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Bill has a small deck of 12 playing cards made up of only 2

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PS: 2 suits of cards  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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 = 1-16/33=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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Cards  [#permalink]

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New post 29 Mar 2009, 09:23
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  [#permalink]

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New post 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 = 1-P. 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  [#permalink]

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New post 31 Mar 2009, 23:31
1
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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 13 Apr 2009, 06:07
1-P(all different)= \(1- C(6,4)*2^4/C(12,4)= 1-16/33=17/33\)
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Re: Cards  [#permalink]

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New post 13 Apr 2009, 10:53
botirvoy wrote:
1-P(all different)= \(1- C(6,4)*2^4/C(12,4)= 1-16/33=17/33\)

Hi botrivoy, Can u explain in language what u did?
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Re: Cards  [#permalink]

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New post 13 Apr 2009, 20:57
bandit wrote:
botirvoy wrote:
1-P(all different)= \(1- C(6,4)*2^4/C(12,4)= 1-16/33=17/33\)

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  [#permalink]

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New post 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  [#permalink]

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New post 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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P&C  [#permalink]

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New post 06 Oct 2009, 07:14
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: P&C  [#permalink]

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New post 06 Oct 2009, 10:06
ansC-17/33..
find prob of none being a pair=12/12*10/11*8/10*6/9=16/33..
ans=1-16/33=17/33
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Re: cards  [#permalink]

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New post 13 Oct 2009, 12:59
1 - (6C4*2C1*2C1*2C1*2C1)/12C4
=1 - 16/33
=17/33

C.
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Re: P&C  [#permalink]

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New post 21 Oct 2009, 14:43
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]
=495-240
=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  [#permalink]

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New post 19 Nov 2009, 07:00
1
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: P&C  [#permalink]

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New post 25 Dec 2009, 06:10
1
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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Re: P&C  [#permalink]

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New post 25 Dec 2009, 07:25
This was simple to understand....
Re: P&C &nbs [#permalink] 25 Dec 2009, 07:25

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