saintneo:

try choosing a marble from the box labeled WB, then you can deduce the rest without taking another pick.

Yeah I figured!

It has to be just from the WB box

The fewest number of marble to make d boxes content correct as laballed is 5marbles
explanation;provided three boxes incorectly labelled(box1 contains 2b labelled 1w&1b,box2 contains 2w labelled 2b and box3 contains 2w labelled 1w&1b)

Jasi7:

Tonzo!

Donzo for u too.....

This is piece of kulikuli. The 'minimum' number of times two look at the marbles in the box is once i.e one time.
Since they are wrongly labelled, just open one of the boxes and determine the content. By eliminating that one, exchange the labels of the other two boxes and 'viola' you get your answer.

MoyoGENERAL: There are three boxes which each contain two marbles: one has two white, one has two black and one has one white and one black marble. Each of the boxes also is labelled as to its content, but each label is incorrect. What is the fewest number of marble you could remove from the boxes and look at in order to definitely determine the content of all the three boxes?
....-


Rules..
The next logic puzzle will only come up when there is an answer to the existing puzzle...
Answers provided may require explanation.
You are allowed to cheat....lol




let the game begin.....

This is a case of Word Problem

The keyword here is ALL.
The question is not asking us to determine what is in EACH of the box but what is in ALL the box. (combined)
Since we know the overall/aggregate content of the 3 boxes, no matter if they were wrongly labeled, the overall/aggregate content in the boxes still remains the same.
Therefore the least number of marble to remove in other determine the content in ALL 3 boxes is 0 (zero).

In other words, I do not need to pick any marble in other to definitely determine what is in ALL the boxes not EACH.


Next puzzle please

U will need to pick one box labeled as having 2 black marble. Pick d first mable and the second marble and see their color. U Wil then know the original label of dat box, then look for anoda box labeled as the color of the the 2 marbles u pick and exchange the label to the original u now found out then interchange then on the remainin two boxes without opening them,
So in conclusion u need to pick 2 marbles from any of d box to determine them.

rattlesnake: hmm.....ehhhhhhhhh

[size=14pt]6[/size] dey this puzzle
[size=14pt]BLACK AND WHITE COLOR[/size] dey this puzzle

GUYS BE CAREFUL

[size=14pt]today is 12/7/2014=12+7+2+1+4=26...2 BALLS IN 3 BOXES =6 TODAYS DATE [/size]

Dont be initiated

AM OUT

mumu y didn't u add 1+2.

The answer is 3.
If you Pick the box with the black & white first,thats two balls then one more box with either b or w to confirm.
So the minimum no of balls are 3.

OP. MoyoGENERAL. Come here and tell everyone the answer is 1. All these wrong answers they are sharing. There is truly God.

[quote author=MissMeiya]



Here's my logic:

Possibility 1:
• You draw one white marble from the box marked "white-black".
• Because we know each label is incorrect, we know that of the two possible identities (white-white and white-black) only white-white can be true.
• The box marked "black-black" cannot be white-white because we have already deduced its location. It cannot be black-black, because its label is false. Therefore, the box marked "black-black" is white-black.
• The box marked "white-white" is left with only one possible identity: black-black.

Possibility 2:
(the inverse of Possibility 1)
• You draw one black marble from the box marked "white-black".
• Because we know each label is incorrect, we know that of the two possible identities (black-black and white-black) only black-black can be true.
• The box marked "white-white" cannot be black-black because we have already deduced its location. It cannot be white-white, because its label is false. Therefore, the box marked "white-white" is white-black.
• The box marked "black-black" is left with only one possible identity: white-white.

***** scribble pad *****
WB (w) = WW, (b) = BB
WW = BB, WB
BB = WB, WW




I respect u madam! U r 100% right.

Like u said one only need to pick 1 from d box labeled BW, if u pick black there den it's BB. Den d one labeled WW must be BW and dat labeled BB must be WW.


Note dat if u pick 2, 3,4,5 and 6 u can also get d ansa but d question said MINIMUM

I hate it when someone starts a thread and abandons it.

Take the first box and pick both balls eg BW, then you know the remaining boxes contains BB and WW. Then pick one ball from one of the remaining two boxes: if you pick a white ball, the you know the last box contains BB, vice versa .

So answer is 3

yeeeeeehhhhhh
I can't believe I made FP
dancing all kind of dance.....
where was I?
I DON MISSSSSSS OOOOOOOO
MEN.......
Now to give you guys the answer......
the answer is one......yes, one
explanation:
practically get three bowls, put two black balls in one, two white balls in one and one black and one white in the other, then intentionally label the boxes wrongly....if ( box containing W W) labelled as B W, then (box containing BB) labelled as WW, and ( box containing BW) as BB....the question is what is the least possible marble you will pick to determine all three content....
guys if you pick from the box wrongly labelled as BW, let's say you pick W, you will know that That box will contain WW and BW since its wrongly labelled....this leaves you with the second box wrongly labelled to be interchanged to BB( since you know the 3rd box is wrongly labelled and we have established the first box as WW,) now the third box is BW...
pronto....

lebesgue: I hate it when someone starts a thread and abandons it.

a beg nor vex bro...
am back

MoyoGENERAL: There are three boxes which each contain two marbles: one has two white, one has two black and one has one white and one black marble. Each of the boxes also is labelled as to its content, but each label is incorrect. What is the fewest number of marble you could remove from the boxes and look at in order to definitely determine the content of all the three boxes?
....


