There are n different size pairs of shoes in the box. One

Question

There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<2n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?

lexis · Answer

Well, no one left comments here

farend · Answer

I get nC1 * 2n-2C2k-2 / 2nC2k 
 = (2*k^2-k)/(2*n-1) .

farend · Answer

lexis; Wats the OA ?

lexis · Answer

u mean nC1=C(n,1)?

U should explain how did U get it. As I see, your answer is not correct.

JohnLewis1980 · Answer

My contribution:

Prob=2k/(2n-1)

Reasoning:

What is the prob of taking the appropriate shoe out once you have taken one out before?

1/(2n-1)

Because you have taken 2k shoes out, thus, 2k/(2n-1)

Regards

RyanDe680 · Answer

all too often, I see a problem like this and just draw a blank.

lexis · Answer

WELL, your answer is not correct. 
-----------

Let me explain more about this statement:
For example, there are 5 pairs of shoes A,B,C,D,E
probability to take only one pair of shoes is correct and 1 incorrect pair of shoes (mean 2 different shoes)?

Mean:
Let A1, A2 is correct pair (pretend)
C1,D2 is incorrect pair or C2, E2 or... (pretend)

Hope it helps you solve the general puzzle.

@ RyanDe680: Your avatar is so interesting.

lexis · Answer

If no one can take the correct answer, I will leave the OA in next week.

JohnLewis1980 · Answer

I'm afraid so :oops:

but why?

Don't we agree in the probability to take one complete pair of shoes? i.e. to take the right shoe once you've already take one out? For me: 1/(2n-1) Explanation: you take one shoe out, therefore, just 2n-1 shoes remain in the box. The probability to take the right one off is 1/(2n-1), doesn't it?

I need to improve my statistic skill. Thanks for the explanation

alex1234 · Answer

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alexsmith · Answer

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cumic · Answer

JohnLewis1980 wrote:
My contribution:

Prob=2k/(2n-1)

Reasoning:

What is the prob of taking the appropriate shoe out once you have taken one out before?

1/(2n-1)

Because you have taken 2k shoes out, thus, 2k/(2n-1)

Regards

===============================================

Shoudn't this be solved like this ->

Probability of taking out 2k shoes from total of 2n shoes = 2k/2n
Probability of taking out second compatible shoe = 1/(2n-1)

Hence probability of both events happening together - 2k/2n(2n-1)

Can someone confirm the OA please?

Regards,
Cumic

walker · Answer

Interesting problem. +1

My attempt:  P=\frac{C^n_k*C^{k}_{1}*(C^2_1)^{k-1}*C^{n-k}_{k-1}}{C^{2n}_{2k}}

when 2k>n+1 P=0

walker · Answer

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alexsmith · Answer

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JingChan · Answer

Let C(i, k) = iCk = i!/(k!(k-i)!)

Total number of ways to grab shoes:
C(2n, 2k)

Total number of ways to grab only 1 pair:
C(n, 1) * C(n-1, 2k-2) * 2^(k-2)

Answer:
C(n, 1) * C(n-1, 2k-2) * 2^(2k-2) / C(2n, 2k)
(unless 2k > n+1, then prob = 0)

This isn't an official question, right? The variables are awkwardly defined.

Either way, can I get the A, B, C, D, E answers/distractors, I would like to give this problem to a friend.

alexsmith · Answer

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lexis · Answer

Congratulation! You're correct. Your math skill is very good!
The tricky is to require ONLY ONE pair of shoes be same size. 
To solve it, we separate two sequences:
1st: There is NO pair of shoes 
2nd: There is ONLY ONE pair of shoes be same size
==> Combine: ONLY ONE pair of shoes be same size + NO pair of shoes in rest shoes (2n-2)

There are n different size pairs of shoes in the box. One

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