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

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= 8/99----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.

Kudos if helped.

Hi Bunnel... need your intervention to explain on the logic of P (None) please

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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??

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

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.

chetan2u Bunuel VeritasKarishma
Gladiator59, generis
I 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?

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?

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 non-outcome, 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.

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.

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.

Lemme quickly touch on why /4! is needed for the # ways to get all unique numbers:
If order matters, # of ways to pick the cards: 12 * 10 * 8 * 6
This includes 1, 2, 3, 4 and 4, 3, 2, 1 and 1, 3, 2, 4... --> 4P4 = 4! ways
Therefore, we need to divide 12 * 10 * 8 * 6 by 4! to remove the duplicate sets.

Hope it helps.

Shall we not select one of the cards from the second pair, giving following result:

6C1*5C1*2C1*8C1= 6*5*2*8 rather than 6*5*8........

Hi,

Your approach is nice but I don't get the "picking all different values" part.

For the 4 pairs,

1st can be 12 out of 12. Fair enough.

But, 2nd should be 9 out of 11.

11 left and 9 can be chosen as one can't be same as the 2nd one and one can't be same as the 1st one. So 2 out and 9 left.

3rd should be 7 out of 10.

10 left and 7 can be chosen as one can't be same as the 3rd one and one can't be same as 2nd and 1st. So 3 out and 7 left.

4th should be 5 out of 9.

Same reasoning.


Please go through my explanation and let me know whether I'm correct or missing anything.


Thank you.

Posted from my mobile device

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

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

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