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The only contents of a container are 10 disks that are each
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Updated on: 01 Feb 2019, 01:16

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The only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7 ?

A) \(\frac{1}{7}\)

B) \(\frac{3}{14}\)

C) \(\frac{2}{7}\)

D) \(\frac{1}{2}\)

E) \(\frac{15}{28}\)

_________________

A) \(\frac{1}{7}\)

B) \(\frac{3}{14}\)

C) \(\frac{2}{7}\)

D) \(\frac{1}{2}\)

E) \(\frac{15}{28}\)

_________________

Originally posted by CAMANISHPARMAR on 21 Jan 2019, 03:08.

Last edited by chetan2u on 01 Feb 2019, 01:16, edited 1 time in total.

Last edited by chetan2u on 01 Feb 2019, 01:16, edited 1 time in total.

Corrected the OA

Re: The only contents of a container are 10 disks that are each
[#permalink]
21 Jan 2019, 21:35

5

This should be a select one multiple choice question.

There are altogether 10 integers namely disk labelled 1 to 10 inclusive therefore total number of outcomes for choosing 4 integers out of 10 is \(10C4 = \frac{10*9*8*7}{4*3*2} = 210\)

Now lets look at the ways in which we can chose 4 integers such that the range becomes 7.

There are only 3 possibilities how a range of 7 could occur when 4 numbers are chosen

1st. \((1,8)\): If among the 4 number chosen 2 of the numbers are 1 and 8 and the rest of the 2 numbers are from {2,3,4,5,6,7}. Notice that 9 cannot be chosen because it results in a range of 8.

2nd. \((2,9)\): Rest of the 2 numbers should be from the set {3,4,5,6,7,8}

3rd \((3,10)\): Rest of the 2 numbers should be from the set {4,5,6,7,8,9}

Now let us take the case for any 1 of 3 possible option.

(2,9)

There are 4 slots to chose the number. First 2 should be (2,9)

Hence 2c2 = 1

Next 2 numbers should be from the set of remaining 6 integers {3,4,5,6,7,8}

Hence \(6c2 = 15\)

Therefore for 1 of the possibility the required outcome = 15 ways

There are 3 possibilities therefore total possible ways = \(15*3 = 45\)

Therefore the probability is \(\frac{45}{210}\) which reduces to \(\frac{3}{14}\)

_________________

There are altogether 10 integers namely disk labelled 1 to 10 inclusive therefore total number of outcomes for choosing 4 integers out of 10 is \(10C4 = \frac{10*9*8*7}{4*3*2} = 210\)

Now lets look at the ways in which we can chose 4 integers such that the range becomes 7.

There are only 3 possibilities how a range of 7 could occur when 4 numbers are chosen

1st. \((1,8)\): If among the 4 number chosen 2 of the numbers are 1 and 8 and the rest of the 2 numbers are from {2,3,4,5,6,7}. Notice that 9 cannot be chosen because it results in a range of 8.

2nd. \((2,9)\): Rest of the 2 numbers should be from the set {3,4,5,6,7,8}

3rd \((3,10)\): Rest of the 2 numbers should be from the set {4,5,6,7,8,9}

Now let us take the case for any 1 of 3 possible option.

(2,9)

There are 4 slots to chose the number. First 2 should be (2,9)

Hence 2c2 = 1

Next 2 numbers should be from the set of remaining 6 integers {3,4,5,6,7,8}

Hence \(6c2 = 15\)

Therefore for 1 of the possibility the required outcome = 15 ways

There are 3 possibilities therefore total possible ways = \(15*3 = 45\)

Therefore the probability is \(\frac{45}{210}\) which reduces to \(\frac{3}{14}\)

_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos

Re: The only contents of a container are 10 disks that are each
[#permalink]
31 Jan 2019, 23:40

2

Expert Reply

In case pure combinatorics is too tough, this is a bit of a longer way, but it explains the how and why

there are only 3 possibilities where the range can be 7.

(1,8) (2,9) (3,10)

If we choose (1,8),

the possibility of getting 1 is 1/10, the possibility of getting 8 afterwards is 1/9

now we cant choose any number higher than 8, so we don't choose 9 or 10, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(6/8)*(5/7)=30/5040= 1/168

But remember, there are several ways to get these two cards out of 4 choices, in fact there are 4C2 =12 ways.

12*1/168= 1/14

If we choose (2,9)

the possibility of getting 2 is 1/10, the possibility of getting 9 after wards is 1/9

now we cant choose any number higher than 9 or lower than 2, so we dont choose 10 and 1, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14

If we choose (3,10)

the possibility of getting 3 is 1/10,the possibility of getting a 10 after 3 is 1/9.

now we cant choose any number lower than 3, so we dont choose 1 and 2, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14

there are 3 instances of 1/14, thus 3*1/14= 3/14, the answer is B

1/90+1/120+1/168= 8/315

Any pointers on how to bridge this gap?

there are only 3 possibilities where the range can be 7.

