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705-805 Level|   Probability|      
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BrentGMATPrepNow
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 13 May 2019, 11:50
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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) \(\frac{25}{33}\)
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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

Cheers,
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 22 Dec 2019, 06:40
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he chance of getting AT LEAST one pair of cards with the same value out of 4 dealt cards should be computed using the 1-x 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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Bill has a set of 6 black cards and a set of 6 red cards. Each card 02 Apr 2021, 03:51
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VeritasKarishma

what is wrong with my solution and why ?

Total number of ways 4 0ut of 12 = 495

Case 1: only one pair the same
Choosing one red from 6 = 6 ways
Choosing one black(of same value as red) = 1 way
Choosing Any of 5 red cards = 5 ways
Choosing Any card of 3 black cards = 3 ways (i excluded Black 4 so as to avoid the second same pair)

So 6*1*5*3 = 90 Total number of arrangements 4!/2! = 12 (so 90*12)

Case 2: Two pairs are the same

Choosing any red of 6 = 6 ways
choosing 1 black (of same value as red) = 1 way
Choosing any red of 5 = 5 ways
Choosing 1 black (of same value as red) = 1 way

So 6*1*5*1 = 30 total # of arrangements = 4!/2!2! = 6 so 60*6 =360

maybe you can help :) BrentGMATPrepNow
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 03 Apr 2021, 07:59
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VeritasKarishma

what is wrong with my solution and why ?

Total number of ways 4 0ut of 12 = 495

Case 1: only one pair the same
Choosing one red from 6 = 6 ways
Choosing one black(of same value as red) = 1 way
Choosing Any of 5 red cards = 5 ways
Choosing Any card of 3 black cards = 3 ways (i excluded Black 4 so as to avoid the second same pair)

So 6*1*5*3 = 90 Total number of arrangements 4!/2! = 12 (so 90*12)

Case 2: Two pairs are the same

Choosing any red of 6 = 6 ways
choosing 1 black (of same value as red) = 1 way
Choosing any red of 5 = 5 ways
Choosing 1 black (of same value as red) = 1 way

So 6*1*5*1 = 30 total # of arrangements = 4!/2!2! = 6 so 60*6 =360

maybe you can help :) BrentGMATPrepNow
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The first part of your solution works:
Case 1: only one pair the same
Choosing one red from 6 = 6 ways
Choosing one black(of same value as red) = 1 way

At this point, we have are guaranteed pair. Great.


Choosing Any of 5 red cards = 5 ways
Choosing Any card of 3 black cards = 3 ways (i excluded Black 4 so as to avoid the second same pair)

Here, your solution assumes that, among the two remaining cards, one card must be red and one card must be black.
So, for example, you are excluding the possibility that the two remaining cards are both red, or the two remaining cards are both black.
That is, one possible outcome is red 1, black 1, red 4 and red 5, but your solution does not allow for this
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 05 Apr 2021, 03:46
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VeritasKarishma

what is wrong with my solution and why ?

Total number of ways 4 0ut of 12 = 495

Case 1: only one pair the same
Choosing one red from 6 = 6 ways
Choosing one black(of same value as red) = 1 way
Choosing Any of 5 red cards = 5 ways
Choosing Any card of 3 black cards = 3 ways (i excluded Black 4 so as to avoid the second same pair)

So 6*1*5*3 = 90 Total number of arrangements 4!/2! = 12 (so 90*12)

Case 2: Two pairs are the same

Choosing any red of 6 = 6 ways
choosing 1 black (of same value as red) = 1 way
Choosing any red of 5 = 5 ways
Choosing 1 black (of same value as red) = 1 way

So 6*1*5*1 = 30 total # of arrangements = 4!/2!2! = 6 so 60*6 =360

maybe you can help :) BrentGMATPrepNow
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dave13

- Arrangements are not required. It is a "selection" question. You are selecting 4 cards - whether they are
{R1, B1, R2, R3} or {R1, R2, B1, R3} - it doesn't matter.

- You should choose both red or both black in the non pair cards.

