A special deck of cards has 3 each of 8 different cards. The deck has

Question

A special deck of cards has 3 each of 8 different cards. The deck has been shuffled so that the cards are randomly distributed. If 3 nonmatching cards are dealt, what is the probability that dealing 2 more cards will result in at least one matching pair of the same cards with the original 3 dealt cards or 3 of the same card?

A) 1/8

B) 1/3

C) 1/5

D) 3/8

E) 1/2

** Saw a long explanation for this problem by adding up the probabilities. Looking for a solution that uses the complement (1-P). Thanks!

A) 1/8

B) 1/3

C) 1/5

D) 3/8

E) 1/2

chetan2u · Accepted Answer

Hi..
The shorter way would be..
We are left with 3*8-3=21 cards.
Ways of picking the other two different from original 3 is 15*14, as we remove the remaining 2 of these original 3, and thus 21-6= 15 are left.
Total ways = ways of picking 2 out of 21=21*20..

Hence probability of not similar to original 3=  \frac{15*14}{21*20} =1/2..
So probability we are looking for = 1-P= 1-1/2=1/2

E

TarPhi · Answer

Easiest way to do this

AAA
BBB
CCC
DDD
EEE
FFF
GGG
HHH

Let's say, ABC are chosen first

Ways of picking at least 1 matching in the next 2 draws are

1 - none = 1 - 15c2/21c2
= 1 - 15*14/21*20
=1/2

AdityaHongunti · Answer

A special deck of cards has 3 each of 8 different cards. The deck has been shuffled so that the cards are randomly distributed. If 3 nonmatching cards are dealt, what is the probability that dealing 2 more cards will result in at least one matching pair of the same cards with the original 3 dealt cards or 3 of the same card?

A. 1/8
B. 1/3
C. 1/5
D. 3/8
E. 1/2

After the first 3 cards are dealt, they are all different.

Since the deck has 3 copies of each card, for each of those 3 original cards, there are 2 matching copies left.

So remaining deck:

Matching cards to the original 3: 3 * 2 = 6
Other cards: 5 types * 3 = 15
Total remaining cards: 21

We want at least one of the next 2 cards to match one of the original 3 cards.

Use the complement:

No match means both new cards come from the 15 nonmatching cards.

Probability of no match:

15/21 * 14/20 = 1/2

Therefore, probability of at least one match:

1 - 1/2 = 1/2

Answer: E.

StuffMob

After selecting the first 4th card with probability of 15/21 - doesnt the probability of selecting 5th card which is a different reduce to 12/20? If we take 14/20 then we have 2 cards in this that is already selected in the first 4. Please help me understand this!

The key point is that, in the complement case, the 4th and 5th cards do not need to be different from each other. They only need to be different from the original 3 cards:

...at least one matching pair of the same cards with the original 3 dealt cards or 3 of the same card?

So, after the 4th card is chosen from the 15 nonmatching cards, there are 14 nonmatching cards left out of 20. This includes the 2 remaining copies of the same type as the 4th card, and those are still allowed because they do not match any of the original 3 cards.

Using 12/20 would incorrectly require the 5th card to be different from the 4th card as well. That is not needed here.

) ...we selected 3 diferent color card. each clolor card has 3 card ...so total number of same color cards which have color similar to earlier three cards is = 9 (YYY/BBB/RRR ...from which we selected Y/B/R ...we dont want this entire group) so 21-9 = 15 cards (colors other than theearlier) we select 2 cardsd from these 15 : 15C2 tyerefore answeer is : E/S = 15C2 /21C2 =1/2

Kinshook · Answer

A special deck of cards has 3 each of 8 different cards. The deck has been shuffled so that the cards are randomly distributed.

If 3 nonmatching cards are dealt, what is the probability that dealing 2 more cards will result in at least one matching pair of the same cards with the original 3 dealt cards or 3 of the same card?

Cards dealt : 3 out of 8 different cards
Cards remaining : 5 out of 8 different cards and 2 each of 8 different cards; 16 + 5 = 21 cards left

Total ways to deal 2 out of remaining 21 cards = 21C2 = 210

Unfavorable ways = 15C2 = 105

Favorable ways = 210 - 105 = 105

The probability that dealing 2 more cards will result in at least one matching pair of the same cards with the original 3 dealt cards or 3 of the same card = 105/210 = 1/2

IMO E

StuffMob · Answer

After selecting the first 4th card with probability of 15/21 - doesnt the probability of selecting 5th card which is a different reduce to 12/20? If we take 14/20 then we have 2 cards in this that is already selected in the first 4. Please help me understand this!

Bunuel · Answer

A special deck of cards has 3 each of 8 different cards. The deck has been shuffled so that the cards are randomly distributed. If 3 nonmatching cards are dealt, what is the probability that dealing 2 more cards will result in at least one matching pair of the same cards with the original 3 dealt cards or 3 of the same card?

