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There are six gamblers (An, Bing, Cheng, Ding, Er, and Fang)

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by fskilnik@GMATH » Wed Feb 16, 2011 12:40 pm
kazemi wrote:There are six gamblers (An, Bing, Cheng, Ding, Er, and Fang). Each of them rolls a fair 6-sided dice with 1,2,3,4,5, and 6 on each of the six sides, respectively. At least three of them got an even number. What is the probability that Bing, Ding and Fang all rolled an even number? Why?
Hi there!

I would certainly use "conditional probability" here (NOT a GMAT-subject)... and the calculations involved (the way I see it) are too lengthly to be GMAT-like.

I will show you how to approach the problem, but I will leave the details (and most calculations) out!

Let X and Y be the events: Y = "at least 3 people got even numbers" , X = "Bing, Ding and Fang got even numbers".

We are looking for Prob(X | Y) = Prob(Y) * Prob (X and Y)

Please note that Prob(Y) = 1 - Prob(not Y) = 1 - Sum of Prob(None got even), Prob(Exactly 1 got even), Prob(Exactly 2 got even).

Now: Prob(None got even) = (1/2)^ 6 , Prob(Exactly 1 got even) = 6*(1/2)*(1/2)^5 , Prob(Exactly 2 got even) = 15*(1/2)^2*(1/2)^4

Prob(X and Y) could be dealt separating in many cases: exactly 3, exactly 4, exactly 5 and exactly 6 got even and, in each case, you should calculate the corresponding probability:

> Prob(X and exactly 3 got even) = Prob(B, D, F even and A, C, E odd) = 1/2^6 ;
> Prob(X and exactly 4 got even) = Prob(B, D, F and one between (A, C, E) even, the other two odd) = 3/2^6 ;
> Prob(X and exactly 5 got even) = Prob(B, D, F and two between (A, C, E) even, the one left odd) = 3/2^6 ;
> Prob(X and exactly 6 got even) = Prob(A, B, C, D, E and F got even) = 1/2^6 ;

I hope you got the picture.

Regards,
Fabio.
Last edited by fskilnik@GMATH on Thu Feb 17, 2011 10:20 am, edited 1 time in total.
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by kazemi » Thu Feb 17, 2011 6:51 am
Thanks, I got 4/21.
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by fskilnik@GMATH » Thu Feb 17, 2011 10:27 am
kazemi wrote:Thanks, I got 4/21.
My pleasure, kazemi.

Well, I did the calculations to check and I got another result, therefore:

1) I edited my first post, explaining and calculating the Prob(X and Y) = 8/2^6 = 1/2^3 = 1/8 ... please check if you agree and if you get the same value ;

2) I calculated Prob(Y) to get 21/2^5 ... please check that! (Take into account that (1/2)^2*(1/2)^4 = (1/2)^6 etc, because that makes most of the calculations easier.)

3) From all that, I guess the answer would be Prob(X|Y) = Prob(Y)*Prob(X and Y) = 21/2^5 * 1/2^3 = 21/2^8 (approx. 8%), different from yours...

I will wait your response!

Regards,
Fabio.
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by GMAT25258 » Thu Feb 17, 2011 12:08 pm
Hi fskilnik,

I followed a similar approach to you, but obtained a different result (close to %40).

Find the probability of each possible even amount (from 2^3 possible events, since E>=3)
P(E=3) =(3C0)/2^3 = 1/8
P(E=4) =(3C1)/2^3 = 3/8
P(E=5) =(3C2)/2^3 = 3/8
P(E=6) =(3C3)/2^3 = 1/8

Find the conditional probability of B,D and F = even, using combinations, for a each possible even amount:
If E=3 -> P(BDF)=3C3/6C3 = 1/20
If E=4 -> P(BDF)= 3C2/6C4 = 3/15
If E=5 -> P(BDF)= 3C1/6C5 = 3/6
If E=6 -> P(BDF)= 3C0/6C6 = 1

Find the joint probability for each event:
For E=3: 1/8 * 1/20
For E=4: 3/8 * 3/15
For E=5: 3/8 * 3/6
For E=6: 1/8 * 1

Sum the 4 events to obtain result:

63/160

Maybe I missed something. What is the OA?
Last edited by GMAT25258 on Thu Feb 17, 2011 1:51 pm, edited 2 times in total.
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by kazemi » Thu Feb 17, 2011 12:32 pm
The stimulus mentioned the numbers 1 to 6. That's a smog. The only thing matters is odd or even among the numbers. Since "At least three of them have rolled an even number", let's do it one step at a time.