- You are double counting when you are selecting 2 pairs. You choose any red from the 6 in 6 ways and then corresponding black in1 way. Say you choose R1 and B1.
Next, you choose a Red from the 5 in 5 ways and then corresponding black in 1 way. Say you choose R3 and B3.
But there will be another case in which you choose R3 and B3 first and then R1 and B1 but essentially they give you the same selection.
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VeritasKarishma

what is wrong with my solution and why ?

Total number of ways 4 0ut of 12 = 495

Case 1: only one pair the same
Choosing one red from 6 = 6 ways
Choosing one black(of same value as red) = 1 way
Choosing Any of 5 red cards = 5 ways
Choosing Any card of 3 black cards = 3 ways (i excluded Black 4 so as to avoid the second same pair)

So 6*1*5*3 = 90 Total number of arrangements 4!/2! = 12 (so 90*12)

Case 2: Two pairs are the same

Choosing any red of 6 = 6 ways
choosing 1 black (of same value as red) = 1 way
Choosing any red of 5 = 5 ways
Choosing 1 black (of same value as red) = 1 way

So 6*1*5*1 = 30 total # of arrangements = 4!/2!2! = 6 so 60*6 =360

maybe you can help :) BrentGMATPrepNow

dave13

- Arrangements are not required. It is a "selection" question. You are selecting 4 cards - whether they are
{R1, B1, R2, R3} or {R1, R2, B1, R3} - it doesn't matter.

- You should choose both red or both black in the non pair cards.

- You are double counting when you are selecting 2 pairs. You choose any red from the 6 in 6 ways and then corresponding black in1 way. Say you choose R1 and B1.
Next, you choose a Red from the 5 in 5 ways and then corresponding black in 1 way. Say you choose R3 and B3.
But there will be another case in which you choose R3 and B3 first and then R1 and B1 but essentially they give you the same selection.
Show more

VeritasKarishma BrentGMATPrepNow

thanks, here is an updated solution but still confused what is wrong :?

Case 1: exactly one pair is the same

-11BB choosing one red and one black of the same value and the rest two are both black - 6*1*5*4 = 120

-11RR choosing one red and one black of the same value and the rest two are both red - 6*1*5*4 = 120

- 1R1B2R3B choosing one red and one black of the same value and the rest two are one black and one red of different values = 6*1*5*3 = 90 (excluded 4 to avoid the second pair to be the same)


Case 2: two pars are the same 6*1*5*1 = 30


120+120+90+30 = 360 :?
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VeritasKarishma
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VeritasKarishma

what is wrong with my solution and why ?

Total number of ways 4 0ut of 12 = 495

Case 1: only one pair the same
Choosing one red from 6 = 6 ways
Choosing one black(of same value as red) = 1 way
Choosing Any of 5 red cards = 5 ways
Choosing Any card of 3 black cards = 3 ways (i excluded Black 4 so as to avoid the second same pair)

So 6*1*5*3 = 90 Total number of arrangements 4!/2! = 12 (so 90*12)

Case 2: Two pairs are the same

Choosing any red of 6 = 6 ways
choosing 1 black (of same value as red) = 1 way
Choosing any red of 5 = 5 ways
Choosing 1 black (of same value as red) = 1 way

So 6*1*5*1 = 30 total # of arrangements = 4!/2!2! = 6 so 60*6 =360

maybe you can help :) BrentGMATPrepNow

dave13

- Arrangements are not required. It is a "selection" question. You are selecting 4 cards - whether they are
{R1, B1, R2, R3} or {R1, R2, B1, R3} - it doesn't matter.

- You should choose both red or both black in the non pair cards.

- You are double counting when you are selecting 2 pairs. You choose any red from the 6 in 6 ways and then corresponding black in1 way. Say you choose R1 and B1.
Next, you choose a Red from the 5 in 5 ways and then corresponding black in 1 way. Say you choose R3 and B3.
But there will be another case in which you choose R3 and B3 first and then R1 and B1 but essentially they give you the same selection.

