Alternate approach:
P(good outcome) = P(one way) * (number of possible ways).

Let K = Kori, J = Jason, and N = a member who is not K or J.

P(one way):
One way to choose K and J is as follows:
KJNN.
P(K is selected first) = 1/7. (Of the 7 members, only one is K.)
P(J is selected second) = 1/6. (Of the 6 remaining members, only one is J.)
P(N is selected third) = 5/5. (Of the 5 remaining members, all 5 are not K or J.)
P(N is selected fourth) = 4/4. (Of the 4 remaining members, all 4 are not K or J.)
To combined these probabilities, we multiply:
1/7 * 1/6 * 5/5 * 4/4 = 1/42.

Total possible ways:
KJNN is only ONE WAY that K and J can be selected.
Now we must account for ALL OF THE WAYS that K and J can be selected..
Any arrangement of the letters KJNN will yield a group with K and J.
Thus, to account for ALL OF THE WAYS that K and J can be selected, the result above must be multiplied by the number of ways to arrange the letters KJNN.
Number of ways to arrange 4 elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical N's:
4!/2! = 12.

Multiplying the results above, we get:
P(K and J are selected) = 12 * 1/42 = 2/7.

More practice:
https://www.beatthegmat.com/select-exact ... 88786.html
https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html

[Matt@VeritasPrep]

I'd think of it as:

(# of groups with Kori and Jason) / (# of total groups)

If Kori and Jason are in the group, there are 5 other spots that we can choose. There are two more people in the groups, so this is (5 choose 2).

If we don't care who's in the group, we have 7 spots for 4 people, so that's (7 choose 4).

That gives us

(5 choose 2) / (7 choose 4) =>

5!/(3!2!) / 7!/(4!3!) =>

2 / 7

This a neat result too, considering we've got 7 people in the group, and Kori and Jason are 2 of them!

[Jeff@TargetTestPrep]

I need help understanding this:

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?

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.