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Re: If 3 different integers are randomly selected from the integ [#permalink]
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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
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Re: If 3 different integers are randomly selected from the integ [#permalink]
GreenlightTestPrep 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) \(\frac{3}{8}\)

B) \(\frac{7}{18}\)

C) \(\frac{19}{44}\)

D) \(\frac{39}{88}\)

E) \(\frac{11}{24}\)

*Kudos for correct solutions


I specifically created this question to illustrate the importance of calculating the denominator first (when using counting techniques to solve a probability question)
There are two reasons why you should calculate the denominator first:

1) The denominator is almost always easier to calculate than the numerator, and while calculating the denominator, you may gain some insight into how to calculate the numerator.
2) Once you know the denominator, you can use this to eliminate answer choices. So, even if you don't know the correct answer, you can still eliminate some answers and increase your chances of guessing correctly.

If you're lucky, you can eliminate 4 of the 5 answer choices, as you have done above!!

For this question, P(a triangle can be constructed with the 3 selected lengths) = (number of triangles with 3 lengths from 1 to 12 inclusive)/(TOTAL number of ways to select 3 numbers)

Let's calculate the denominator.

TOTAL number of ways to select 3 numbers
Since the order in which we select the 3 numbers doesn't matter, we can use COMBINATIONS.
We can select 3 numbers from 12 numbers in 12C3 ways.
12C3 = (12)(11)(10)/(3)(2)(1) = 220 ways

So, the correct answer will either be in the form k/220, OR some equivalent fraction in which the denominator is a FACTOR of 220.
For example, IF we calculate the numerator and get 110, then the answer = 110/220 = 1/2 (notice that 2 is a FACTOR of 220)
IF we calculate the numerator and get 15, then the answer = 15/220 = 3/44 (notice that 44 is a FACTOR of 220) And so on.

When we check the answer choices, we see that only one answer choice (C) has a denominator that's a FACTOR of 440.
So, C must be the correct answer.

We can answer the question without having to calculate the numerator (which is a time-consuming task)

On test day, it's unlikely that this technique will allow you to eliminate 4 answer choices. HOWEVER, if you're pressed for time, or you can't calculate the numerator, this technique may allow you to eliminate some of the answer choices and increase your likelihood of a correct guess.

Cheers,
Brent


With this logic, D should also work as 88 (*10) may be a multiple of 220 as well?
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Re: If 3 different integers are randomly selected from the integ [#permalink]
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