VeritasKarishma BrentGMATPrepNow

thanks, here is an updated solution but still confused what is wrong :?

Case 1: exactly one pair is the same

-11BB choosing one red and one black of the same value and the rest two are both black - 6*1*5*4 = 120

-11RR choosing one red and one black of the same value and the rest two are both red - 6*1*5*4 = 120

- 1R1B2R3B choosing one red and one black of the same value and the rest two are one black and one red of different values = 6*1*5*3 = 90 (excluded 4 to avoid the second pair to be the same)


Case 2: two pars are the same 6*1*5*1 = 30


120+120+90+30 = 360 :?
Show more

You are again arranging them.
When you want to pick 2 red cards and you pick them as 5 * 4, you are rearranging them into first pick and second pick. Say you pick R3 and R4. In another selection, you pick R4 and R3. For you they are distinct but actually they are same.

So instead, you need to use 5C3 = 5*4/2 (effectively, we have un-arranged them)

Also, when you select 2 pairs in 6*5 ways, you are again arranging them into pair 1 and pair 2.

Easier method:

Select 1 pair out of the 6 in 6C1 ways = 6 ways (say you selected (R1, B1))

Of the leftover 10 cards, select any two in 10C2 = 45 ways.
But of these 45 ways, 5 are such that you have another pair (R2, B2), (R3, B3) ... (R6, B6)
So you have 40 ways of selecting exactly 1 pair.
Total = 6 * 40 = 240 ways

Selecting 2 pairs can be done in 6C2 = 6*5/2 = 15 ways.

Total = 255 ways
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 06 Apr 2021, 09:57
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VeritasKarishma thanks, is it possible that you use my method for solving question ? :)
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VeritasKarishma thanks, is it possible that you use my method for solving question ? :)
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That method involves multiple levels of calculations which makes it cumbersome at test time.

A pair and two Reds:
6 * 5C2 = 6 * 10 = 60

A pair and two Blacks:
6 * 5C2 = 6 * 10 = 60

A pair and a red and a black non pair:
6 * 5 * 4 = 120
5 ways to pick a red and 4 ways to pick a non matching black.

(All I have done in my method above is combined these three steps together to get 240)

Two pairs:
6C2 = 15

Total = 255
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 07 Apr 2021, 04:10
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dave13
VeritasKarishma thanks, is it possible that you use my method for solving question ? :)

That method involves multiple levels of calculations which makes it cumbersome at test time.

A pair and two Reds:
6 * 5C2 = 6 * 10 = 60

A pair and two Blacks:
6 * 5C2 = 6 * 10 = 60

A pair and a red and a black non pair:
6 * 5 * 4 = 120
5 ways to pick a red and 4 ways to pick a non matching black.

(All I have done in my method above is combined these three steps together to get 240)

Two pairs:
6C2 = 15

Total = 255
Show more


many thanks VeritasKarishma , i almost understand it :) just have a few questions to make it clear for me :grin:

as per you method (see highlighted portion below):

"Select 1 pair out of the 6 in 6C1 ways = 6 ways (say you selected (R1, B1))
Of the leftover 10 cards, select any two in 10C2 = 45 ways.
But of these 45 ways, 5 are such that you have another pair (R2, B2), (R3, B3) ... (R6, B6)
So you have 40 ways of selecting exactly 1 pair.
Total = 6 * 40 = 240 ways"


you say that out of 10 cards, you still can have pairs so to avoid another same pair you exclude 5 cards (R2, B2), (R3, B3) ... (R6, B6)

and you are left with 40 ways ... this is how i understand it ...though a bit confusing

but when i wrote this:

" - 1R1B2R3B choosing one red and one black of the same value and the rest two are one black and one red of different values = 6*1*5*3 = 90 (excluded 4 to avoid the second pair to be the same) "

here i excluded 4 to avoid the second pair to be the same.... but you say its not correct. is it not correct at all, or just in this case? in which would it be correct ? do i arrange here too ? i m curious to know :)


Btw, here "A pair and a red and a black non pair: 6 * 5 * 4 = 120" if third card is red which is fifth, in the rest of 4 cards there is a card of identical value among black ones left, so my question is why you didnt exclude that black card .... i mean i would do it 6*5*3 .... thats kinda confuses me
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 08 Apr 2021, 00:32
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dave13
VeritasKarishma thanks, is it possible that you use my method for solving question ? :)

That method involves multiple levels of calculations which makes it cumbersome at test time.

