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Bill has a set of 6 black cards and a set of 6 red cards. Each card

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Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post Updated on: 29 Dec 2017, 01:06
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Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

(A) \(\frac{8}{33}\)

(B) \(\frac{62}{65}\)

(C) \(\frac{17}{33}\)

(D) \(\frac{103}{165}\)

(E) \(\frac{25}{33}\)

Originally posted by gmatter0913 on 05 Aug 2013, 22:47.
Last edited by Bunuel on 29 Dec 2017, 01:06, edited 2 times in total.
Edited the question, renamed the topic.
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 06 Aug 2013, 00:23
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Probability at least one pair equal \(= 1-P(none)\).

Now I am gonna find the probability of picking all different values.
You can chose the first card is 12 ways ( every card is OK ), the second in 10 ways (there are 11 cards remaining, and 1 has the same value as the first one), the third in 8 ways (there are 10 cards remaining, 2 with the same value as the first ones), and the fourth in 6 ways (same reasoning).

The first is chosen out of a pool of 12(all cards), the second out of a pool of 11, the third out of a pool of 10, the fourth out of a pool of 9.

\(P=1-\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=1-\frac{16}{33}=\frac{17}{33}.\)

Hope it's clear
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 06 Aug 2013, 05:48
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I solved this question as follows and I know I am wrong. The problem is: I don't know why am I wrong?

First card can be picked in 12C1 ways
Second card can be picked in 10C1 ways
Third card can be picked in 8C1 ways
Fourth card can be picked in 6C1 ways
All possibilities to pick 4 cards out of 12 = 12C4

Probability to have atleast one pair = 1 - prob to have no pairs = 1 - 12c1*10c1*8c1*6c1/12c4

It looks like I have done a permutation on the numerator. Could you help me with the error in the above?
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 06 Aug 2013, 05:59
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gmatter0913 wrote:
I solved this question as follows and I know I am wrong. The problem is: I don't know why am I wrong?

First card can be picked in 12C1 ways
Second card can be picked in 10C1 ways
Third card can be picked in 8C1 ways
Fourth card can be picked in 6C1 ways
All possibilities to pick 4 cards out of 12 = 12C4

Probability to have atleast one pair = 1 - prob to have no pairs = 1 - 12c1*10c1*8c1*6c1/12c4

It looks like I have done a permutation on the numerator. Could you help me with the error in the above?


Your denominator is correct \(12C4=495\). The problem is the numerator: the probability of picking 4 different values among 6 possible values: \(6C4=15\). But because each of the four cards can be black or red, each of the four slots can have 2 colors , so we have to multiply 15 by \(2*2*2*2=16\) (each of the four cards can be B or R).

Hope I've explained myself well.

So now \(P=\frac{495-15*16}{495}=\frac{17}{33}\).
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 06 Aug 2013, 07:58
Thanks Zarrolou

I understood the solution, but I would need a little more help on some concepts.

PERMUTATIONS OF 'n' items having some identical elements:
Number of permutations of 'n' items out of which p1 are alike, p2 are alike, p3 are alike and the rest are different is n!/p1!p2!p3!

Suppose, we have a set of 8 alphabets - {A, A, B, B, C, C, D, D}

The number of ways we can arrange the 8 alphabets is 8!/2!*2!*2!*2! (where we divide by 2! because we cannot distinguish between AA and its combination reversed, AA :) )

So, just the way we have the above formula, what is the formula to get all the possible combinations of 3 elements in the above case.

Will it be? - 4c3*(2)*(2)*(2)/8c4

where 4c3 is choosing combination of 3 elements among 4 unique elements.
and (2)*(2)*(2) is for each of the 3 elements in the combination there are two options (two identical elements)

Is there a general formula used to determine combinations in cases such as above?

The other doubt that I have is in the above set of alphabets {A, A, B, B, C, C, D, D}

how do I get the number of permutations of 3 elements? Is it 3!/{what?} because I don't know how many elements are repeated among those 3.
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 06 Aug 2013, 09:22
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gmatter0913 wrote:
Thanks Zarrolou

I understood the solution, but I would need a little more help on some concepts.

PERMUTATIONS OF 'n' items having some identical elements:
Number of permutations of 'n' items out of which p1 are alike, p2 are alike, p3 are alike and the rest are different is n!/p1!p2!p3!

Suppose, we have a set of 8 alphabets - {A, A, B, B, C, C, D, D}

The number of ways we can arrange the 8 alphabets is 8!/2!*2!*2!*2! (where we divide by 2! because we cannot distinguish between AA and its combination reversed, AA :) )

So, just the way we have the above formula, what is the formula to get all the possible combinations of 3 elements in the above case.

Will it be? - 4c3*(2)*(2)*(2)/8c4

where 4c3 is choosing combination of 3 elements among 4 unique elements.
and (2)*(2)*(2) is for each of the 3 elements in the combination there are two options (two identical elements)

Is there a general formula used to determine combinations in cases such as above?

