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Siblings , Probabilities

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Siblings , Probabilities

by Giorgio » Wed Dec 16, 2009 2:00 pm
n 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

OA: E

Does anyone has any good explanation for this problem? This is from MGMAT but i don't really like the way they solve it
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by papgust » Wed Dec 16, 2009 7:05 pm
Total possibilities are selecting 2 ppl out of 7 ==> 7C2 = 21.

First we'll select siblings. Consider 2 scenarios here as we have 2 different groups with numbers.

1. You can select 2 people from a group of 4 people ==> 2C2 + 2C2 = 2 [Reason: There can be 2 different pairs of siblings in the group of 4 - say "AB" or "CD"]
OR
2. You can select 2 people from a group of 3 people ==> 3C2 = 3

Probability of selecting siblings = 2/21 + 3/21 = 5/21

Now, probability of NOT selecting siblings = 1 - 5/21 = 16/21
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by viju9162 » Wed Dec 16, 2009 9:01 pm
Hi papgust,

Thanks for the explanation. I couldn't get the first part : select 2 people from a group of 4 people.. There are 2 different pair of siblings.

But, the question says 4 people have exactly 1 sibling in the room..

I didn't really understand that. Can you please explain?

Thank you,
Viju
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by Testluv » Wed Dec 16, 2009 10:57 pm
Hi viju,

Among the seven people, we have two pairs and a trio. In each of the pairs, the two people are siblings to each other--that's 4 people each of whom have exactly 1 sibling. In the trio, each person is sibling to the other two--that's 3 people, each of whom have exactly 2 siblings.

EDIT: And, just to elaborate on papgust's solution, whenever a probability problem says "at least" it is almost always easier to think about the undesired outcomes (what you don't want) than the desired outcomes, and so use the formula:

Prob total = 1 = prob desired + prob undesired or

prob desired = 1 - prob undesired (this is the approach papgust used)

Again, solve for undesired first and subtract from total.

Similarly, when dealing with combinatorics, if there is a restriction imposed use:

Total number of selections = permitted number + restricted number (we can also use this approach here)

Again, in combinatorics, where there is a restriction imposed and if the question is asking you to solve for permitted, it is almost always easiest to solve for total and restricted, and then to subtract restricted from total to compute permitted.

Here, we are asked for the probability of NOT getting siblings. So, after figuring out the total number of ways (7C2), it will be easier to think about the ways we CAN get the siblings, and then subtract this from total. As papgust and I wrote above, we can deduce that there are two different pairs, and then a trio of siblings. How can we select two siblings then? Either by selecting either of the two pairs--that's 2 ways...OR by selecting any 2 siblings from the trio, which is 3C2, or another 3 ways. So there are 5 ways we CAN select siblings. There are 21 ways to select 2 people in total. So, there are 21 - 5= 16 (out of 21) selections of 2 people that do NOT have siblings.
Last edited by Testluv on Wed Dec 16, 2009 11:40 pm, edited 2 times in total.
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by getso » Wed Dec 16, 2009 11:35 pm
Hi Viju,

For easy understanding,

Lets say 7 people are 1,2,3,4,5,6,7.

1,2,3,4 have 1 sibling each where (1,2 ) one pair siblings and (3,4 )are another pair of siblings.

5,6,7 have 2 siblings each.

Now we need to select 2 people such that ,the selected people are not siblings, which is equal to

1 - [prob of selecting 2 people who are siblings].........Eq (1)

Now prob of selecting 2 people who are siblings is as follows:

Probability of selecting 2 sibling from pair (1,2) is 2c2=1

Probability of selecting 2 siblings from pair (3,4) is 2c2=1

Probability of selecting 2 siblings from (5,6,7)=3C2=3

Probability of selecting 2 people from given 7 is 7C2= 21

Therefore total probability of selecting 2 people who are siblings is 1+1+3/21=5/21

Subsiting the value obtained in EQ(1)

1-5/21=16/21
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by viju9162 » Wed Dec 16, 2009 11:42 pm
Thanks getso and Testluv for the explanation.

Regards,
Viju
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by valleeny » Sun Dec 27, 2009 12:30 am
Hi I have another way of solving.

First, select the first person.
Probability of selecting someone with 1 sibling = 4/7
Probability of selecting someone with 2 siblings = 3/7

Then, as we select the second person
Probability of selecting someone not a sibling of the first person with 1 sibling = 5/6
Probability of selecting someone not a sibling of the first person with 2 sibling = 4/6

Total probability = (4/7)(5/6) + (3/7)(4/6) = 16/21
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by viidyasagar » Sat Jan 23, 2010 7:32 am
whenever a probability problem says "at least" it is almost always easier to think about the undesired outcomes (what you don't want) than the desired outcomes, and so use the formula
Although this is right, i don't see an "at least" in the Question.

Another way to solve is as follows

1. n(A)/ n(P) = formula used

2. n(P) = quite obviously 7C2 = 21

3. n(A) = ?

there are 3 groups of siblings, G1 has 2siblings, G2 has 2 siblings and G3 has 3 siblings

2 non-siblings can be chosen when both selected members belong to different groups. i.e.one from G1 and another G2 OR one from G1 and another from G3 OR one from G2 and another from G3...

Mathematically that means

n(A) = 2C1*2C1 + 2C1*3C1 + 2C1*3C1 = 4+6+6 = 16

4. Required probability = 16/ 21, choose E
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by Testluv » Sat Jan 23, 2010 4:31 pm
Although this is right, i don't see an "at least" in the Question.
No, I realize that; re-read my post--I was drawing an analogy between probability questions that ask about "at least" and combinatorics problems that impose a restriction.
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