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

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
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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

Please kudos

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?

First lets understand the question :
- A deck of cards has 3 cards of EACh of 8 DIFFERENT cards >> assume 8 different colors (you can color code the cards ,ie, 3 cards of yellow color...3 cards of green color...)
so total number of cards is 24 (3*8)
-action already done: 3 cards of NON MATCHING cards are dealth (this mean 1 card of yellow 1 card of red and 1 card of green are selected)
- 2 cards are RANDOMLY DRAWN (please not that we are not given that the cards are replaced ..so assume that the remaining cards are 21) . From this 21 cards we have to select such 2 cards that atleast 1 of the cards selected will match with color of any of the earlier (3 cards) . OR we select such 2 cards from the 21 that when considered with earlier 3 cards will form 3 same colors.
For eg : first case - earlier 3 cards : yellow /blue/red
now i select 2 acrds and from that atleast one the cards will be of same color : yellow/blue/red .....note that we can get 2 pairs of colors also
second case : earlier 3 cards : Y/B/R
i seleted 2 : these 2 will be of same color ie Y/Y ..and these two will match with one of the earlier colors Y

SO we have to find the probability of case 1 +case 2
soliution :
Here going the conventional form is tidious ...so we use (1-P)

We will slect such 2 cards that they are not from the GROUP of any of the arlier thre cards...ie we will not select any cards from Y/B/R...we want any color but these...

Sample space : we have 21 cards left and we will slect 2 cards from these. (here we cannot exempt the group that we do not want as these are also the possibilities ) = 21C2 = 210

event : as we do not wajt any card from the same group of earlier 3 cards we will subtrat those groups from 21 cards. Now we selected 3 cards of different color. All colors have 3 cards ...so these 3 cards will have 2 MORE same color cards ( i just got confused wait ) ...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

That makes perfect sense. When I tried to do it I for got that I need to reduce 15 to 14 and 21 to 20 because there was not replacement. Thanks!

Instead of picking the other 2 different from original why don't we pick the same directly ? We have 6 cards that match the original cards dealt, so why cannot it be 6*5 ?

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