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Manager
Joined: 11 Oct 2013
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In a room filled with 7 people, 4 people have exactly 1
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11 Dec 2015, 02:50
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
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12 Aug 2017, 10:49
lagomez wrote: You have people 1234567
4 have exactly 1 sibling which can mean: 12 are siblings 34 are siblings
3 have exactly 2 siblings which can mean: 567 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 1234 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 nonsibling pair is 1234 Therefore the probability is 4/42. Multiply that probability by 3, which represent 567 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
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02 Oct 2017, 21:38
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
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01 Jan 2018, 03:32
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
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14 Jul 2018, 23:47
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
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11 Aug 2018, 04:28
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
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03 Oct 2018, 16:53
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)




Re: In a room filled with 7 people, 4 people have exactly 1 &nbs
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