In a room filled with 7 people, 4 people have exactly 1
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- varun289
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233. 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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Take the case as 1-2(one sibling each) and 3-4(one sibling each) and 5-6-7(two siblings each) ..so 1-2-3-4 have one sibling each and 5-6-7 have two siblings each.
Total ways 7C2 = 21
Favorable ways 2C1*2C1 + 3C1*2C1 + 3C1*2C1 = 16
Answer is [spoiler]16/21[/spoiler]
Total ways 7C2 = 21
Favorable ways 2C1*2C1 + 3C1*2C1 + 3C1*2C1 = 16
Answer is [spoiler]16/21[/spoiler]
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First we need to recognize that the given information tells us that the 7 people consist of:varun289 wrote:233. 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 sibling trio
- a sibling pair
- and another sibling pair
Using counting techniques:
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
= [spoiler]16/21[/spoiler]
Cheers,
Brent
Can u explain the highlighted partpuneetkhurana2000 wrote:Take the case as 1-2(one sibling each) and 3-4(one sibling each) and 5-6-7(two siblings each) ..so 1-2-3-4 have one sibling each and 5-6-7 have two siblings each.
Total ways 7C2 = 21
Favorable ways 2C1*2C1 + 3C1*2C1 + 3C1*2C1 = 16
Answer is [spoiler]16/21[/spoiler]
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2C1*2C1(selecting 1 from (1,2) and selecting 1 from (3,4).puneetkhurana2000 wrote:
Take the case as 1-2(one sibling each) and 3-4(one sibling each) and 5-6-7(two siblings each) ..so 1-2-3-4 have one sibling each and 5-6-7 have two siblings each.
Total ways 7C2 = 21
Favorable ways 2C1*2C1 + 3C1*2C1 + 3C1*2C1 = 16
Answer is 16/21
ritind wrote:
Can u explain the highlighted part
3C1*2C1(selecting 1 from (5,6,7) and selecting 1 from (3,4).
3C1*2C1(selecting 1 from (5,6,7) and selecting 1 from (1,2).
Hope this helps!!!