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Re: If 3 different integers are randomly selected from the integ [#permalink]
2
harishsridharan wrote:
GreenlightTestPrep wrote:
If 3 different integers are randomly selected from the integers from 1 to 12 inclusive, what is the probability that a triangle can be constructed so that its 3 sides are the lengths of the 3 selected numbers?

A) 3/8
B) 7/18
C) 19/44
D) 39/88
E) 11/24

*Kudos for correct solutions

Trusting my gut without calculation;
\(12C3\) will give \(220\).
Option C divisible by 220. Hence C :)


I did the same thing. Not sure if it's a reliable approach.
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Re: If 3 different integers are randomly selected from the integ [#permalink]
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170896 wrote:
harishsridharan wrote:
GreenlightTestPrep wrote:
If 3 different integers are randomly selected from the integers from 1 to 12 inclusive, what is the probability that a triangle can be constructed so that its 3 sides are the lengths of the 3 selected numbers?

A) 3/8
B) 7/18
C) 19/44
D) 39/88
E) 11/24

*Kudos for correct solutions

Trusting my gut without calculation;
\(12C3\) will give \(220\).
Option C divisible by 220. Hence C :)


It is, the problem would arise if we have 2 option choices with a number in denominator and factor of 220.
I did the same thing. Not sure if it's a reliable approach.
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Re: If 3 different integers are randomly selected from the integ [#permalink]
1
Let's be systematic and arrange the lengths in descending order

KEY CONCEPT: The longest side must be less than the sum of the other two sides

Triangle lengths with 12 as the longest side
12, 11, 10
12, 11, 9
12, 11, 8
12, 11, 7
12, 11, 6
12, 11, 5
12, 11, 4
12, 11, 3
12, 11, 2
Total outcomes in the form 12, 11, _ = 9

12, 10, 9
12, 10, 8
12, 10, 7
12, 10, 6
12, 10, 5
12, 10, 4
12, 10, 3
Total outcomes in the form 12, 10, _ = 7

12, 9, 8
12, 9, 7
12, 9, 6
12, 9, 5
12, 9, 4
Total outcomes in the form 12, 9, _ = 5


12, 8, 7
12, 8, 6
12, 8, 5
Total outcomes in the form 12, 8, _ = 3

12, 7, 6
Total outcomes in the form 12, 7, _ = 1

So, the total number of outcomes with 12 as the longest side = 9 + 7 + 5 + 3 + 1= 25

Triangle lengths with 11 as the longest side
11, 10, 9
11, 10, 8
11, 10, 7
11, 10, 6
11, 10, 5
11, 10, 4
11, 10, 3
11, 10, 2
Total outcomes in the form 11, 10, _ = 8

11, 9, 8
11, 9, 7
11, 9, 6
11, 9, 5
11, 9, 4
11, 9, 3
Total outcomes in the form 11, 9, _ = 6

11, 8, 7
11, 8, 6
11, 8, 5
11, 8, 4
Total outcomes in the form 11, 8, _ = 4

11, 7, 6
11, 7, 5
Total outcomes in the form 11, 7, _ = 2
Total number of outcomes with 11 as the longest side = 8 + 6 + 4 + 2= 20

Let's do one more round!

Triangle lengths with 10 as the longest side
10, 9, 8
10, 9, 7
10, 9, 6
10, 9, 5
10, 9, 4
10, 9, 3
10, 9, 2
Total outcomes in the form 10, 9, _ = 7
Total outcomes in the form 10, 8, _ = 5
Total outcomes in the form 10, 7, _ = 3
Total outcomes in the form 10, 6, _ = 1
Total number of outcomes with 10 as the longest side = 7 + 5 + 3 + 1 = 16

-------------------------------------------------

Let's summarize what we have so far:
Total number of outcomes with 12 as the longest side = 9 + 7 + 5 + 3 + 1= 25
Total number of outcomes with 11 as the longest side = 8 + 6 + 4 + 2 = 20
Total number of outcomes with 10 as the longest side = 7 + 5 + 3 + 1 = 16

See the patterns of ODDS and EVENS?
Keep going to get:
The total number of outcomes with 9 as the longest side = 6 + 4 + 2 = 12
The total number of outcomes with 8 as the longest side = 5 + 3 + 1 = 9
The total number of outcomes with 7 as the longest side = 4 + 2 = 6
The total number of outcomes with 6 as the longest side = 3 + 1 = 4
The total number of outcomes with 5 as the longest side = 2 = 2
The total number of outcomes with 4 as the longest side = 1

At this point we're done.

So, the total number of triangles possible = 25 + 20 + 16 + 12 + 9 + 6 + 4 + 2 + 1
= 95

Since we already learned (from earlier posts) that the denominator = 220

So, P(creating a triangle) = 95/220 = 19/44

Answer: C

Cheers

Daksh Kumar
Prep Club for GRE Bot
Re: If 3 different integers are randomly selected from the integ [#permalink]
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