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Re: The only contents of a container are 10 disks that are each [#permalink]
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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) \(\frac{15}{28}\)


!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
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Re: The only contents of a container are 10 disks that are each [#permalink]
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) \(\frac{15}{28}\)



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?
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Re: The only contents of a container are 10 disks that are each [#permalink]
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?


4C2 is 6. How come it is 12?
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Re: The only contents of a container are 10 disks that are each [#permalink]
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Where is pointed out that is 12 ??

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Re: The only contents of a container are 10 disks that are each [#permalink]
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Carcass wrote:
Where is pointed out that is 12 ??
Regards

Attachments

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
correction.PNG [ 29.5 KiB | Viewed 31620 times ]

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Re: The only contents of a container are 10 disks that are each [#permalink]
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.

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Re: The only contents of a container are 10 disks that are each [#permalink]
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
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Re: The only contents of a container are 10 disks that are each [#permalink]
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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

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.
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Re: The only contents of a container are 10 disks that are each [#permalink]
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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



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 [#permalink]
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Re: The only contents of a container are 10 disks that are each [#permalink]
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