A pair and two Reds:
6 * 5C2 = 6 * 10 = 60

A pair and two Blacks:
6 * 5C2 = 6 * 10 = 60

A pair and a red and a black non pair:
6 * 5 * 4 = 120
5 ways to pick a red and 4 ways to pick a non matching black.

(All I have done in my method above is combined these three steps together to get 240)

Two pairs:
6C2 = 15

Total = 255


many thanks VeritasKarishma , i almost understand it :) just have a few questions to make it clear for me :grin:

as per you method (see highlighted portion below):

"Select 1 pair out of the 6 in 6C1 ways = 6 ways (say you selected (R1, B1))
Of the leftover 10 cards, select any two in 10C2 = 45 ways.
But of these 45 ways, 5 are such that you have another pair (R2, B2), (R3, B3) ... (R6, B6)
So you have 40 ways of selecting exactly 1 pair.
Total = 6 * 40 = 240 ways"


you say that out of 10 cards, you still can have pairs so to avoid another same pair you exclude 5 cards (R2, B2), (R3, B3) ... (R6, B6)

and you are left with 40 ways ... this is how i understand it ...though a bit confusing

but when i wrote this:

" - 1R1B2R3B choosing one red and one black of the same value and the rest two are one black and one red of different values = 6*1*5*3 = 90 (excluded 4 to avoid the second pair to be the same) "

here i excluded 4 to avoid the second pair to be the same.... but you say its not correct. is it not correct at all, or just in this case? in which would it be correct ? do i arrange here too ? i m curious to know :)


Btw, here "A pair and a red and a black non pair: 6 * 5 * 4 = 120" if third card is red which is fifth, in the rest of 4 cards there is a card of identical value among black ones left, so my question is why you didnt exclude that black card .... i mean i would do it 6*5*3 .... thats kinda confuses me
Show more


Yes, out of 10 when you pick 2 cards, they can be anything (R2, B3), (R4, R5), (B2, B6)... etc. You get 45 such cases. But 5 of these cases will be (R2, B2), (R3, B3), (R4, B4), (R5, B5), (R6, B6)
which give us another pair. So you need to remove them.

Quote:

here i excluded 4 to avoid the second pair to be the same.... but you say its not correct. is it not correct at all, or just in this case? in which would it be correct ? do i arrange here too ? i m curious to know :)
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I don't understand what you mean by - "I excluded 4".

If you want 1 pair and 1 red, 1 black of different values, you will select these as:

6C1 (6 ways to select a pair. Say you got R1, B1)

Now there are 5 ways to select R (R2, R3, R4, R5, R6). Say you got R2.
Now there are 4 ways to select B (B3, B4, B5, B6) to ensure that you get different values. You have already removed B2 to ensure that you do not get a pair. You are still left with 4 different values of B.

So it will be 6 * 5 * 4
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VeritasKarishma thanks, i now tried to approach this problem with probability method (but not using 1-all non desirable outcomes) just regular ...

so i did this


case 1: one pair is the same

12/12 * 1/11 * 10/10 * 9/9 = 1/11 *2 =2/11 (since it can be 11RR or 11BB i multiplied by two)

case 2: two pairs are same

12/12 * 1/11* 10/10 *9/9 = 1/99

1/99+ 2/11 = 19/99

whats wrong with my approach ?
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VeritasKarishma thanks, i now tried to approach this problem with probability method (but not using 1-all non desirable outcomes) just regular ...

so i did this


case 1: one pair is the same

12/12 * 1/11 * 10/10 * 9/9 = 1/11 *2 =2/11 (since it can be 11RR or 11BB i multiplied by two)

case 2: two pairs are same

12/12 * 1/11* 10/10 *9/9 = 1/99

1/99+ 2/11 = 19/99

whats wrong with my approach ?
Show more

You have not arranged them. But arranging these probabilities is a bit tricky since you need to know the cases that you have covered in your probability and those that you haven't.

