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well, that certainly works, but i wouldn't call it "easier". look at the enormous size of the numbers in your calculations, vs. the tiny size of the numbers in the other posters' calculations.analyst218 wrote:well a easier way , IMO , is to
just 1 - [getting non-pair/ total number]
= 1 - [12C4 / 12C4]
however the numerator has to be 12x10x8x6/4! since say you pick 1 for the first card,
you have to eliminate the other 1 so you have 10, same with the 2nd 3rd cards..
so 1-[12x10x8x6/4!]/[12C4] = 1 - 240/495 = 17/33
Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, 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) 8/33
(B) 62/165
(C) 17/33
(D) 103/165
(E) 25/33
But combinations itself means that the order does not matter right? why divide by 4! then?? can u please enlighten me on this?? if we choose 2 members from a team of 5 we say it there are 5C2 ways of doing it and we do not divide by 2! factorial. PLease help?logitech wrote:1 2 3 4 5 6
1 2 3 4 5 6
So we can choose 12x10x8x6 ways but we need to divide it by 4!
= 240 ways ( NO PAIR )
C(12:4) =495
So : (495-240)/495
That's right! But 5C2 is not 5P2 at the same time. What do we do when we need to get nC2 from a given nP2?lakshmi86 wrote:But combinations itself means that the order does not matter right? why divide by 4! then?? can u please enlighten me on this?? if we choose 2 members from a team of 5 we say it there are 5C2 ways of doing it and we do not divide by 2! factorial. PLease help?logitech wrote:1 2 3 4 5 6
1 2 3 4 5 6
So we can choose 12x10x8x6 ways but we need to divide it by 4!
= 240 ways ( NO PAIR )
C(12:4) =495
So : (495-240)/495
yeah, but the poster here is not using a combination formula. he's using consecutive multiplication, which always implies that the positions for which items are being chosen are distinguishable (i.e., that "order matters").lakshmi86 wrote:But combinations itself means that the order does not matter right? why divide by 4! then?? can u please enlighten me on this?? if we choose 2 members from a team of 5 we say it there are 5C2 ways of doing it and we do not divide by 2! factorial. PLease help?logitech wrote:1 2 3 4 5 6
1 2 3 4 5 6
So we can choose 12x10x8x6 ways but we need to divide it by 4!
= 240 ways ( NO PAIR )
C(12:4) =495
So : (495-240)/495
The issue is that you're not counting all of the ways you might get exactly one pair, or two pairs. When you calculate:pkonduri wrote:
It is evident that I am missing something , its a shame that i am not able to figure it out after looking into so many discussions that have happened. can someone help me out.Please!
you are assuming that the 1st and 2nd cards are pairs, and that the 3rd and 4th cards are paired. There are other possibilities - it might be that the 1st and 3rd cards are paired, and the 2nd and 4th cards are paired, for example.pkonduri wrote:
A) Probability of two pairs:
first chance:12/12
Second:1/11
Third:10/10
Fourth:1/9
So (12/12)*(1/11)*(10/10)*(1/9) = 1/99
shulapa wrote:Hi,
I got to the same answer as Brent but through a different method. Tell me whether this works for you guys.
It is easier to find out the chances of getting 4 card without any couple of them with the same value. This can be deducted from 1 to get the probability of at least one couple.
Card 1: can be any number at any suit. So we can chose 12 out of 12.
Card 2: can be any card from the remaining 11 besides the card, which we already hold its number. So it is 10 out of 11.
Card 3: can be any card from the remaining 10 besides the 2 cards, which we already hold the numbers. So it is 8 out of 10.
Card 4: the same idea, only now we have to deduct 3 invalid cards. So it is 6 out of 9.
Combining the option gives us:
Probability of no couples = 1*(10/11)*(8/10)*(6/9) = 16/33.
Now we can say that:
Probability of at least one couple = 1 - 16/33 = 17/33
shulapa wrote:Hi,
I got to the same answer as Brent but through a different method. Tell me whether this works for you guys.
It is easier to find out the chances of getting 4 card without any couple of them with the same value. This can be deducted from 1 to get the probability of at least one couple.
Card 1: can be any number at any suit. So we can chose 12 out of 12.
Card 2: can be any card from the remaining 11 besides the card, which we already hold its number. So it is 10 out of 11.
Card 3: can be any card from the remaining 10 besides the 2 cards, which we already hold the numbers. So it is 8 out of 10.
Card 4: the same idea, only now we have to deduct 3 invalid cards. So it is 6 out of 9.
Combining the option gives us:
Probability of no couples = 1*(10/11)*(8/10)*(6/9) = 16/33.
Now we can say that:
Probability of at least one couple = 1 - 16/33 = 17/33
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