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3 numbers are randomly selected, with replacement, from the set of

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3 numbers are randomly selected, with replacement, from the set of  [#permalink]

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New post 02 Sep 2015, 22:49
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3 numbers are randomly selected, with replacement, from the set of integers {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. If the first number selected is w, the second number selected is x, and the third number is y, what is the probability that w < x < y ?

A. 3/40
B. 28/243
C. 3/25
D. 33/100
E. 64/125

Kudos for a correct solution.

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Re: 3 numbers are randomly selected, with replacement, from the set of  [#permalink]

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New post 03 Sep 2015, 06:26
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Bunuel wrote:
3 numbers are randomly selected, with replacement, from the set of integers {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. If the first number selected is w, the second number selected is x, and the third number is y, what is the probability that w < x < y ?

A. 3/40
B. 28/243
C. 3/25
D. 33/100
E. 64/125

Kudos for a correct solution.


Total number of ways to select 3 numbers from the given set will be 10 x 10 x 10 = 1000
The number of combinations in which w < x < y will be

1) when w = 0
let w= 0 and x=1 then y can have 8 values
let w= 0 and x=2 then y can have 7 values

This pattern will continue till y can only 1 value i.e 9
So for w=0 we will have (8+7+6+5+4+3+2+1) combinations
or 8(9)/2 = 36 ... ( sum of first n natural number is given by n(n+1)/2 )

2) when w=1
let w=1 and x=2 then y can have 7 values
let w=1 and x=3 then y can have 6 values
This pattern will again continue till y can only 1 value i.e 9
number of combinations in this case will be 7(8)/2
= 28

Notice that for each increasing value of w, the maximum number of values which y can have, decreases by 1. This can be seen in highlighted statements.
So, we just have to subtract the maximum value of y from the sum in each case to get the sum of next case. This will save time as we do not have to calculate the number of combinations for every case.

So the total number of combinations will be

36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120

So the probability will be
120/1000
= 3/25

Answer:- C
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Re: 3 numbers are randomly selected, with replacement, from the set of  [#permalink]

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New post 06 Sep 2015, 07:35
Solution:
Total possible ways = 10*10*10 = 1000
Case1: w=0.
if x = 1, then y can have 8 ways.
if x =2 , then y can have 7 ways and so on.
So, no. of ways = 8+7+..+1

Case2 : w=1.
if x = 2, then y can have 7 ways.
if x =3 , then y can have 6 ways and so on.
So, no. of ways = 7+6 +..+1

There will be 8 cases in this way and they will follow a pattern which looks like this:
No. of ways for Case1 : 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8
No. of ways for Case2 : 1 + 2 + 3 + 4 + 5 + 6 + 7
No. of ways for Case3 : 1 + 2 + 3 + 4 + 5 + 6
No. of ways for Case4 : 1 + 2 + 3 + 4 + 5
No. of ways for Case5 : 1 + 2 + 3 + 4
No. of ways for Case6 : 1 + 2 + 3
No. of ways for Case7 : 1 + 2
No. of ways for Case8 : 1
____________________________________________________________
Total number of ways : 8(1) + 7(2) + 6(3) + 5(4) + 4(5) + 3(6) + 2(7) + 1(8) = 2(4(5) + 3(6) + 2(7) + 1(8)) = 2(60) = 120

Therefore, ans is 120/1000 = 3/25.
Option c.
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Re: 3 numbers are randomly selected, with replacement, from the set of  [#permalink]

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New post 29 Nov 2015, 08:13
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solution :

total number of ways 3 numbers can be selected: 10*10*10 = 1000

Number of ways 3 different numbers can be selected = 10C3= 120
out of each selection there is only 1 way in which w<x<y.

so probability = 120/1000 = 3/25 .. c is the Answer
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Re: 3 numbers are randomly selected, with replacement, from the set of  [#permalink]

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New post 07 Jul 2017, 09:52
Why do we consider it to be 10*10*10?? Because out of 10 we have to select 3 and so it has to be 10 things taken 3 at a time
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Re: 3 numbers are randomly selected, with replacement, from the set of  [#permalink]

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New post 07 Jul 2017, 10:02
longhaul123 wrote:
Why do we consider it to be 10*10*10?? Because out of 10 we have to select 3 and so it has to be 10 things taken 3 at a time



Hi,
we require 10*10*10 as we are talking of REPLACEMENT..

so the first can be any of the 10.... 10 ways
the second can be again any of the 10 as we are talking of replace ment... 10
similarly third time 10..

total ways 10*10*10

hope it helps
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3 numbers are randomly selected, with replacement, from the set of  [#permalink]

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New post 07 Jul 2017, 10:07
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Bunuel wrote:
3 numbers are randomly selected, with replacement, from the set of integers {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. If the first number selected is w, the second number selected is x, and the third number is y, what is the probability that w < x < y ?

A. 3/40
B. 28/243
C. 3/25
D. 33/100
E. 64/125

Kudos for a correct solution.



Hi..

the MOST time saving and proper method will be..

1)TOTAL ways 3 can be chosen... 10*10*10

2)Now when will be three W<X<Y
when all three are different 10*9*8
But ONLY one way of selection will give us the desired result...
example ABC, ACB,BCA,BAC,CAB,CBA... these 6 have only ONE ABC in the required format

so ways = \(\frac{10*9*8}{3!}\)=10*3*4

Probability = \(\frac{10*3*4}{10*10*10}=\frac{3}{25}\)

C
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Re: 3 numbers are randomly selected, with replacement, from the set of  [#permalink]

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Re: 3 numbers are randomly selected, with replacement, from the set of   [#permalink] 03 Nov 2018, 05:00
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