For example: The case of 2 pairs

1 * (1/11) * 1 * (1/9)

Pick anything * Pick its pair * Pick anything else * Pick its pair

Say the two pairs you pick are (3, 3, 4, 4). So you have already taken (4, 4, 3, 3) into account too because when you pick anything first, you could have picked 4 first too. But you haven't taken (3, 4, 3, 4) and (3, 4, 4, 3) into account.
So the entire thing gets multiples by 3 to make up for these cases.

Hence, the better approach is combinatorics or that in which arrangement is irrelevant (e.g. when you pick no pair - pick anything and follow it up by picking anything other than what you have already picked in each turn)
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dave13
VeritasKarishma thanks, i now tried to approach this problem with probability method (but not using 1-all non desirable outcomes) just regular ...

so i did this


case 1: one pair is the same

12/12 * 1/11 * 10/10 * 9/9 = 1/11 *2 =2/11 (since it can be 11RR or 11BB i multiplied by two)

case 2: two pairs are same

12/12 * 1/11* 10/10 *9/9 = 1/99

1/99+ 2/11 = 19/99

whats wrong with my approach ?

You have not arranged them. But arranging these probabilities is a bit tricky since you need to know the cases that you have covered in your probability and those that you haven't.

For example: The case of 2 pairs

1 * (1/11) * 1 * (1/9)

Pick anything * Pick its pair * Pick anything else * Pick its pair

Say the two pairs you pick are (3, 3, 4, 4). So you have already taken (4, 4, 3, 3) into account too because when you pick anything first, you could have picked 4 first too. But you haven't taken (3, 4, 3, 4) and (3, 4, 4, 3) into account.
So the entire thing gets multiples by 3 to make up for these cases.

Hence, the better approach is combinatorics or that in which arrangement is irrelevant (e.g. when you pick no pair - pick anything and follow it up by picking anything other than what you have already picked in each turn)
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VeritasKarishma as per the same logic as case of two pairs the same ... i did this

for one pair the same = 12/12 * 1/11 *10/10 * 8/9 = 8/99

since there can be arrangements such as 3345 , 4533, 3453, 3435 and 4353 multiply 8/99 by 5

so i get 40/99

1/33+40/99 = 43/99

now whats wrong ? :?

update: i googled this question and for one pair the same found this solution (12/12)(1/11)(10/10)(8/9) * (4!/2!2!)


why here we are divide 4! by 2! 2! ? if for example i have this arrangement 1134 then total number of arrangements is 4!/2!

curious to know :)
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 22 Apr 2021, 03:40
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hello :) GMATGuruNY IanStewart AndrewN

"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. "


in the solution (not by me) above

the case when one pair is the same,

there are there scenarios

First Scenario: 1134

Second Scenario: 1314

Third Scenario: 1341


i understand that 1314 is the same as 1413, similarly 1341 is the same as 1431


But there are ALSO scenarios such as 3411 and 3141 these are two distinct scenarios that are not counted via probability method above, Why?

to elaborate further i ask myself question :)

but in the solution (not by me) below for case 1

(12/12)(1/11)(10/10)(8/9) * (4!/2!2!)

the result is multiplied by (4!/2!2!)

from here i can deduce that since among four letters (1134) two are the same one must divide by 2! Also since 1314 is the same as 1413 one should divide by another 2!

BUT (4!/2!2!) yields 6 total arrangements

Question 1: why when using probability we use only 3 scenarios whereas there are total SIX scenarios/arrangements ?

