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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card
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21 Oct 2015, 03:21
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 = 116/33 = 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
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19 Apr 2017, 16:37
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 1p. When translating the question into an equation, and translating the various probabilities for each card draw, we should use the 1x 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 DrawThe 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 DrawHere is where we start to apply the 1x 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 16 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/tonyownssi ... 34352.html



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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card
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01 May 2017, 01:02
mcwoodhill wrote: Zarrolou wrote: Probability at least one pair equal \(= 1P(none)\).
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
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04 Jun 2017, 01:52
russ9 wrote: Hi, Maybe Bunuel can shed some light on this? I didn't use the 1Not 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 MethodTheory: 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= 8/99assuming 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 = 24/99assuming 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 13 * 24 similar and 14 & 23 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
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05 Mar 2018, 12:16
Hi Bunnel... need your intervention to explain on the logic of P (None) please Sent from my SMG935F using GMAT Club Forum mobile app



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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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02 Feb 2019, 06:58
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
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13 May 2019, 10:50
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 nonmatching card 2nd AND select any nonmatching card 3rd AND select any nonmatching card 4th) = P(select any 1st card) x P(select any nonmatching card 2nd) x P(select any nonmatching card 3rd) x P(select any nonmatching card 4th) = 1 x 10/11 x 8/10 x 6/9 = 16/33So, P(at least one pair) = 1  16/33= 17/33 Answer: C Cheers, Brent
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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card
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19 Jun 2019, 07:25
VeritasKarishma wrote: 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. Very well, then, it at least contains 1 pair and also includes cases with two pairs. So what is wrong with this expression? I am missing something here. Please help.



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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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10 Aug 2019, 08:28
chetan2u Bunuel VeritasKarishma Gladiator59, generisI chose a rather long method. There are 5 ways of selecting different cards: 0R 4B : (6*5*4*3)/(12*11*10*9) 4R 0B : (6*5*4*3)/(12*11*10*9) 1R 3B : (6*5*4*3)/(12*11*10*9) 3R 1B : (6*5*4*3)/(12*11*10*9) 2R 2B : (6*5*4*3)/(12*11*10*9) Total probability of getting cards of different value=5*((6*5*4*3)/(12*11*10*9))=5/33 Required probability=1(5/33)=28/33 What's wrong with my approach?



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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card
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15 Aug 2019, 00:36
Hi,
So what i did:
possibility of 1 pair: 12*1*10*8 = 960 possibility of two pairs: 12*1*10*1 = 120
Total = 1080
Total possibilities = 12*11*10*9
so, ans =1080/12*11*10*9
but im not getting the right answer. Where am i going wrong?



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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card
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15 Aug 2019, 16:07
Hello, I am lost on how to calculate the numerator. Here is my logic and please let me know if you can help me see where I am going wrong. Since it is easier to find the nonoutcome, we want 0 card pairs out of the 6 in total. 12C1*10C1 would get us the number of ways of selecting the first card pair while getting different values for each card. Once we picked two cards, we have 10 cards left. Wouldn't the next selection be 10C1*8C1 since we can pick from the 10 remaining cards and just exclude the other matching card? I do not understand why we go from 12, 10, 8, 6. I thought it would be 12,10,10,8. Would really appreciate the help. Thank you.



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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card
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22 Dec 2019, 05:40
OE
he chance of getting AT LEAST one pair of cards with the same value out of 4 dealt cards should be computed using the 1x technique. That is, you should figure out the probability of getting NO PAIRS in those 4 cards (an easier probability to compute), and then subtract that probability from 1.
First card: The probability of getting NO pairs so far is 1 (since only one card has been dealt).
Second card: There is 1 card left in the deck with the same value as the first card. Thus, there are 10 cards that will NOT form a pair with the first card. there are 11 cards left in the deck.
Probability of NO pairs so far = 10/11.
Third card: Since there have been no pairs so far, there are two cards dealt with different values. There are 2 cards in the deck with the same values as those two cards. Thus, there are 8 cards that will not form a pair with either of those two cards. There are 10 cards left in the deck.
Probability of turning over a third card that does NOT form a pair in any way, GIVEN that there are NO pairs so far = 8/10.
Cumulative probability of avoiding a pair BOTH on the second card AND on the third card = product of the two probabilities above = (10/11) (8/10) = 8/11.
Fourth card: Now there are three cards dealt with different values. There are 3 cards in the deck with the same values; thus, there are 6 cards in the deck that will not form a pair with any of the three dealt cards. There are 9 cards left in the deck.
Probability of turning over a fourth card that does NOT form a pair in any way, GIVEN that there are NO pairs so far = 6/9.
Cumulative probability of avoiding a pair on the second card AND on the third card AND on the fourth card = cumulative product = (10/11) (8/10) (6/9) = 16/33.
Thus, the probability of getting AT LEAST ONE pair in the four cards is 1 − 16/33 = 17/33.
The correct answer is C.



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Re: Bill has a set of 6 black cards and a set of 6 red cards. Each card
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01 Feb 2020, 21:58
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) \(\frac{25}{33}\) The chance of getting AT LEAST one pair of cards with the same value out of 4 dealt cards should be computed using the 1x technique. That is, you should figure out the probability of getting NO PAIRS in those 4 cards (an easier probability to compute), and then subtract that probability from 1. First card: The probability of getting NO pairs so far is 1 (since only one card has been dealt). Second card: There is 1 card left in the deck with the same value as the first card. Thus, there are 10 cards that will NOT form a pair with the first card. there are 11 cards left in the deck. Probability of NO pairs so far = 10/11. Third card: Since there have been no pairs so far, there are two cards dealt with different values. There are 2 cards in the deck with the same values as those two cards. Thus, there are 8 cards that will not form a pair with either of those two cards. There are 10 cards left in the deck. Probability of turning over a third card that does NOT form a pair in any way, GIVEN that there are NO pairs so far = 8/10. Cumulative probability of avoiding a pair BOTH on the second card AND on the third card = product of the two probabilities above = (10/11) (8/10) = 8/11. Fourth card: Now there are three cards dealt with different values. There are 3 cards in the deck with the same values; thus, there are 6 cards in the deck that will not form a pair with any of the three dealt cards. There are 9 cards left in the deck. Probability of turning over a fourth card that does NOT form a pair in any way, GIVEN that there are NO pairs so far = 6/9. Cumulative probability of avoiding a pair on the second card AND on the third card AND on the fourth card = cumulative product = (10/11) (8/10) (6/9) = 16/33. Thus, the probability of getting AT LEAST ONE pair in the four cards is 1 − 16/33 = 17/33. The correct answer is C.




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