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MGMAT Question

by fangtray » Tue Sep 13, 2011 4:07 am
Hi Guys, I came across this MGMAT Question and I did not understand the explanation they provided. Could someone please clarify? thanks

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?
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by cans » Tue Sep 13, 2011 4:13 am
total 7.
4 people have exactly 1 sibling and 3 people have exactly 2 siblings.
let people be a,b,c,d,e,f,g.
if a has 'b' as sibling, then 'b' will have 'a' as sibling.
similarly 'c' and 'd' are siblings and 'e','f','g' are siblings.
We have 3 groups of siblings.
total number of ways to select 2 persons = 7C2 = 21.
total number those two individuals are NOT siblings: select any 2 groups and select one from each.
if you select (a,b) and (c,d) -> 2*2 ways = 4
if you select (e,f,g,) and any of the 2 sibling group: 2*3*2 = 12
Thus prob = (12+4)/21 = 16/21
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by knight247 » Tue Sep 13, 2011 5:06 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=[spoiler]16/21[/spoiler]


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.
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by fangtray » Tue Sep 13, 2011 5:07 am
sorry i still don't understand. what is 7c2=21?

select any 2 groups...i thought there were 3 groups?

if i select a,b and c,d? ...

total number those two individuals are NOT siblings: select any 2 groups and select one from each.
if you select (a,b) and (c,d) -> 2*2 ways = 4
if you select (e,f,g,) and any of the 2 sibling group: 2*3*2 = 12
Thus prob = (12+4)/21 = 16/21

i lost you there... if you could clarify? sorry.
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by cans » Tue Sep 13, 2011 5:56 am
fangtray wrote:sorry i still don't understand. what is 7c2=21?

select any 2 groups...i thought there were 3 groups?

if i select a,b and c,d? ...

total number those two individuals are NOT siblings: select any 2 groups and select one from each.
if you select (a,b) and (c,d) -> 2*2 ways = 4
if you select (e,f,g,) and any of the 2 sibling group: 2*3*2 = 12
Thus prob = (12+4)/21 = 16/21

i lost you there... if you could clarify? sorry.
7c2 is total number of ways we can select any 2 people. We have total 7 people and we have to select 2 people.

There are 3 groups. 2 groups have 2 people each and one group has 3 people.
So for people not siblings, they should be from different group.
Same group people are siblings.
And as we want 2 people, we have to select 2 groups.
If we select groups with 2 people each, one can select one person from each group: 2*2 ( 2 options for selecting from 1st group and 2 for selecting from other group)
if we select a 3 people group and combine it with 2 people group we have 2*3*2 (first 2 means: select one group from 2 groups of 2 people each... 3 means: one member from 3 people group and other 2 means: 1 member from 2 people group)
Total = 2*2 + 2*3*3 = 16
prob = 16/21
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by cans » Tue Sep 13, 2011 5:58 am
Alternate method: Total ways=21 (=7C2)
As we know there are 3 groups (a,b) (c,d) (e,f,g)
no of ways they are sibling = either select group (a,b) or select group (c,d) or select any 2 members from group (e,f,g)
= 1 + 1 + 3C2 = 5
Prob (siblings)=5/21
prob(not siblings) = 1 - 5/21 = 16/21
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