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In a room filled with 7 people, 4 people have exactly 1

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In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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New post 11 Dec 2015, 03:50
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reply2spg wrote:
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


This is how I solved.

Case 1: The group of 4 have exactly 1 sibling. So maximum 2 people can be picked from that group who are not siblings.
Case 2: The group of 3 has 2 siblings each. So maximum only 1 person from the group can be chosen.
Total number of ways in which we can pick people is 7C2 = 21.


Case 1 -
\(2C1 * 2C1\) ways

Case 2 -
\(3C1 * 4C1\) ways

Probability -
\(\frac{2C1 * 2C1 + 4C1 * 3C1}{21} = \frac{16}{21}\)

Experts, please confirm if my approach is correct :)

+Kudos if this helped!
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In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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New post 12 Aug 2017, 11:49
lagomez wrote:
You have people 1-2-3-4-5-6-7

4 have exactly 1 sibling which can mean:
1-2 are siblings
3-4 are siblings

3 have exactly 2 siblings which can mean:
5-6-7 are siblings

Let's start with the group of 4:
The probability of picking 1 is (1/7)
The probability of not getting a sibling pair is (5/6) because the only other sibling is 2
Therefore, if 1 is selected first the probability of not getting a sibling pair is 5/42
Multiply that by 4 because the probability is the same whether you start with 1-2-3-4 so you get 20/42 for the first 4 people

Now let's go to the group of 3:
The probability of picking 5 is (1/7)
The probability of not getting a sibling pair is (4/6) which the non-sibling pair is 1-2-3-4
Therefore the probability is 4/42. Multiply that probability by 3, which represent 5-6-7 so the probability is 12/42

Now you have two probabilities: 12/42 and 20/42
add both and you get 32/42 or 16/21



It's easier to use the complement. Not sure why people are trying so hard here to use the choose formula when it makes the question 1000X harder

4/7 chance to pick a person with 1 sibling * 1/6 chance you get a sibling or 4/42

3/7 change to pick a 2 person sibling * 2/6 or 6/42

Total of 10/42 of getting a sibling.

We get to use the simple or rule not the generalized since the things are mutually exclusive. And we aren't drawing twice just once. So either a 4 person is picked first or 3 person.

So that simplifies to 5/21 then 1 minus this is 16/21 for the odds that they are not siblings. 5/21 they are.
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Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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New post 02 Oct 2017, 22:38
lagomez Can you explain how you got 5/6?
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Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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New post 01 Jan 2018, 04:32
My solution but wrong

4 siblings have one sibling each (A1, A2, B1, B2) and 3 have 2 each (C1, C2, C3). Thus total possibilities of selecting 2 from 7 = 7C2 = 21

Now, when the two selected are not siblings, therefore, 1 from one group and another from the other group. Thus, selection = 4C1*3C1 = 12.

Therefore, solution = 12/21 = 4/7
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Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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New post 15 Jul 2018, 00:47
Hi,

Have a quick question on whether my approach is right - I got the answer of 16/21, but just wanted to double check.

P(pair selected are siblings) = (4/7)(1/6) + (3/7)(2/6) = 10/42 = 5/21
P(pair is not siblings) = 1 - 5/21 = 16/21

(4/7) = from 7, select 1 person from the first group of 4
(1/6) = select his sibling from the rest of the group

(3/7) = from 7, select 1 person from the second group of 3
(2/6) = select either of his sibling from the rest of the group

Thanks in advance!
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In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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New post 11 Aug 2018, 05:28
I have used alphabet analogy to solve this Q.

Consider three groups, each containing same alphabets.

Group 1: AA
Group 2: BB
Group 3: CCC

We need to select 2 distinct alphabets.

TOTAL OUTCOMES:
ways of selecting 2 alphabets out of 7 without any restriction is 7C2= 21

FAVORABLE OUTCOMES:
based on restriction of selecting distinct alphabets, we have following possibilities:

1 each from G1&G2: 2C1 x 2C1= 4
1 each from G1&G3: 2C1 x 3C1= 6
1 each from G2&G3: 2C1 x 3C2= 6
Total favorable outcomes: 16

Probability: 16/21

Ans E

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Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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New post 03 Oct 2018, 17:53
PiyushK wrote:
\(2*\frac{1}{7}*\frac{5}{6}+2*\frac{1}{7}*\frac{5}{6}+3*\frac{1}{7}*\frac{4}{6}\)
\(\frac{32}{42} = \frac{16}{21}\)


this one made the most sense to me.

We have two groups of ppl:

A B C D
E F G

starting from A, chance of picking the first person:
1/7
after A, the chance of not picking a sibling:
5/6

Repeat this for B, C, and D
2*(1/7)(5/6)

If we pick E F G first
then, we know the first chance is 1/7, but second chance is 4/6
hence (1/7)(4/6)
repeat this for E F G
hence 3x

so total should
2*(1/7)(5/6)+2*(1/7)(5/6)+3(1/7)(4/6)
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In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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New post 10 Feb 2019, 19:31
I think this question can be a lot easier if we realize that within the 4 people (A,B,C,D) who has exactly 1 sibling, there are two 'sets' of siblings -- any selection two people within the group of 2 could be siblings with each other. The three people(E,F,G) with exactly 2 siblings, are siblings with each other.

So instead of breaking the 7 people into three groups, we can solve it by dividing the people into two groups, making up 3 cases.

Case 1: ABCD, Pick 2
4C2 = 6. But of the 6 potential sets of siblings, we already know that there are siblings - so subtract 2, we have 4.

Case 2: EFG, pick 2
3C2=3. Again, we already know that the three people are siblings with each other, so all three potential sets will be siblings - so subtract 2, we get 0 from this group.

Case 3: Pick one from ABCD, pick another from EFG.
No restrictions here, people from 2 sibling group are not the same people from 1 sibling group.
4C1+3C1=12

Non-sibling selections: 4+0+12=16
Total selections: 7C2=21
Part/All = 16/21
Answer E
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Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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New post 13 May 2019, 11:36
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reply2spg wrote:
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


First we need to recognize that the given information tells us that the 7 people consist of:
- a sibling trio
- a sibling pair
- and another sibling pair

Let's use counting techniques to answer this question

For this question, it's easier to find the complement.
So P(not siblings) = 1 - P(they are siblings)

P(they are siblings) = [# of ways to select 2 siblings] / [total # of ways to select 2 people]

# of ways to select 2 siblings
Case a) 2 siblings from the sibling trio: from these 3 siblings, we can select 2 siblings in 3C2 ways (3 ways)
Case b) 2 siblings from first sibling pair: we can select 2 siblings in 2C2 ways (1 way)
Case c) 2 siblings from second sibling pair: we can select 2 siblings in 2C2 ways (1 way)

So, total number of ways to select 2 siblings = 3+1+1 = 5

total # of ways to select 2 people
We have 7 people and we want to select 2 of them
We can accomplish this in 7C2 ways (21 ways)

So, P(they are siblings) = 5/21

This means P(not siblings) = 1 - 5/21
= 16/21

Answer: E

Cheers,
Brent
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Re: In a room filled with 7 people, 4 people have exactly 1   [#permalink] 13 May 2019, 11:36

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