BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Picking friends

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Fri Nov 19, 2010 8:07 am

Picking friends

by varun7nurav » Fri Sep 23, 2011 8:35 am
Hi, came across this question in MGmat test. The solution provided isn't clear. If anybody knows an easier method- plz share. Tq!

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?
A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1309
Joined: Mon Apr 04, 2011 5:34 am
Location: India
Thanked: 310 times
Followed by:123 members
GMAT Score:750

by cans » Fri Sep 23, 2011 9:11 am
3 people have exactly 2 sibling. thus a,b,c are siblings.
d,e are siblings and f,g are siblings
total ways of selecting 2 people = 7C2=21
not siblings = total - siblings
siblings = select either of the 2 two persons group (either (d,e) or (f,g)) thus 2 ways
or select 2 people from 3 person group (a,b,c) thus 3C2 = 3 ways
total 5.
thus non siblings = 21-5 = 16
prob = 16/21
IMO E
If my post helped you- let me know by pushing the thanks button ;)

Contact me about long distance tutoring!
[email protected]

Cans!!
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 158
Joined: Sat Sep 03, 2011 10:31 am
Thanked: 29 times
Followed by:2 members

by gmatclubmember » Fri Sep 23, 2011 9:15 am
varun7nurav wrote:Hi, came across this question in MGmat test. The solution provided isn't clear. If anybody knows an easier method- plz share. Tq!

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?
A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21
Lets find the probability of selecting 2 individuals who are siblings:
1. If we choose one of the person who have exactly 1 sibling in the room then the probability would be 4/7 * 3/6
2. If we choose one of the peson who have exactly 2 siblings in the room then the probability would be 3/7 * 2/6
Total probability of choosing 2 individuals who are siblings = (12+6)/42 = 3/7.
Probability of choosing 2 individuals who are NOT siblings = 1-3/7=4/7.

Is C the OA?


Cheers
Ami/-
Top
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pm
Thanked: 114 times
Followed by:11 members

by knight247 » Fri Sep 23, 2011 9:20 am
We have a group of 7 people. After careful analysis of the problem we can figure out the following...
We have one sibling trio x1 x2 x3
we have one sibling pair a1 and a2
we have another sibling pair b1 and b2

Out of seven when we pull out a person we are obviously not going to replace him. Also when two people are pulled out we have two possible outcomes i.e. either both are siblings of each other or both are not siblings of each other. So the two outcomes are complementary meaning either one can occur but both can't occur. So we have a probability problem without replacement with complementary outcomes.

P(Both are siblings of each other)+P(Both are not siblings of each other)=1

First we'll find the answer using the direct method. For this we have to consider 3 different scenarios.

1.Our first pick is from x1 x2 and x3 and the second pick can be from any 4 of the remaining 6 so we have
3/7*4/6=6/21

2. Out first pick is from a1 a2 and our second pick can be from any 5 of the remaining 6 so we have
2/7*5/6=5/21

3. This scenario is same as 2 so 5/21

Adding all three we have 6/21+5/21+5/21=16/21


Next, we can solve by finding the probability where the two ppl picked will always be siblings
Again we consider 3 different scenarios

1.Our first pick is from the sibling group x1 x2 x3 and the second pick has to be from the same group
So we have. 3/7*2/6=1/7=3/21

2. Our first pick is from the sibling group a1 a2 and the second pick also has to be from the same group. So we have 2/7*1/6=1/21

3. Same as scenario 2 so 1/21

Adding all the three ways we have 1/21+1/21+3/21=5/21 .This is the probability that in the two picks both will be siblings. So 1-5/21=16/21 which is the probability that in the two picks neither will be siblings. Hope this clarifies everything.
Top
Senior | Next Rank: 100 Posts
Posts: 33
Joined: Fri Nov 19, 2010 8:07 am

by varun7nurav » Fri Sep 23, 2011 9:45 am
OMG... lol.. I just figured the dumbest mistake that I ever committed in the history on GMAT!!! haha..

I mistook the meaning of siblings to be child (lol) and wondering.. how can it be possible.. the question isn't making any sense... now it does... lmao!!!

Anyway, the solution:

Say ABCDEFG are the people.
4 people having exactly 1 sibling implies 2 sets of 2 siblings each: AB and CD
3 people having exactly 2 siblings implies 1 set of 3 siblings: EF, GH & FG.

Now, as always, for 'ATLEAST' / 'NOT' probability questions--- using the 1- X rule.

i.e., 1 - (no. of ways in which the two individuals picked are siblings)

no. of ways in which the two individuals picked are siblings= 5 sets/ 7C2

Therefore, 1 - (5/21) = 16/21.
:)
Top
Post new topic Post Reply
• Page 1 of 1