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PS: 2 suits of cards
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22 Mar 2008, 18:06
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value. Bill likes to play a game in which he shuffles the deck, 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. 8/33 B. 62/165 C. 17/33 D. 103/165 E. 25/33
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Re: PS: 2 suits of cards
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22 Mar 2008, 18:30
chineseburned wrote: Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.
Bill likes to play a game in which he shuffles the deck, 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. 8/33 B. 62/165 C. 17/33 D. 103/165 E. 25/33 prob that at least one pair of cards that have the same value = P(one pair with same value) + P(both pair with same value) P(one pair with same value) = 6 x 9c1 = 54 P(both pair with same value) = 6c2 = 15 so the prob = [(6x9c1)+(6c2)]/12c4 = (54 + 15)/495 = 23/165 do not know what i miss that i did not get the OA.
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Re: PS: 2 suits of cards
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22 Mar 2008, 19:10
C
I tried two ways..in one way I don't know what went wrong but arrived at an obviously wrong answer...
Approach 1:
Probability of second card not matching the first card = 10/11 Doing similar math for other 2 cards left.. Probability that no card matces other three is 10/11 * 8/10 * 6/9 = 16/33 Probability that a card matches one of other 3 = 116/33=17/33
Approach 2:
Total number of combinations=12C4 Combinations that have no pair = 12*10*8*6 Probability that no card matces other three = 12*10*8*6/12C4 which is greater than 1  How is this possible?
ANY HELP WITH CORRECTING APPROACH 2 ?



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Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.
Bill likes to play a game in which he shuffles the deck, 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?
8/33 62/165
17/33 103/165 25/33



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Re: Cards
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30 Mar 2009, 21:09
What is OA? I could get through upto half part of this problem but cant find the answer in the choices! Prob of finding all cards diff = P P = 12 * 10 * 8 * 6 * 4 / 12C4. So prob of finding atleast one pair = 1P. But I cant find the ans in the choices. Can anyone explain? ritula wrote: Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.
Bill likes to play a game in which he shuffles the deck, 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?
8/33 62/165
17/33 103/165 25/33



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Re: Cards
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31 Mar 2009, 23:31
this is "at least" probability problem => p(at least)= 1  p(event not occuring) = 1  12/12*10/11*8/10*6/9 = 1  16/33 = 17/33



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Re: Cards
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31 Mar 2009, 23:58
Perfect! OA is 17/33 alpha_plus_gamma wrote: this is "at least" probability problem => p(at least)= 1  p(event not occuring) = 1  12/12*10/11*8/10*6/9 = 1  16/33 = 17/33



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Re: Cards
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13 Apr 2009, 03:47
What would be the answer if you would ask "What is the probability that EXACTLY one pair of cards will have the same value"?
Label cards: 1 2 3 4
1 & 2 a pair: 12/12 * 1/11 * 10/10 * 8/9 = 8/99
2 & 3 a pair: 12/12 * 10/11 * 1/10 * 9/9 = 1/11
3 & 4 a pair: 12/12 * 10/11 * 8/9 * 1/8 = 10/99
(1 & 2 a pair AND 3 & 4 a pair): 12/12 * 1/11 * 10/10 * 1/9 = 1/99
Thus, chance of EXACTLY one pair:
8/99 + 1/11 + 10/99  1/99 = 26/99
Is that correct? Is there an easier way of doing this?



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Re: Cards
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13 Apr 2009, 06:07
1P(all different)= \(1 C(6,4)*2^4/C(12,4)= 116/33=17/33\)



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Re: Cards
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13 Apr 2009, 10:53
botirvoy wrote: 1P(all different)= \(1 C(6,4)*2^4/C(12,4)= 116/33=17/33\) Hi botrivoy, Can u explain in language what u did?



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Re: Cards
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13 Apr 2009, 20:57
bandit wrote: botirvoy wrote: 1P(all different)= \(1 C(6,4)*2^4/C(12,4)= 116/33=17/33\) Hi botrivoy, Can u explain in language what u did? 1  all diff i.e. All diff prob= 6/12*5/11*4/10*3/9*2^4 as it can any number from 1 to 6.



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Re: Cards
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13 Apr 2009, 21:00
i will give a try to explain
12C4 = Number of ways of taking four cards out of 12.
for selecting all different cards
6C4 = there are 6 pairs of cards. Select 4 pairs from six pairs
2C1 = each position call be filled in 2 ways.
total ways of selecting all different cards = 6c4 * 2C1
prob of all diffrent cards = 6c4 * 2C1 / 12C4
Prob of atleast one pair = 1  (6C4 *2c1)/12C4



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Re: cards
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27 Sep 2009, 22:19
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.
Bill likes to play a game in which he shuffles the deck, 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 8/33
B 62/165
C 17/33
D103/165
E 25/33
Soln: Chance that Bill finds at least one pair of cards with same value is = 1  None of the cards are same = 1  (12/12) * (10/11) * (8/10) * (6/9) = 1  16/33 = 17/33
Soln is C



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Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value. Bill likes to play a game in which he shuffles the deck, 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. 8/33 b. 62/165 c. 17/33 d. 103/165 e. 25/33



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ansC17/33.. find prob of none being a pair=12/12*10/11*8/10*6/9=16/33.. ans=116/33=17/33
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Re: cards
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13 Oct 2009, 12:59
1  (6C4*2C1*2C1*2C1*2C1)/12C4 =1  16/33 =17/33
C.



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Nice job chetan2u! That's how I did it as well.
But for those of you who more readily understand combinations.
12C4 = 495 is the total number of possible hands.
To find the number of hands that contain AT LEAST one pair, one need only find the Total number of hands LESS the number of hands with NO PAIRS.
[6C4][2C1][2C1][2C1][2C1] = 15*16 = 240 is the number of hands with NO PAIRS. where 6C4 is the number of ways to choose 4 cards of different ranks [2C1][2C1][2C1][2C1] = 16 is the number of ways to choose the suit for each of the 4 cards
Thus, the number of hands with at least one pair is: Total Hands  Hands with no pairs =12C4  [6C4][2C1][2C1][2C1][2C1] =495240 =255
So the chances of being dealt at least one pair is: = 255/495 = 17/33



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probability  Bill has a small deck of 12
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19 Nov 2009, 07:00
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.
Bill likes to play a game in which he shuffles the deck, 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?
8/33 62/165 17/33 103/165 25/33



Intern
Joined: 22 Sep 2009
Posts: 34

I used different logic
let the cards be like 1,1 2,2 3,3 4,4 5,5 6,6
total 12 cards
we are going to select 4 cards hence
first card can be selected from 12 cards, second from 11, third from 10, and fourth from 9
hence total no. of events = 12*11*10*9
=========================
lets find the condition where no pair is selected hence first one can be selected from 12 cards now the second one cannot be selected from the same pair hence for second the events are 10 for third 8 and for forth 6 Hence favorable events = 12*10*8*6
hence probability that no pair is selected = 12*10*8*6/12*11*10*9 = 16/33
Hence the required probability = 1  16/33 = 17/33



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This was simple to understand....







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