Probability: Mark and Anna

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vivibcn: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Mon Aug 27, 2012 8:28 am

by vivibcn » Sat Oct 13, 2012 8:06 am

Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from this group, what is the probability that Mark and not Anna will be selected?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7

OA:B

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Source: Beat The GMAT — Problem Solving | Sat Oct 13, 2012 8:06 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Oct 13, 2012 8:21 am

vivibcn wrote:Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from this group, what is the probability that Mark and not Anna will be selected?
A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7
OA:B

We can solve this question using counting methods.

So, P(Mark but not Anna) = [# of ways to select Mark but not Anna]/[# of ways to select 4 people]

I always begin with the denominator.

# of ways to select 4 people
There are 8 people altogether, and we must select 4 of them.
Since the order in which we select the 4 people does not matter, this can be solved using combinations.
We can select 4 people in 8C4 ways (70 ways)

# of ways to select Mark but not Anna
First, put Mark in the group.
Next, select the other 3 people from the remaining 6 eligible people (we are excluding Anna from the remaining people, because don't want her to be selected)
Since the order in which we select the 3 other people does not matter, this can be solved using combinations.
We can select 3 people in 6C3 ways (20 ways)

So, P(Mark but not Anna) = 20/70 = [spoiler]2/7 = B[/spoiler]

Cheers,
Brent

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

Brent Hanneson - Creator of GMATPrepNow.com

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Whitney Garner: GMAT Instructor; Posts: 273; Joined: Tue Sep 21, 2010 5:37 amFollowed by: 74 members; Location: Durham, NC; Thanked: 154 times; GMAT Score:770

by Whitney Garner » Sat Oct 13, 2012 8:28 am

vivibcn wrote:Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from this group, what is the probability that Mark and not Anna will be selected?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7

OA:B

Hi vivibcn!

One way I like to solve these weird "constraint" problems is to just try to imagine that I'm the one doing the selecting. Since it isn't just combinations, but probability, I like to use the fundamental probability formula where the probability of an event occurring is:

# Outcomes I Want
------------------
Total # of Outcomes

So the total # of Outcomes is usually the easiest, and this would just be the number of ways I can select a team of 4 people from 8, or 8c2. This translates to 8!/(4!4!)...(a total of 8 and I divide these into 4 I want and 4 I don't want).

Now, the outcomes I want just takes a bit of logic. If I know I'm selecting Mark, then let me now assume that I really only have 3 seats left on my team and 7 people waiting on the bench. But I'm NOT allowed to select Anna, so in reality, I only have 6 people from which to choose my team of 3. So this is just 6c3, or 6!(3!3!).

Put this all together to get...

The answer is B.

Hope this helps!

Whit

Whitney Garner
GMAT/GRE/EA Instructor & Anxiety/Accommodations Coach
www.whitneygarner.com

Contributor to Beat The GMAT!

Math is a lot like love - a simple idea that can easily get complicated

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vivibcn: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Mon Aug 27, 2012 8:28 am

by vivibcn » Sat Oct 13, 2012 9:36 am

Thank you both so much!

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Oct 13, 2012 3:14 pm

vivibcn wrote:Anna and Mark are included in a group of 8 people. If 4 people are selected at random and without replacement from this group, what is the probability that Mark and not Anna will be selected?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 5/7

Case 1: The FIRST person selected is Mark and Anna is NOT among the other 3 selected
P(1st person is Mark) = 1/8. (Of the 8 people, Mark must be selected.)
P(2nd person is not Anna) = 6/7. (Of the 7 remaining people, anyone but Anna.)
P(3nd person is not Anna) = 5/6. (Of the 6 remaining people, anyone but Anna.)
P(2nd person is not Anna) = 4/5. (Of the 5 remaining people, anyone but Anna.)
Since we want all of these events to happen, we multiply the probabilities:
1/8 * 6/7 * 5/6 * 4/5.

Remaining cases:
Since Mark could be the 1st, 2nd, 3rd or 4th person selected -- for a TOTAL OF 4 WAYS to select Mark -- the result above must be multiplied by 4:
4 * 1/8 * 6/7 * 5/6 * 4/5 = 2/7.

The correct answer is B.

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