A) 1/8

B) 1/3

C) 1/5

D) 3/8

E) 1/2

** Saw a long explanation for this problem by adding up the probabilities. Looking for a solution that uses the complement (1-P). Thanks!

A. 1/8
B. 1/3
C. 1/5
D. 3/8
E. 1/2

After the first 3 cards are dealt, they are all different.

Since the deck has  3 copies of each card , for each of those 3 original cards, there are  2 matching copies left .

So remaining deck:

Matching cards to the original 3: 3 * 2 = 6  
 Other cards: 5 types * 3 = 15  
 Total remaining cards: 21

We want at least one of the next 2 cards to match one of the original 3 cards.

Use the complement:

No match means both new cards come from the 15 nonmatching cards.

Probability of no match:

15/21 * 14/20 = 1/2

Therefore, probability of at least one match:

1 - 1/2 = 1/2

Answer: E.

The key point is that, in the complement case, the 4th and 5th cards do  not  need to be different from each other. They only need to be different from the original 3 cards:

...at least one matching pair of the same cards  with the original 3 dealt cards  or 3 of the same card?

So, after the 4th card is chosen from the 15 nonmatching cards, there are 14 nonmatching cards left out of 20. This includes the 2 remaining copies of the same type as the 4th card, and those are still allowed because they do not match any of the original 3 cards.

Using 12/20 would incorrectly require the 5th card to be different from the 4th card as well. That is not needed here.

SVaidyaraman · Answer

it's a complement problem. "At least one match" is the classic signal: computing the many ways to succeed is a nightmare; computing the single way to fail is easy. Complement is the whole engine here.

Key idea: P(at least one match) = 1 − P(no match at all). The engine is that the "no match" event is a single clean counting task, while "at least one" fans out into cases.

Set up the deck state.
Total deck: 8 different cards × 3 copies = 24 cards.
3 non-matching cards have been dealt — three different card types, one copy each.

Remaining in deck: 24 − 3 = 21 cards.

Of those 21, split them by what a match would mean. The 3 dealt types have 2 copies each still in the deck:
- Matching cards (would pair with a dealt card): 3 types × 2 copies = 6
- Non-matching cards (the other 5 types, untouched): 5 × 3 = 15

Check: 6 + 15 = 21

Compute the complement — draw 2 more, neither is a match.
"No match" means both new cards come from the 15 non-matching cards.

$$P(	ext{no match}) = \frac{\binom{15}{2}}{\binom{21}{2}} = \frac{105}{210} = \frac{1}{2}$$

Flip it:

$$P(	ext{at least one match}) = 1 - \frac{1}{2} = \frac{1}{2}$$

Answer: (E) 1/2

Note on the "3 of the same card" clause: it's a red herring — drawing 2 copies of a dealt type gives you 3 of a kind, but that's already inside the "at least one match" event. The complement handles every success case at once without enumerating them, which is exactly why complement is the right tool.

The only move that mattered: turn "at least one" into "one minus none." Once the deck was split into 6 matching / 15 non-matching, the answer was two combinations away.

alamin8 · Answer

I like this way because the explanation is very easy.

Kinshook · Answer

A special deck of cards has 3 each of 8 different cards. The deck has been shuffled so that the cards are randomly distributed.

If 3 nonmatching cards are dealt, what is the probability that dealing 2 more cards will result in at least one matching pair of the same cards with the original 3 dealt cards or 3 of the same card?

Let the cards be 3 each of A, B, C, D, E, F, G & H cards, total 3*8=24 cards

Let the 3 dealt cards be A, B, C. 
Remaining matching cards = 2*3 = 6
Remaining mon-matching cards = 3*5 = 15
Remaining total cards = 24-3=21

Total ways (Select 2 cards out of remaining 21 cards) = 21C2 = 210

Unfavorable ways (2 different type of non-matching cards are dealt) = 15C2 = 15*7 = 105

Favorable ways = 210 - 105 = 105

The probability that dealing 2 more cards will result in at least one matching pair of the same cards with the original 3 dealt cards or 3 of the same card = 105/210 = 1/2

IMO E

culpavoluptatum · Answer

As per me, I think the easiest way to solve and understand this would be. Because the language is a little tricky.

You have to first understand that there are 8 Different categories of card and each category have 3 cards. This states  Total Cards is 24.

Total Cards= 8 x 3 = 24
Already picked card = 3 (each from 3 different categories)

Remaining Cards = 21
Numbers of cards that can match the picked card's categories =  2  (as 2 cards are left from each 3 category) x  3  =  6

So we have 15 fresh card from 5 unpicked categories.
=>The Probability of picking  both  matching cards would be: P = 6/21 x 5/20 = 1/14

Probability of  only  1 fresh card = 15/21
=> The probability of picking  only 1 Matching card.  P = 6/21 x 15/20 = 3/14

=>  3/14 x 2 (2 different cases because we can pick the matching card first or the non matching card 1st) 
=>  3/7
 
 Total P = 1/14 + 3/7 = 1 +6/14 = 7/14

Answer E. 1/2

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