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

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

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)

=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

P(Andrew is selected but Karen is not selected) = (number of 4-person groups with Andrew but not Karen)/(total # of 4-person groups possible)

number of 4-person groups with Andrew but not Karen
Take the task of creating groups and break it into stages.

Stage 1: Place Andrew in the 4-person group
We can complete this stage in 1 way

Stage 2: Send Karen out of the room and, from the remaining 6 volunteers, select 3 more people.
Since the order in which we select the 3 volunteers does not matter, we can use combinations.
We can select 3 people from 6 volunteers in 6C3 ways (20 ways).

the video below shows you how to calculate combinations (like 6C3) in your head

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a 4-person group) in (1)(20) ways (= 20 ways)

total # of 4-person groups possible
We can select 4 people from all 8 volunteers in 8C4 ways ( = 70 ways).

So, P(Andrew is selected but Karen is not selected) = (20)/(70)
= 2/7

Answer: D

Cheers,
Brent

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Hi kusena,

could you please elaborate how you approach this kind of questions? what do 6C3 & 8C4 mean?
thanks

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.

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

Therefore the answer is 20/70= 2/7

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.

Thanks

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.

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

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.

courtesy of Reinfrank2011:

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, \((\frac{1}{14})*4=\frac{2}{7}\)

Happy studies!

The probability that we’re looking for can be expressed as a follows:

# ways the group can be selected with Andrew but not Karen/total # of ways to select the group

Let’s begin with the numerator. If Andrew makes the team but Karen does not, we can mentally remove Andrew and Karen from the pool of people available for the group. This leaves 6 people available for the full 3 spots (remember Andrew must be selected). 3 people can be chosen from a group of 6 people in 6C3 ways:

(6 x 5 x 4)/3! = 20 ways

For the denominator, we’re looking for the total number of ways in which the 4-person team can be made. The team can be made in 8C4 ways or:

(8 x 7 x 6 x 5)/4! = 70 ways

Thus, the probability that Andrew is selected for the group but Karen is not is 20/70 = 2/7.

Answer: D

Bunuel

Definitely not my strong area but need to understand conceptually why this is wrong.

for andrew to be included - 8C1
For the rest three positions left 6C3

why is the 8C1 not required?

There are not 8 ways to choose Andrew. The probability of choosing Andrew out of 8 people is 1/8 but the number of ways to choose Andrew is 1 (1 out of 1 Andrew)..

Hi All,

There's another way to "do the math" behind this question, although it takes a little longer - we can treat this question as a straight probability question and keep track of 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...

(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

Final Answer:

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

Hi there, really helpful responses here.

Just wondering why you are dividing 20/70 to find the probability? How do we know this is what you have to do to find the probability?