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Probability: Mark and Anna

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Probability: Mark and Anna

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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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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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...

Image

The answer is B.

Hope this helps!
:)
Whit
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by vivibcn » Sat Oct 13, 2012 9:36 am
Thank you both so much! :)
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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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