If all 6 dices are even, there is only one way to do it (all six gamblers get even numbers) and there is one way in which the three specific gamblers have all even.
If 5 dices are even, there are 6 ways to do it (only one gambler gets an odd number) and there are 3 ways in which the three gamblers have all even.
If 4 dices are even, there are 15 ways to do it and there are 3 ways in which the three gamblers have all even.
If 3 dices are even, there are 20 ways to do it and only one way in which the three gamblers have all even.

8/42 = 4/21
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by fskilnik@GMATH » Thu Feb 17, 2011 2:09 pm
kazemi wrote:The stimulus mentioned the numbers 1 to 6. That's a smog. The only thing matters is odd or even among the numbers.
That´s right. We all used this fact in our arguments.
kazemi wrote:If all 6 dices are even, there is only one way to do it (all six gamblers get even numbers) and there is one way in which the three specific gamblers have all even.
That´s correct, too. That´s what GMAT25258 wrote as "If E=6 -> P(BDF)= 3C0/6C6 = 1" , by the way.
kazemi wrote: If 5 dices are even, there are 6 ways to do it (only one gambler gets an odd number) and there are 3 ways in which the three gamblers have all even.
That´s correct, too. That´s what GMAT25258 wrote as "If E=5 -> P(BDF)= 3C1/6C5 = 3/6" , by the way.
kazemi wrote: If 4 dices are even, there are 15 ways to do it and there are 3 ways in which the three gamblers have all even.
That´s correct, too. That´s what GMAT25258 wrote as "If E=4 -> P(BDF)= 3C2/6C4 = 3/15" , by the way.
kazemi wrote: If 3 dices are even, there are 20 ways to do it and only one way in which the three gamblers have all even.
8/42 = 4/21
That´s correct, too. That´s what GMAT25258 wrote as "If E=3 -> P(BDF)=3C3/6C3 = 1/20" , by the way.

kazemi wrote: 8/42 = 4/21
Uau... now I see that (1+3+3+1)/(1+6+15+20) is really right (I didn´t think so at first)!

I thought you were not dealing with the "weighted average" of each scenario, but in fact you are using the proportions 1:1, 3:6, 3:15, 1:20... and the proportion 1:6:15:20 is itself correct! That´s really perfect, kazemi, congrats!!!


Regards,
Fabio.
Last edited by fskilnik@GMATH on Thu Feb 17, 2011 3:05 pm, edited 2 times in total.
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by fskilnik@GMATH » Thu Feb 17, 2011 2:57 pm
fskilnik wrote:We are looking for Prob(X | Y) = Prob(Y) * Prob (X and Y)
Silly mistake (sorry)... the correct equation (from the direct definition of conditional probability) is:

-------------------------------------------------
Prob (X and Y) = Prob(Y) * Prob(X | Y)
-------------------------------------------------

Therefore (from all correct calculations I presented in my previous post) we get:

P(X|Y) = 1/8 over 21/32 = 4/21, and that agrees with kazemi´s final answer!

@GMAT25258: I will see your arguments with care till Monday, but now it is 20h57min (I am really tired) and tomorrow I guess I will not have time to come here! Please wait a bit, okay?

Regards,
Fabio.
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by sanju09 » Fri Feb 18, 2011 1:30 am
kazemi wrote:There are six gamblers (An, Bing, Cheng, Ding, Er, and Fang). Each of them rolls a fair 6-sided dice with 1,2,3,4,5, and 6 on each of the six sides, respectively. At least three of them got an even number. What is the probability that Bing, Ding and Fang all rolled an even number? Why?

When it's given that at least three of them got an even number, we won't consider 6^6 to be the sample space now, we would rather look for its subset which is not very big now. Now the question is evidently asking for the probability that Bing, Ding and Fang were common among all rolling an even number.

Let's find out all possible cases where at least three of them got an even number is seen. At least three in this case means six or five or four or three of them got an even number.