The other doubt that I have is in the above set of alphabets {A, A, B, B, C, C, D, D}

how do I get the number of permutations of 3 elements? Is it 3!/{what?} because I don't know how many elements are repeated among those 3.


I) I do not know a general formula for the above case (and probably there is none).

II)You cannot use n!/p1!p2!p3! for your second problem. You have to try to work with the other formulas you know.
This is how I personally would do it:

First I would consider the case in which all three elements are different: [A,B,C]-[B,C,D]-... and since here we are taking about permutations, the total number of cases is \(4*3*2=24\) (you can pick any letter for the first slot, three (four-the previous chosen) for the second slot and so on).
Then I would consider the case in which a pair of letters is repeated: [A,A,B]-[A,A,C]-...
You can pick the pair of identical letters in 4 ways, the remaining letter is 3 ways. But each combination can be arranged in 3 ways: example
A,A,B-A,B,A-B,A,A.
So the total number is \(4*3*3=36\). Overall the possible permutations are 60.
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 07 Aug 2013, 04:14
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\frac{17}{33} is the answer.
You cant also do it using combinations.
select any pair(i.e. 2 cards) and the other two + select any two pairs and divide the total by 12C4.
this thought process leads to :-

(6C1*8*5 + 6C2)/12C4

for the first term in the Numerator - first select any one pair(6C1 ways), the we are left with 5 of black and 5 of red. Now select any one from one of these groups (5 ways). Lastly we are left with 9 cards, but in this group, one of them is the one with the same value in another color of the previously selected. So we are left with 8 to select one from(8 ways). Multiplying together -> 6C1*8*5.
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 07 Aug 2013, 12:04
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First find the probability of no pair P(no pair) : There are 6 pair. Choose any 4 pair and then select any one from each pair. 6C4*2C1*2C1*2C1*2C1 = 240
Total possibilities = 12C4 = 495
P (no pairs) = 240/495
P(at least one pair) = 1 - p(no pair) = 1 - 240/495 = 255/495 = 17/33. Option C
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 13 Mar 2014, 15:48
Here is my calculation:

Answer= 1- ( number of ways to select 4 different cards/ total number of ways to select 4 cards)


( number of ways to select 4 different cards/ total number of ways to select 4 cards)=

denominator= 12C4 so I haven't got any problem with calculating denominator

numerator= 12*10*8*6 Obviously I have a problem in calculating the numerator. My logic is that for the first card we can select one of the 12 cards. Then there are 10 cards left because we have already selected one of them and have discarded the pair of what we have selected previously. So there are 10 different cards for our second selection. Regarding previous way, we have 8 cards left to select for our third selection. Finally, we have 6 cards left to select four our fourth selection. Therefore total number of ways to select 4 different cards is 12*10*8*6.

I don't know what is wrong with my calculations but numerator should not be 12*10*8*6. Can anybody help please ?
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 11 May 2014, 14:33
Hi,

Maybe @Bunuel can shed some light on this? I didn't use the 1-Not probable approach, rather I carried out the tedious calc:

Probability Method:

Theory: Prob of at least 1 pair + Prob of 2 pairs

\(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{9}{9}\) * \(\frac{4!}{2!2!}\)

+

\(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{1}{9}\) * \(\frac{4!}{2!2!}\) = 20/33.

Where am I going wrong? EDIT: maybe in the permutation part?

Combinatorics Method

Theory: Prob of at least 1 pair + Prob of 2 pairs

(6c1)(6c1)/(12c4)

+

(6c2)(6c2)/(12c4) I get a completely wrong answer.

Any thoughts on what I'm doing wrong here?
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 16 Jun 2014, 08:24
Probability Method:

Theory: Prob of at least 1 pair + Prob of 2 pairs

\(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{9}{9}\) * \(\frac{4!}{2!2!}\)

+

\(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{1}{9}\) * \(\frac{4!}{2!2!}\) = 20/33.

Can you explain why r we multiplying by 4C2 ?

I get that we have 4 outcomes to be arranged. What is it divided by in the form of 2! * 2! ?
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 19 Jul 2014, 03:23
cumulonimbus wrote:
Probability Method:

Theory: Prob of at least 1 pair + Prob of 2 pairs

\(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{9}{9}\) * \(\frac{4!}{2!2!}\)

+

\(\frac{12}{12}*\frac{1}{11}*\frac{10}{10}*\frac{1}{9}\) * \(\frac{4!}{2!2!}\) = 20/33.

Can you explain why r we multiplying by 4C2 ?

I get that we have 4 outcomes to be arranged. What is it divided by in the form of 2! * 2! ?


