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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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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, 04:21
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 19 Apr 2017, 17:37
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) \(\frac{25}{33}\)


This question can be solved using complementary probability or in other words 1-p. When translating the question into an equation, and translating the various probabilities for each card draw, we should use the 1-x technique which is a concept relating to pairs in probability.

So

We need to find the complementary probability of not drawing a pair on each draw. There are four draws, so when we find the probability of not drawing a pair for each draw we will multiply those four probabilities and subtract from one.

First Draw

The probability of not drawing a pair on the first draw is 12/12, or 1, because you cannot have a pair with two items of course


Second Draw

Here is where we start to apply the 1-x technique

10 ( we subtract 2 from the numerator ) / 11 (we subtract 1 from the numerator

Third Draw

8 (again subtract 2 from the numerator)/ 10 (again subtract 1 from the numerator)

Fourth Draw

6 (again subtract 2 from the numerator)/9 (again subtract 1 from the numerator)

1- ( [1] X [10/11] x [8/10] x [6/9] )
1-(16/33)= 17/33

Possibly, the most fundamental example of using the complementary probability is a dice roll problem- on a dice numbered 1-6 what is the chance of rolling five at least once. *note the words at least and exact are important in probability problems

1- (5^3/6^3)

Though a probability question that is more on par with this question is one regarding a pair of socks

https://gmatclub.com/forum/tony-owns-si ... 34352.html
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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 01 May 2017, 02:02
mcwoodhill wrote:
Zarrolou wrote:
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


nice way. my way was a bit more complicated.


Hi All,

In the same line of above approach of finding probability (without permutation or combination), can we find solution without subtract by 1.
I mean in above approach we are finding no pair then to find at least one pair by subtracting by 1, so I am looking for a solution where directly we can find at least one pair(no subtraction and without finding no pair probability).

Thanks
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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 04 Jun 2017, 02:52
russ9 wrote:
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?


Hi,
The way to solve with probablity method without using 1- probablity of(none) is
Using the full understanding of all cases:

Probablity of selecting one pair :
(12/12) * (1/12)* 10/10 *8/9= 8/99----assuming ist two cards are same.
12/12 * 10/11* 2/10 * 8/9 = 16/99 -------assuming 1st and third card are same .
12/12 * 10/11 * 8 /10 * 3/9 = 24/99-------assuming ist and fourth card are same.
Total probablity of only one card same is therefore : (8+16+24)/99 = 48/99 =16/33-----(1)
Probablity of two pair same in similar way is
12/12 *1/12* 10/10*1/9 = 1/99 ....ist two same and third and fourth same.
12/12*10/11*2/10*1/9 = 2/99----- This includes both 1-3 * 2-4 similar and 1-4 & 2-3 similar
Therefore total probablity of any two same is 3/99=1/33----(2)

therefore total probablity = (1) + (2) = 17/33.

Kudos if helped.
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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 05 Mar 2018, 13:16
Hi Bunnel... need your intervention to explain on the logic of P (None) please

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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 02 Feb 2019, 07:58
Hi Bunuel,
I did this question using the following aproach
1) For one pair: 6C1*10C2
2) For 2 pairs : 6C2

P(E)= (6C1*10C2 + 6C2)/12C4

This gives me 19/33

Can you please tell me where I am wrong??
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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 May 2019, 11:50
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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) \(\frac{25}{33}\)


We can solve this question using probability rules.

First, recognize that P(at least one pair) = 1 - P(no pairs)

P(no pairs) = P(select ANY 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33

So, P(at least one pair) = 1 - 16/33
= 17/33

Answer: C

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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card   [#permalink] 13 May 2019, 11:50

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