Question 2: (4!/2!2!) yields 6 total arrangements , What are these 6 total arrangements ? i can only see five arrangements: 1134, 1314, 1341, 3411, 3141

thank you! :grin: :)
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 22 Apr 2021, 04:25
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VeritasKarishma thanks, i now tried to approach this problem with probability method (but not using 1-all non desirable outcomes) just regular ...

so i did this


case 1: one pair is the same

12/12 * 1/11 * 10/10 * 9/9 = 1/11 *2 =2/11 (since it can be 11RR or 11BB i multiplied by two)

case 2: two pairs are same

12/12 * 1/11* 10/10 *9/9 = 1/99

1/99+ 2/11 = 19/99

whats wrong with my approach ?

You have not arranged them. But arranging these probabilities is a bit tricky since you need to know the cases that you have covered in your probability and those that you haven't.

For example: The case of 2 pairs

1 * (1/11) * 1 * (1/9)

Pick anything * Pick its pair * Pick anything else * Pick its pair

Say the two pairs you pick are (3, 3, 4, 4). So you have already taken (4, 4, 3, 3) into account too because when you pick anything first, you could have picked 4 first too. But you haven't taken (3, 4, 3, 4) and (3, 4, 4, 3) into account.
So the entire thing gets multiples by 3 to make up for these cases.

Hence, the better approach is combinatorics or that in which arrangement is irrelevant (e.g. when you pick no pair - pick anything and follow it up by picking anything other than what you have already picked in each turn)


VeritasKarishma as per the same logic as case of two pairs the same ... i did this

for one pair the same = 12/12 * 1/11 *10/10 * 8/9 = 8/99

since there can be arrangements such as 3345 , 4533, 3453, 3435 and 4353 multiply 8/99 by 5

so i get 40/99

1/33+40/99 = 43/99

now whats wrong ? :?

update: i googled this question and for one pair the same found this solution (12/12)(1/11)(10/10)(8/9) * (4!/2!2!)


why here we are divide 4! by 2! 2! ? if for example i have this arrangement 1134 then total number of arrangements is 4!/2!

curious to know :)
Show more

Here are all the cases:

No Pair:

1 * (10/11) * (8/10) * (6/9) = 48/99
Pick any * Pick any except the one you picked previously * Pick any except two already picked * Pick any except the three already picked

This will give you all arrangements such 1234, 2134, 2341 etc.


Only one pair:

1 * (1/11) * 1 * (8/9)
Pick any * Pick the same again * Pick any * Pick any other

This will give you 1123, 1132, 2238, 2283.. etc
But, for 1123 we will not get 1213, 1231, 2113, 2131, 2311. So you need to multiply it by 6 (there are total 6 cases attached to each)

1 * (1/11) * 1 * (8/9) * 4!/2!*2! = 48/99
Why do we multiply by 4!/2!*2! (which is same as 6 we saw above)? Arrange all 4 digits in 4! ways. But two digits are same so divide by 2!. Also, distinct two digits have already been arranged (1123 and 1132 are different cases that we get because our third pick is also "pick any". So divide by 2! again.


Two pairs:

1 * (1/11) * 1 * (1/9)
Pick any * Pick its pair * Pick any * Pick its pair
This gives us 1122, 2211, 1155... etc

1 * (1/11) * 1 * (1/9) * 3 = 3/99
We multiply by 3 to account for all 3: 1122, 1212, 1221
Or we can say that we multiply by 4!/2!*2!*2! because 2 pairs are same also arrangement of smaller/greater pairs is already accounted for i.e. we count 1122 and 2211 as separate cases.
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 22 Apr 2021, 06:22
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dave13

"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.

in the solution (not by me) above

But there are ALSO scenarios such as 3411 and 3141 these are two distinct scenarios that are not counted via probability method above, Why?
Show more

I'm not sure where that solution is from, but it's not explaining its cases correctly. The last case, for example, calculates the probability that it's the fourth card that makes the pair (with the first, second, or third card). It's not the probability that the "1st and fourth card are same". So the last case is considering all of the possibilities ABCA, BACA and BCAA. Similarly the second case covers two possibilities, when the first and third cards are pairs, and when the second and third cards are pairs.

