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Probability; Combinatrics- Choosing from a group of 8

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by Brent@GMATPrepNow » Mon Dec 01, 2014 12:57 pm
Here's the full question with answer choices:
From a group of 8 volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 selected and Karen will not?

A 3/7
B 5/12
C 27/70
D 2/7
E 9/35
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).

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)
= [spoiler]2/7 = D[/spoiler]

------------------------------

Cheers,
Brent

Aside: If anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
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by GMATGuruNY » Mon Dec 01, 2014 1:33 pm
From a group of 8 volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 selected and Karen will not?

A 3/7
B 5/12
C 27/70
D 2/7
E 9/35
Alternate approach:

From the 8 people, 4 will be selected.
Thus, P(A is selected) = 4/8.
From the 7 remaining people, 4 will NOT be selected.
Thus, P(K is not selected) = 4/7.
To combine these probabilities, we multiply:
4/8 * 4/7 = 2/7.

The correct answer is D.
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by MartyMurray » Sat Dec 20, 2014 11:40 am
From a group of 8 volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 selected and Karen will not?

A 3/7
B 5/12
C 27/70
D 2/7
E 9/35
Here's even another approach.

First the probability that Andrew is selected is 1 - the probability that Andrew is not selected.

The probability that Andrew is not selected can work like this.

The first person chosen has 7/8 chance of being not Andrew, the second has 6/7, the third, 5/6, and the fourth 4/5. If that's how the four are selected, then no Andrew.

Multiply them out to get (7/8)(6/7)(5/6)(4/5) = 4/8 = 1/2. So half the time Andrew is not selected, and 1 - 1/2 = 1/2 of the time Andrew is selected.

So we know the chances of Andrew being selected, 1/2.

Now what is the chance that with Andrew selected Karen is not.

With Andrew already selected, we have three more slots to fill, with seven people left to fill them. The first has a 6/7 chance of being not Karen, the second, 5/6, and the third, 4/5. Multiply them to get (6/7)(5/6)(4/5) = 4/7 chance of Karen not being selected from the other 7.

So we have the chance that Andrew will be selected, 1/2, and the chance that from the rest Karen will not be selected, 4/7. So to get the probability that Andrew will be selected and Karen will not be selected, multiply those to get (1/2)(4/7) = 2/7.

Choose D.

P.S. It's kinda bugging me that with Andrew selected there is not a 1/2 chance that Karen is not selected. I mean originally I figured that with Andrew selected, Karen would be selected half the time and not selected the other half of the time. So this would be an easy problem. But that's not the case, and I was saved by their not having put a trap answer that would have worked that way.

P.P.S. Gotta love the way Brent did it because it is so basic and is applicable to so many situations. No need to have all kinds of different strategies at hand. Just figure out how many ways one can get the favorable outcome and divide it by the total number of possible outcomes. End of story.

So cool to see so many cool methods and ways of thinking here.
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by mbawisdom » Sun Dec 21, 2014 8:39 am
I approach this problem by first working how many combinations of 4 people there are: this is 8C4, which is 8!/(4!.4!) or 70.
Now I need to work out how many combinations there are with Adam in and Karen out. We can set Andrew in and then see how many combinations of 3 people there without being able to chose Karen. This is 6C3, 6!/(3!.3!) or 20.

Thus, the answer is 20/70 or 2/7.
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by [email protected] » Sun Dec 21, 2014 7:57 pm
Hi arvin825,

If you find that trying to use the Combination Formula is confusing on this question, then there are other (slightly longer) ways to get to the correct answer.

We're told to select 4 people from a group of 8. We're asked for the probability that the group include Andrew but NOT include Karen. Since the order of the group members does NOT matter, this is clearly a Combination Formula question, BUT you COULD solve it using permutations. Here's how:

To get Andrew into the group and exclude Karen, we COULD order the possibilities; we'd just have to consider 4 calculations:

(Andrew)(Not Karen)(Not Karen)(Not Karen) = (1/8)(6/7)(5/6)(4/5) = ....

While this calculation "looks" crazy, a number of the terms cancel out. Notice the 5s and 6s in both the numerator and denominator? We'd be left with...

(1/8)(4/7) = 4/56 = 1/14

This is the probability that Andrew is the FIRST one chosen AND Karen is not chosen with any of the other picks.

If Andrew was to be the SECOND one chosen, we'd have...

(Not Karen and Not Andrew)(Andrew)(Not Karen)(Not Karen) = (6/8)(1/7)(5/6)(4/5) = ....

Simplifying this, we get...

4/56 = 1/14

This is the SAME result we got when Andrew was picked first! This means that the "math" will be the same no matter what position Andrew is picked:

Picked 1st = 1/14
Picked 2nd = 1/14
Picked 3rd = 1/14
Picked 4th = 1/14

Total = 4(1/14) = 4/14 = 2/7

Final Answer: D

While not the fastest way to get to the correct answer, it does get you there.

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by Matt@VeritasPrep » Mon Dec 22, 2014 12:06 pm
Brent's way is safe - here's another quick way.

Let's start with the probability that Andrew is selected: 1/2. (Half the people are selected, half are not.)

Now that Andrew is in, we need to worry about Karen. Since we know Andrew is in, 4 of the 7 remaining people will be out. The probability that Karen is one of them is thus 4 out of 7.

Since we need BOTH of these to occur, our probability = 1/2 * (4/7), or 2/7.
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