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?
Can you please provide an answer to this?
Probability Question
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- shovan85
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Let's the 7 people be ABCDEFG.chrisjim5 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?
Can you please provide an answer to this?
4 people have exactly 1 sibling:
Let A and B are siblings and C and D are siblings.
So,
A has 1 sibling B
B has 1 sibling A
C has 1 sibling D
D has 1 sibling C
3 people have exactly 2 siblings:
Let E, F and G are all siblings of each other.
So,
E has 2 siblings (F and G).
F has 2 siblings (E and G).
G has 2 siblings (E and F).
Total number of sibling pairs = 5 [AB, CD, EF, EG, FG]
Total number of pairs that can be formed from 7 people C(7,2) = 7!/5!*2! = 21
P(sibling pair) = 5/21
P(not sibling pair) = 1 - 5/21 = 16/21.
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- waltz2salsa
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Can someone please tell me where am i going wrong!!!
7 people,
4 have 1 siblings i.e. (a1, a2) and (b1, b2)
3 have 2 siblings i.e. (c1,c2,c3)
total ways of selecting two people out of 7 = 7c2
favorable ways:
(ways of selecting one from Ci *(ways of selecting one from left 4people) + (ways of selecting one from Ai *(ways of selecting one from left 5 people) + (ways of selecting one from Bi *(ways of selecting one from left 5 people)
= 3c1*4c1 + 2c1*5c1 + 2c1*5c1
P(no siblings) = fav ways/total ways = (3c1*4c1 + 2c1*5c1 + 2c1*5c1) / 7c2
= 2( 16/21)
7 people,
4 have 1 siblings i.e. (a1, a2) and (b1, b2)
3 have 2 siblings i.e. (c1,c2,c3)
total ways of selecting two people out of 7 = 7c2
favorable ways:
(ways of selecting one from Ci *(ways of selecting one from left 4people) + (ways of selecting one from Ai *(ways of selecting one from left 5 people) + (ways of selecting one from Bi *(ways of selecting one from left 5 people)
= 3c1*4c1 + 2c1*5c1 + 2c1*5c1
P(no siblings) = fav ways/total ways = (3c1*4c1 + 2c1*5c1 + 2c1*5c1) / 7c2
= 2( 16/21)
- GMATGuruNY
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You're overcounting the number of non-sibling pairs:waltz2salsa wrote:Can someone please tell me where am i going wrong!!!
7 people,
4 have 1 siblings i.e. (a1, a2) and (b1, b2)
3 have 2 siblings i.e. (c1,c2,c3)
total ways of selecting two people out of 7 = 7c2
favorable ways:
(ways of selecting one from Ci *(ways of selecting one from left 4people) + (ways of selecting one from Ai *(ways of selecting one from left 5 people) + (ways of selecting one from Bi *(ways of selecting one from left 5 people)
= 3c1*4c1 + 2c1*5c1 + 2c1*5c1
P(no siblings) = fav ways/total ways = (3c1*4c1 + 2c1*5c1 + 2c1*5c1) / 7c2
= 2( 16/21)
Number of ways to combine 1 choice from (c1,c2,c3) and 1 choice from (a1,a2,b1,b2) = 3*4 = 12.
We've now counted all the non-sibling pairs that include (c1, c2, c3). We need to count only the non-sibling pairs that don't include (c1,c2,c3).
Number of ways to combine 1 choice from (a1,a2) and 1 choice from (b1,b2) = 2*2 = 4.
Total non-sibling pairs = 12+4 = 16.
Total possible pairs from 7 people = 7C2 = 21.
P(non-sibling pair) = 16/21.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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