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

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?

(A) \(\frac{8}{33}\)

(B) \(\frac{62}{65}\)

(C) \(\frac{17}{33}\)

(D) \(\frac{103}{165}\)

(E) \(\frac{25}{33}\)

(A)  \frac{8}{33}

(B)  \frac{62}{65}

(C)  \frac{17}{33}

(D)  \frac{103}{165}

(E)  \frac{25}{33}

Zarrolou · Accepted Answer

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

Zarrolou · Answer

Your denominator is correct  12C4=495 . The problem is the numerator: the probability of picking 4 different values among 6 possible values:  6C4=15 . But because each of the four cards can be black or red, each of the four slots can have 2 colors , so we have to multiply 15 by  2*2*2*2=16  (each of the four cards can be B or R).

Hope I've explained myself well.

So now  P=\frac{495-15*16}{495}=\frac{17}{33} .

gmatter0913 · Answer

I solved this question as follows and I know I am wrong. The problem is: I don't know why am I wrong?

First card can be picked in 12C1 ways
Second card can be picked in 10C1 ways
Third card can be picked in 8C1 ways
Fourth card can be picked in 6C1 ways
All possibilities to pick 4 cards out of 12 = 12C4

Probability to have atleast one pair = 1 - prob to have no pairs = 1 - 12c1*10c1*8c1*6c1/12c4

It looks like I have done a permutation on the numerator. Could you help me with the error in the above?

gmatter0913 · Answer

Thanks Zarrolou I understood the solution, but I would need a little more help on some concepts. PERMUTATIONS OF 'n' items having some identical elements: Number of permutations of 'n' items out of which p1 are alike, p2 are alike, p3 are alike and the rest are different is n!/p1!p2!p3! Suppose, we have a set of 8 alphabets - {A, A, B, B, C, C, D, D} The number of ways we can arrange the 8 alphabets is 8!/2!*2!*2!*2! (where we divide by 2! because we cannot distinguish between AA and its combination reversed, AA

) So, just the way we have the above formula, what is the formula to get all the possible combinations of 3 elements in the above case. Will it be? - 4c3*(2)*(2)*(2)/8c4 where 4c3 is choosing combination of 3 elements among 4 unique elements. and (2)*(2)*(2) is for each of the 3 elements in the combination there are two options (two identical elements) Is there a general formula used to determine combinations in cases such as above? The other doubt that I have is in the above set of alphabets {A, A, B, B, C, C, D, D} how do I get the number of permutations of 3 elements? Is it 3!/{what?} because I don't know how many elements are repeated among those 3.

Zarrolou · Answer

I) I do not know a general formula for the above case (and probably there is none).

II)You cannot use n!/p1!p2!p3! for your second problem. You have to try to work with the other formulas you know.
This is how I personally would do it:

First I would consider the case in which all three elements are different:  - -... and since here we are taking about permutations, the total number of cases is  4*3*2=24  (you can pick any letter for the first slot, three (four-the previous chosen) for the second slot and so on).
Then I would consider the case in which a pair of letters is repeated:  - -...
You can pick the pair of identical letters in 4 ways, the remaining letter is 3 ways. But each combination can be arranged in 3 ways: example
A,A,B-A,B,A-B,A,A.
So the total number is  4*3*3=36 . Overall the possible permutations are 60.

ShashankDave · Answer

\frac{17}{33} is the answer.
You cant also do it using combinations.
select any pair(i.e. 2 cards) and the other two + select any two pairs and divide the total by 12C4.
this thought process leads to :-
 
 (6C1*8*5 + 6C2)/12C4

for the first term in the Numerator - first select any one pair(6C1 ways), the we are left with 5 of black and 5 of red. Now select any one from one of these groups (5 ways). Lastly we are left with 9 cards, but in this group, one of them is the one with the same value in another color of the previously selected. So we are left with 8 to select one from(8 ways). Multiplying together -> 6C1*8*5.

binary01 · Answer

First find the probability of no pair P(no pair) : There are 6 pair. Choose any 4 pair and then select any one from each pair. 6C4*2C1*2C1*2C1*2C1 = 240
Total possibilities = 12C4 = 495
P (no pairs) = 240/495
P(at least one pair) = 1 - p(no pair) = 1 - 240/495 = 255/495 = 17/33. Option C

mohit2491 · Answer

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?

(A) \(\frac{8}{33}\)

(B) \(\frac{62}{65}\)

(C) \(\frac{17}{33}\)

(D) \(\frac{103}{165}\)

(E) \(\frac{25}{33}\)

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

No. of ways of selecting two pairs (AABB)=  6C2

No. of arrangements= 6C2 *  6

Total number of possible favorable arrangements=  (6C1 *10C2 *12) + (6C2 * 6)

Total number of arrangements=  12C4

Hence probability = {  (6C1 *10C2 *12) + (6C2 * 6)  } /  12C4

= 37/132

Can someone please explain me the flaw in this.

Thanks and Regards

KarishmaB · Answer

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.

BrentGMATPrepNow · Answer

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,
Brent

KaranB1 · Answer

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 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.

dave13 · Answer

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

BrentGMATPrepNow · Answer

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

KarishmaB · Answer

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.

dave13 · Answer

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

KarishmaB · Answer

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

dave13 · Answer

VeritasKarishma thanks, is it possible that you use my method for solving question ?

KarishmaB · Answer

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

dave13 · Answer

many thanks VeritasKarishma , i almost understand it

just have a few questions to make it clear for me

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

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

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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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