(1,8) (2,9) (3,10)

If we choose (1,8),

the possibility of getting 1 is 1/10, the possibility of getting 8 afterwards is 1/9

now we cant choose any number higher than 8, so we don't choose 9 or 10, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(6/8)*(5/7)=30/5040= 1/168

But remember, there are several ways to get these two cards out of 4 choices, in fact there are 4C2 =12 ways.

12*1/168= 1/14

If we choose (2,9)

the possibility of getting 2 is 1/10, the possibility of getting 9 after wards is 1/9

now we cant choose any number higher than 9 or lower than 2, so we dont choose 10 and 1, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14

If we choose (3,10)

the possibility of getting 3 is 1/10,the possibility of getting a 10 after 3 is 1/9.

now we cant choose any number lower than 3, so we dont choose 1 and 2, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14

there are 3 instances of 1/14, thus 3*1/14= 3/14, the answer is B

1/90+1/120+1/168= 8/315

Any pointers on how to bridge this gap?

Re: The only contents of a container are 10 disks that are each
[#permalink]
01 Feb 2019, 03:46

1

Expert Reply

CAMANISHPARMAR wrote:

The only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7 ?

A) \(\frac{1}{7}\)

B) \(\frac{3}{14}\)

C) \(\frac{2}{7}\)

D) \(\frac{1}{2}\)

E) \(\frac{15}{28}\)

A) \(\frac{1}{7}\)

B) \(\frac{3}{14}\)

C) \(\frac{2}{7}\)

D) \(\frac{1}{2}\)

E) \(\frac{15}{28}\)

!0 numbers.. 1 to 10..

range of 7 is possible in 3 cases only..

1) the extremes are 1 and 8... so other to can be chose from any of 6 from 2 to 7, ways to choose these 2 is 6C2

2) the extremes are 2 and 9... so other to can be chose from any of 6 from 3 to 8, ways to choose these 2 is 6C2

3) the extremes are 3 and 10... so other to can be chose from any of 6 from 4 to 9, ways to choose these 2 is 6C2

Thus total 3*6C2=3*6*5/2=45

Total ways = 10C4=\(\frac{10*9*8*7}{4*3*2*1}=210\)

Probability = 45/210=3/14

_________________

Some useful Theory.

1. Arithmetic and Geometric progressions : https://gre.myprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048

2. Effect of Arithmetic Operations on fraction : https://gre.myprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825

3. Remainders : https://gre.myprepclub.com/forum/remainders-what-you-should-know-11524.html

4. Number properties : https://gre.myprepclub.com/forum/number-property-all-you-require-11518.html

5. Absolute Modulus and Inequalities : https://gre.myprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

1. Arithmetic and Geometric progressions : https://gre.myprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048

2. Effect of Arithmetic Operations on fraction : https://gre.myprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825

3. Remainders : https://gre.myprepclub.com/forum/remainders-what-you-should-know-11524.html

4. Number properties : https://gre.myprepclub.com/forum/number-property-all-you-require-11518.html

5. Absolute Modulus and Inequalities : https://gre.myprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Re: The only contents of a container are 10 disks that are each
[#permalink]
19 May 2019, 11:49

CAMANISHPARMAR wrote:

The only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7 ?

A) \(\frac{1}{7}\)

B) \(\frac{3}{14}\)

C) \(\frac{2}{7}\)

D) \(\frac{1}{2}\)

E) \(\frac{15}{28}\)

A) \(\frac{1}{7}\)

B) \(\frac{3}{14}\)

C) \(\frac{2}{7}\)

D) \(\frac{1}{2}\)

E) \(\frac{15}{28}\)

It is mentioned in the question that disks are picked without replacement. How can we use nCk here?

In the numerator instead of 6C2 it should be 6x5 (Choose 1 fom 6 then 1 from remaining 5,ie 6C1 X 5C1)

Similarly, in the denominator instead of 10C4 it should be 10x9x8x7

The answer comes out to be 3/28 by this approach

Where am i going wrong?

Re: The only contents of a container are 10 disks that are each
[#permalink]
19 Jun 2019, 21:10

HappyMathtutor wrote:

In case pure combinatorics is too tough, this is a bit of a longer way, but it explains the how and why

there are only 3 possibilities where the range can be 7.

(1,8) (2,9) (3,10)

If we choose (1,8),

the possibility of getting 1 is 1/10, the possibility of getting 8 afterwards is 1/9

now we cant choose any number higher than 8, so we don't choose 9 or 10, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(6/8)*(5/7)=30/5040= 1/168

But remember, there are several ways to get these two cards out of 4 choices, in fact there are 4C2 =12 ways.

12*1/168= 1/14

If we choose (2,9)

the possibility of getting 2 is 1/10, the possibility of getting 9 after wards is 1/9

now we cant choose any number higher than 9 or lower than 2, so we dont choose 10 and 1, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14

If we choose (3,10)

the possibility of getting 3 is 1/10,the possibility of getting a 10 after 3 is 1/9.

now we cant choose any number lower than 3, so we dont choose 1 and 2, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14

there are 3 instances of 1/14, thus 3*1/14= 3/14, the answer is B

1/90+1/120+1/168= 8/315

Any pointers on how to bridge this gap?

there are only 3 possibilities where the range can be 7.

