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Counting and probability

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Counting and probability

by cypherskull » Tue Aug 28, 2012 12:24 am
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


I got the solution but then I don't fully understand. My initial approach to this problem was as follows:

P(at least 1 pair) = 1 - P(no pair)

P(no pair) = (12*10*8*6)/12C4

However, as it turns out, 12C4 yields a wrong answer. If I use 12P4 instead, I get the right answer.
Can some one please explain the gap in my understanding here? Thanks!

[Moderator edit: Moved the post to a relevant forum]
Regards,
Sunit

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by mehaksal » Tue Aug 28, 2012 1:51 am
how can we use 12P4 here? permutations are only for arrangements and not for selections. correct me if I am wrong.
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by cypherskull » Tue Aug 28, 2012 1:58 am
You're right by all means. And that's precisely my question here too. I first tried 12C4. Got an incorrect answer. While doing a hit n trial, I observed that 12P4 worked. That's my question...HOW and WHY did it work?

I researched a little further and it turns out that my numerator (12*10*8*6) is incorrect. It should be - (12*10*8*6)/4! while the denominator should be 12C4. Again...am not sure why.
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by Brent@GMATPrepNow » Tue Aug 28, 2012 5:56 am
cypherskull wrote: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
We can also 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
= C

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by Brent@GMATPrepNow » Tue Aug 28, 2012 6:06 am
cypherskull wrote: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
Here's one approach using counting techniques.
First recognize that P(at least one pair) = 1-P(no pairs)

We'll find P(no pairs)

First, the number of possible outcomes.
We have 12 cards and we select 4.
This is accomplished in 12C4 ways = 495

Now we count the number of ways to select 4 different cards with no pairs. In other words, we want 4 different card values.
First look at how many different cards we can select (WITHOUT considering the suits)
There are 6 card values and we want to select 4.
This is accomplished in 6C4 = 15 ways.
For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
So, for example, in how many ways can we have card values of 1,2,4,5?
For the 1-card we have 2 suits from which to choose.
For the 2-card we have 2 suits from which to choose.
For the 4-card we have 2 suits from which to choose.
For the 5-card we have 2 suits from which to choose.
So, for the selection 1,2,4,5 there are 2x2x2x2=16 possibilities.

So, the number of ways to select 4 cards such that there are no pairs is 15x16=240

So, the probability that there are no pairs = 240/495 = 16/33

So, P(at least one pair) = 1- 16/33 = 17/33 = C

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Brent
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by cypherskull » Tue Aug 28, 2012 6:20 am
Thanks a lot Brent! I understood both the approaches you explained. But I'm not sure where I went wrong in my approach. Here's how I selected the number of ways to choose non-pairs -

1st card - 12 ways
2nd card - 10 ways (since 11 cards remaining one of which has the same value as the 1st card)

Similarly,
3rd card - 8 ways
4th card - 6 ways


Therefore, P(no pairs) = 12*10*8*6/12C2. But this isn't correct. The correct solution using this approach would be -

P(no pairs) = [(12*10*8*6)/4!]/12C2

I'm unable to understand why do I need to divide the numerator with 4!. It'd be great if you can help me out with this part. Thanks!
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by Brent@GMATPrepNow » Tue Aug 28, 2012 6:40 am
cypherskull wrote:Thanks a lot Brent! I understood both the approaches you explained. But I'm not sure where I went wrong in my approach. Here's how I selected the number of ways to choose non-pairs -

1st card - 12 ways
2nd card - 10 ways (since 11 cards remaining one of which has the same value as the 1st card)

Similarly,
3rd card - 8 ways
4th card - 6 ways

Therefore, P(no pairs) = 12*10*8*6/12C2. But this isn't correct.
Your solution is close, but you have a bit of a mismatch here.

The portion in green treats each ordering of the cards differently. For example, your solution (in green) says that selecting 1 then 3 then 6 then 2 is different from selecting 6 then 2 then 1 then 3.

That's totally fine. You have successfully counted the number of ways to select 4 cards such that there are no pairs.

However, the portion in red has a few problems.
Here we want to count the total number of ways to select 4 cards from the 12 cards. More importantly, we need to treat the card ordering as mattering since we already treated the card ordering as mattering when we calculated the green portion.

So, if the card order matters, in how many ways can we select 4 cards?
Stage 1: Select the first card? We can do this in 12 ways.
Stage 2: Select the second card? We can do this in 11 ways.
Stage 3: Select the third card? We can do this in 10 ways.
Stage 4: Select the fourth card? We can do this in 9 ways.
So, we can select 4 cards in 12x11x10x9 ways

Aside: Some may recognize this as 12P4 ways, but I'm not a big fan of permutations on the GMAT.

Now we can see that P(no pairs) = (12*10*8*6)/(12x11x10x9) = 16/33

So, P(at least one pair) = 1- 16/33 = [spoiler]17/33 = C [/spoiler]
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by cypherskull » Tue Aug 28, 2012 6:45 am
Thanks a ton Brent! I get the picture now.
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by GaneshMalkar » Wed Aug 29, 2012 10:09 am
cypherskull wrote:Thanks a ton Brent! I get the picture now.
Let me simplify this a bit :)

let bill has two colored cards one of RED and second BLACK

Red Cards - 1,2,3,4,5,6
Black Cards - 1,2,3,4,5,6


Now,

P(atleast 2 cards having same value) = 1 - P(cards not having same value) ....1

P(cards not having same value) is to find first


P(cards not having same value) = # of ways in which cards having different value / total # of ways of selecting 4 cards from 12

total # of ways of selecting 4 cards from 12 = 12C4


Card 1 can be selected in 12 ways from the available 12 cards = 12
Card 2 can be selected in 10 ways from the available 10 cards = 10 (should not take as 11 as we are removing the possibility of repetition of first card)
Card 3 can be selected in 8 ways from the available 8 cards = 8
Card 6 can be selected in 6 ways from the available 12 cards = 6

So total no of ways = 12 * 10 * 8 * 6

This is permutation. It takes for eg :- 1,2,2,4 and 1,4,2,2 as two different things...They are same :)

so to convert it to combination nPr = nCr * r!

we need to divide this by 4! since 4! is the no of ways of arranging 4 cards within themseleves....

so, # of ways in which cards having different value = (12 * 10 * 8 * 6 / 4!)

total # of ways of selecting 4 cards from 12 = 12C4

so P(cards not having same value) = (12 * 10 * 8 * 6 / 4!) / 12C4

P(cards not having same value) = 16/33


From 1,

P(atleast 2 cards having same value) = 1 - P(cards not having same value)
= 1 - 16/33 = 17/33

[/u]
If you cant explain it simply you dont understand it well enough!!!
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