dave13

Question 1: why when using probability we use only 3 scenarios whereas there are total SIX scenarios/arrangements ?
Show more

I don't think the solution you quote is the best one to follow -- there's no reason to even consider three scenarios here. Each of the six scenarios is equally likely, so you just need to consider one of them, then multiply by 6. The solution you quote collapses some of the scenarios into a single case (it makes one case of the three situations ABCA, BACA and BCAA, for example) which is why it has three instead of six cases. But solving this problem directly is longer than solving the opposite problem -- getting no pairs -- and subtracting from 1.

dave13

Question 2: (4!/2!2!) yields 6 total arrangements , What are these 6 total arrangements ? i can only see five arrangements: 1134, 1314, 1341, 3411, 3141
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If you want to list arrangements in situations like this, it's best to proceed systematically, by changing only one thing at a time. I'd first list all of the arrangements with 'A' or '1' first position, then all of the additional ones where 'A' or '1' is in the second position, and so on. Here the six arrangements are:

AABC
ABAC
ABCA
BAAC
BACA
BCAA

In your notation, you missed 3114.
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 23 Apr 2021, 04:25
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hello :) GMATGuruNY IanStewart AndrewN
thank you! :grin: :)
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Hello, dave13. I see that both VeritasKarishma and IanStewart have answered your questions above. For my part, I solved the question using a hybrid method of logic and combinations. I knew that I had to get rid of the cases in which no pairs were selected, so I worked that out in the following manner:

The first card selected has 12 possibilities.

The second card selected has 10 possibilities, since it cannot match the first card selected.

The third card selected has 8 possibilities, since it cannot match either of the first two.

The fourth card selected has 6 possibilities, since it cannot match any of the first three.

Thus, the total number of ways to draw these cards would be

\(12*10*8*6\)

But I did not work that product out. Since the first card selected could just as easily have been the second, third, or fourth selected, we need to divide out the repetitions for each card selected. Thus,

\(\frac{12*10*8*6}{4*3*2*1}=240\)

There are 240 ways to draw these cards. We just need to know how many ways there are to draw 4 cards from 12. A quick 12C4 will get us that answer:

\(\frac{12!}{4!8!}=495\)

Finally, we need to figure out our desired cases and divide by the total we just derived.

\(\frac{495-240}{495}\)

\(\frac{255}{495}\)

Here, I divided the numerator and denominator by 10 to get more workable numbers, 25.5 and 49.5, and then multiplied by 2 to make them integers again. This gave me

\(\frac{51}{99}\)

I did not bother reducing it again. I just looked at the answers and chose (C), since 51 and 99 are both divisible by 3.

I am not advocating the above method over more efficient methods outlined earlier in the thread, but it can work in a pinch. (Sometimes I start working a problem one way before I realize I could have done it more efficiently in another.)

I hope this information is of some interest to you. Thank you for thinking to ask me about the question.

- Andrew
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 10 Sep 2021, 01:48
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EASIEST SOLUTION

1. Ways of selecting 4 cards out of 12 cards : 12C4

2. Ways of selecting 4 cards so that no pair is selected : 6C4 x 2C1 x 2C1 x 2C1 x 2C1 (Select 4 different pairs out of 6 pairs then select 1 card from each pair)

3. Probability of selecting 4 cards out of 12 so that no pair is selected : 6C4 x 2C1 x 2C1 x 2C1 x 2C1 / 12C4 = 16/33

4. Probability of selecting 4 cards out of 12 so that at least one pair is selected = 1 - 16/33 = 17/33
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Bill has a set of 6 black cards and a set of 6 red cards. Each card 07 Aug 2022, 08:03
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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) \(\frac{25}{33}\)
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The chance of getting AT LEAST one pair of cards with the same value out of 4 dealt cards should be computed using the 1-x 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.

Alternative

Total no of ways to choose 4 card with zero duplicate numbers = 6C4
Total no of ways to choose 4 cards = 12C4
Required ways = 12C4-6C4x16
Prob = (12C4-6C4 x 16)/12C4
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