(1,8) (2,9) (3,10)

If we choose (1,8),

the possibility of getting 1 is 1/10, the possibility of getting 8 afterwards is 1/9

now we cant choose any number higher than 8, so we don't choose 9 or 10, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(6/8)*(5/7)=30/5040= 1/168

But remember, there are several ways to get these two cards out of 4 choices, in fact there are 4C2 =12 ways.

12*1/168= 1/14

If we choose (2,9)

the possibility of getting 2 is 1/10, the possibility of getting 9 after wards is 1/9

now we cant choose any number higher than 9 or lower than 2, so we dont choose 10 and 1, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14

If we choose (3,10)

the possibility of getting 3 is 1/10,the possibility of getting a 10 after 3 is 1/9.

now we cant choose any number lower than 3, so we dont choose 1 and 2, that means that out of the eight remaining disks, we can only choose 6 more,

so the next fraction is 6/8.

now there are 5 out of 7 disks left to choose from, so the next fraction is 5/7

(1/10)*(1/9)*(7/8)*(6/7)=42/5040= 1/168 * 4C2= 1/14

there are 3 instances of 1/14, thus 3*1/14= 3/14, the answer is B

1/90+1/120+1/168= 8/315

Any pointers on how to bridge this gap?

4C2 is 6. How come it is 12?

Re: The only contents of a container are 10 disks that are each
[#permalink]
20 Jun 2019, 02:10

Expert Reply

Where is pointed out that is 12 ??

Regards

_________________

Regards

_________________

Re: The only contents of a container are 10 disks that are each
[#permalink]
20 Jun 2019, 14:10

1

Carcass wrote:

Where is pointed out that is 12 ??

Regards

Regards

**File comment:** Its the first para second last line.

I really loved your explanation but got befuddled at this minor point. Could you please explain that

correction.PNG [ 29.5 KiB | Viewed 29867 times ]

Re: The only contents of a container are 10 disks that are each
[#permalink]
20 Jun 2019, 14:49

Expert Reply

Yes. Now gotcha.

True \(\frac{4\times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6\)

Clearly, 12 is wrong

Refer to the chetan explanation, much more consistent.

Regards

_________________

True \(\frac{4\times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1} = 6\)

Clearly, 12 is wrong

Refer to the chetan explanation, much more consistent.

Regards

_________________

Re: The only contents of a container are 10 disks that are each
[#permalink]
05 Sep 2019, 06:38

Hello,

This question is really hard.

I didn't understand the way we got 210. Why we didn't do the factorial way, 10*9*8*7 to find the total ways desks can be selected?

because at the first trial we have 10 then 9 then 8 then 7...

Kindly, help

This question is really hard.

I didn't understand the way we got 210. Why we didn't do the factorial way, 10*9*8*7 to find the total ways desks can be selected?

because at the first trial we have 10 then 9 then 8 then 7...

Kindly, help

Re: The only contents of a container are 10 disks that are each
[#permalink]
06 Sep 2019, 02:14

1

Asmakan wrote:

Hello,

This question is really hard.

I didn't understand the way we got 210. Why we didn't do the factorial way, 10*9*8*7 to find the total ways desks can be selected?

because at the first trial we have 10 then 9 then 8 then 7...

Kindly, help

This question is really hard.

I didn't understand the way we got 210. Why we didn't do the factorial way, 10*9*8*7 to find the total ways desks can be selected?

because at the first trial we have 10 then 9 then 8 then 7...

Kindly, help

Hello

Answering to your question, since 4 disks are chosen at random from the 10 disks; this clearly implies that it would be 10C4 => 210.

Re: The only contents of a container are 10 disks that are each
[#permalink]
06 Sep 2019, 03:26

1

Asmakan wrote:

Hello,

This question is really hard.

I didn't understand the way we got 210. Why we didn't do the factorial way, 10*9*8*7 to find the total ways desks can be selected?

because at the first trial we have 10 then 9 then 8 then 7...

Kindly, help

This question is really hard.

I didn't understand the way we got 210. Why we didn't do the factorial way, 10*9*8*7 to find the total ways desks can be selected?

because at the first trial we have 10 then 9 then 8 then 7...

Kindly, help

Hi there,

You ve been confused with the criteria of :: ORDER MATTERS and ORDER DOESN'T MATTERS

When u get "what order could 16 pool balls be" - In this case order matters and then we use the Factorial approach i.e. 16 X 15 X 14 ...

But if we need to choose 3 object out of 10 randomly - In that case the order doesn't matter and then we use Combination : i.e 10C3

Probability = \(\frac{{No. of possible outcomes}}{{Total no. Outcomes}}\)

So total no. of out comes will be without any restriction and hence we use the combination

You can refer the link https://www.mathsisfun.com/combinatorics/combinations-permutations.html

Hope that helps.

_________________

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Re: The only contents of a container are 10 disks that are each
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