All six read even = 3^6 ways

Any five read even = 6C5 × 3^5 ways

Any four read even = 6C4 × 3^4 ways

Any three read even = 6C3 × 3^3 ways

Hence, the diminished sample space is

= 6C6 × 3^6 + 6C5 × 3^5 + 6C4 × 3^4 + 6C3 × 3^3

= 3^3 (3^3 + 6C5 × 3^2 + 6C4 × 3^1 + 6C3)

= 3^3 (27 + 54 + 45 + 20)

= 27 × 146

Now the favorable ways are those in which Bing, Ding and Fang are common, it means we only need to select any 3 to fill six, any 2 to fill 5, any 1 to fill 4, Bing, Ding and Fang to fill 3, and the favorable ways are

= 3C3 ×3^3 + 3C2 × 3^2 + 3C1 × 3^1 + 3C0 × 3^0

= 27 + 27 + 9 + 1

= 64

Required probability = [spoiler]64/ (27 × 146)[/spoiler]

I am feeling very sleepy now, please find an error in concept or calculation of this NOT GMAT stuff...
:|
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by fskilnik@GMATH » Sun Feb 20, 2011 7:51 am
GMAT25258 wrote:Maybe I missed something.
Hi there, GMAT25258!

I´m back to spot your mistake, as promised some days ago.
GMAT25258 wrote: Find the probability of each possible even amount (from 2^3 possible events, since E>=3)
P(E=3) =(3C0)/2^3 = 1/8
P(E=4) =(3C1)/2^3 = 3/8
P(E=5) =(3C2)/2^3 = 3/8
P(E=6) =(3C3)/2^3 = 1/8
That´s correct. Please note that these are exactly the 1, 3, 3, 1 "numerators" present in kazemi´s solution, and in your solution each of them is part of the whole (8), nothing left (1+3+3+1 = 8), so to speak.
GMAT25258 wrote: Find the conditional probability of B,D and F = even, using combinations, for a each possible even amount:
If E=3 -> P(BDF)=3C3/6C3 = 1/20
If E=4 -> P(BDF)= 3C2/6C4 = 3/15
If E=5 -> P(BDF)= 3C1/6C5 = 3/6
If E=6 -> P(BDF)= 3C0/6C6 = 1
This is also correct by itself, but please note that the same 1, 3, 3 and 1 are obtained (also) here as numerators.

This is not a coincidence... in fact, at this moment you are ALREADY taken into account that from all 42 (=20+15+6+1) equiprobable scenarios (all of them with at least 3 even amounts, as you put in the beginning of your solution), there are 1, 3, 3, and 1 (respectively) cases that are "weighted" according to theirs specific portion on the whole.

Explicitly:

The "E=3" scenario represents weight = 20/42 of the "(at least) 3 even amounts reality" (as the whole sample space, your idea!) and, on it, 1 case is favourable to us (that is, the one in which BDF are the even's ones) ;

The "E=4" scenario represents weight = 15/42 of the "(at least) 3 even amounts reality" (as the whole sample space, your idea!) and, on it, 3 cases are favourable to us (that is, the 3 ones in which BDF are 3 of the 4 even's ones) ;

etc.

From that, I guess we understand that:
GMAT25258 wrote:Find the joint probability for each event:
For E=3: 1/8 * 1/20
For E=4: 3/8 * 3/15
For E=5: 3/8 * 3/6
For E=6: 1/8 * 1
is wrong (you "double weighted" the corresponding scenarios, in an informal way of putting it), but the "naive" (no pun intended) calculation: (1+3+3+1)/(20+15+6+1), used by kazemi, it is now (seems to me) absolutely clear/justified!

Regards,
Fabio.

P.S.: the careful reader may note how the "conditional probability" definition (used in my solution) gets control over the whole sample space (not only the E>=3 ones), over the "weighted" proportions of the scenarios AND over the favourable cases in each one of them. As a rule, when we use a "more powerful weapon" (as this concept) we are able to "let" mathematics itself "manage" the difficulties/subtleties of the problem; on the other hand, kazemi´s (almost trivial) approach is marvellously simpler (theoretically speaking), but must be used with care, because there are many subtle details that are involved, as the ones I mentioned related to GMAT25258´s approach. (GMAT25258 and kazemi´s approaches are really the very same, by the way!)
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