This was my approach and i got it wrong :( need some help here...ohh and btw i did not use the
famous 1-x approach as many have already mentioned above.
ways to select 4 cards : 6c1 x 1 x 10c2
total ways to select 4 cards : 12c4
probability = 270/495

Explanation for choosing the 4 cards : I am first ensuring that atleast 1 pair is present. So for this one card can be chosen in 6c1 ways and the other card in only 1 way since pairing has to be done. Now the remaining 2 cards need to be chosen from a pool of 10 cards. The choice of the remaining 2 cards may or may not result in a pair but will cover all possible cases i.e 1 pair (from first 2 cards) + 0 (from remaining 2 cards) and 1pair(from first 2 cards) and 1 pair (from remaining 2 cards).

where am i wrong here ? :?
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 18 Sep 2014, 16:22
Hi Guys:

Had a doubt with the following question:

Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

Correct Answer: 17/33

I used the following way:

No. of ways of selecting one pair (AAXY)= 6C1 * 10C2

No. of arrangements= 12 * 6C1 * 10C2


No. of ways of selecting two pairs (AABB)= 6C2

No. of arrangements= 6C2 * 6

Total number of possible favorable arrangements= (6C1 *10C2 *12) + (6C2 * 6)

Total number of arrangements= 12C4

Hence probability = { (6C1 *10C2 *12) + (6C2 * 6) } / 12C4

= 37/132

Can someone please explain me the flaw in this.

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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 18 Sep 2014, 22:10
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There's a great explanation here: http://gmatclub.com/forum/bill-has-a-set-of-6-black-cards-and-a-set-of-6-red-cards-157417.html

Please try looking there first, and if it still doesn't make sense, we can look into your thought process in more detail.
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 18 Sep 2014, 23:11
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mohit2491 wrote:
Hi Guys:

Had a doubt with the following question:

Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

Correct Answer: 17/33

I used the following way:

No. of ways of selecting one pair (AAXY)= 6C1 * 10C2

No. of arrangements= 12 * 6C1 * 10C2




This is not correct. 6C1 * 10C2 includes the cases in which you get both pairs. Say you select the 3 black - 3 red pair from 6C1. Then when you select 2 of the remaining 10 cards, you could get both 4s. In that case, you will have two pairs.
Also, no arrangement is necessary. You just have to select the 4 cards.

Check out the correct solution at the link given above.
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 07 Oct 2014, 19:56
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Hi Guys. My strategy is:

First i am calculating no of ways to select no pairs:

first card 12 ways
second car 10 ways
third car 8 ways
fourth car 6 ways

Total 12*10*8*6 = 5760

But we need to divide this by 4! that is 24

5760/24 = 240

Total no of ways of selecting 4 cards out of 12 is 12c4 = 495

So probability of selecting no pairs = 240/495

and probability of selecting at least one pair is 1-240/495 = 255/495 = 17/33.

But, can someone please explain in a simple way why we divided by 4!. I have, by far just memorized this rule for such questions but never understood the logic.
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 07 Oct 2014, 20:15
Hi there, when you multiply 10*12*8*6 you are also including the arrangements of the four cards. The number of arrangements for 'each' possible selection is 4!. Since in this question we only need to select cards, and not arrange them, we divide the total number of arrangements by 4! so that only the selections remain.

Hope that helps. :)

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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 08 Oct 2014, 03:56
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aadikamagic wrote:
Hi Guys. My strategy is:

First i am calculating no of ways to select no pairs:

first card 12 ways
second car 10 ways
third car 8 ways
fourth car 6 ways

Total 12*10*8*6 = 5760

But we need to divide this by 4! that is 24

5760/24 = 240

Total no of ways of selecting 4 cards out of 12 is 12c4 = 495

So probability of selecting no pairs = 240/495

and probability of selecting at least one pair is 1-240/495 = 255/495 = 17/33.

But, can someone please explain in a simple way why we divided by 4!. I have, by far just memorized this rule for such questions but never understood the logic.


Dear aadikamagic,

Here's my take to clarify your doubt.

Your way to collect the number of possible options for no pairs is absolutely correct = 12*10*8*6 = 5760.

After this, you can find out the total number of possible outcomes. They'll be, 12*11*10*9

Prob. = (Number of favorable outcomes) / (Total Number of Outcomes)
= 12*10*8*6 / 12*11*10*9
= 16 / 33.

Now we need probability of getting at least one pair = 1 - 16/33 = 17 / 33.

Hope that helps you.
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 21 Oct 2015, 04:09
gmatter0913 wrote:
Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

(A) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)



My Solution:

We can solve it by find the opposite probability when bill finds no pair of cards that have the same value:

P = 1 -(none)

We have total 12 cards, then we have

12/12 * 10/11* 8/10* 6/9 = 16/33

Now subtract this from 1 we get

P = 1-16/33 = 17/33 Answer is Option C

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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card  [#permalink]

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New post 21 Oct 2015, 04:11
gmatter0913 wrote:
Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

(A) \(\frac{8}{33}\)
(B) \(\frac{62}{65}\)
(C) \(\frac{17}{33}\)
(D) \(\frac{103}{165}\)
(E) \(\frac{25}{33}\)




My Solution:

We can solve it by find the opposite probability when bill finds no pair of cards that have the same value:

P = 1 -(none)

We have total 12 cards, then we have

12/12 * 10/11* 8/10* 6/9 = 16/33

Now subtract this from 1 we get

P = 1-16/33 = 17/33 Answer is Option C

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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card &nbs [#permalink] 21 Oct 2015, 04:11

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