Rules..
The next logic puzzle will only come up when there is an answer to the existing puzzle...
Answers provided may require explanation.
You are allowed to cheat....lol




let the game begin.....

Puzzle 2...
this is a trickier one....

"Mrs Joy's five daughters each gave books for Christmas to one or more of her sisters. Each presented four books and each received four books, but no two girls allocated her book the same way. That is, only one gave two books to one sister and two to another. Beth gave all her books to Alice; Christy gave three to Edith; which sisters gave the four books to Deborah??

and again....you guys are allowed to cheat...lol
next question comes up when we have a correct answer...

MissMeiya:

[size=14pt]Answer: 1 marble, from the box marked "white-black"[/size]

Here's my logic:

Possibility 1:
• You draw one white marble from the box marked "white-black".
• Because we know each label is incorrect, we know that of the two possible identities (white-white and white-black) only white-white can be true.
• The box marked "black-black" cannot be white-white because we have already deduced its location. It cannot be black-black, because its label is false. Therefore, the box marked "black-black" is white-black.
• The box marked "white-white" is left with only one possible identity: black-black.

Possibility 2:
(the inverse of Possibility 1)
• You draw one black marble from the box marked "white-black".
• Because we know each label is incorrect, we know that of the two possible identities (black-black and white-black) only black-black can be true.
• The box marked "white-white" cannot be black-black because we have already deduced its location. It cannot be white-white, because its label is false. Therefore, the box marked "white-white" is white-black.
• The box marked "black-black" is left with only one possible identity: white-white.

***** scribble pad *****
WB (w) = WW, (b) = BB
WW = BB, WB
BB = WB, WW

miss meîya...thumbs up....NYC oñe...

MoyoGENERAL:
miss meîya...thumbs up....NYC oñe...

There you are!!!! Welcome back

Lol yayyyy I win
Give me points joor I like competition

deadie:

Come OP, I don give you the answer wey you want. Better prove me wrong or give me my price

wrong bro......

Fit2Rule: OP where is your next puzzle? I've solved it. If I didn't, tell me how! Fewest number of marbles is 4!

my Oga na one oooo...
BTW next puzzle don dey oooo

[quote author=MoyoGENERAL]There are three boxes which each contain two marbles: one has two white, one has two black and one has one white and one black marble. Each of the boxes also is labelled as to its content, but each label is incorrect. What is the fewest number of marble you could remove from the boxes and look at in order to definitely determine the content of all the three boxes?
....


Rules..
The next logic puzzle will only come up when there is an answer to the existing puzzle...
Answers provided may require explanation.
You are allowed to cheat....lol




let the game begin.....




So first 8 winners of puzzle one include....

Mc whillion
abes
miss meiya
adekayo 1234
rapture ready
lilpervert
lobell
dukila......




......congrats.....
you have been awarded 4 points each.....
NB.... starting frôm next puzzle, the first 8 winners will be awarded points based on their positions ( 8 points to 1st and 1 point to the 8th position) .....I have 30 puzzles in all...........
this is going to be fun.

MoyoGENERAL:

Puzzle 2...
this is a trickier one....

"Mrs Joy's five daughters each gave books for Christmas to one or more of her sisters. Each presented four books and each received four books, but no two girls allocated her book the same way. That is, only one gave two books to one sister and two to another. Beth gave all her books to Alice; Christy gave three to Edith; which sisters gave the four books to Deborah??

and again....you guys are allowed to cheat...lol
next question comes up when we have a correct answer...

Puzzle 2
....we need first 8 winners...

MoyoGENERAL:

Puzzle 2...
this is a trickier one....

"Mrs Joy's five daughters each gave books for Christmas to one or more of her sisters. Each presented four books and each received four books, but no two girls allocated her book the same way. That is, only one gave two books to one sister and two to another. Beth gave all her books to Alice; Christy gave three to Edith; which sisters gave the four books to Deborah??

and again....you guys are allowed to cheat...lol
next question comes up when we have a correct answer...

So, to avoid duplication in their distribution patterns this arrangement has to be followed. One sister gave all her books to one sister(4;0). The second sister gave three and one books respectively to two sisters(3;1). The third sister gave two and two books respectively to two sisters(2;2). The fourth sister must give two and one and one books respectively to three sisters(2;1;1) while the last sister can only give one and one and one and one books respectively to four sisters(1;1;1;1)!

EXPLANATION (edited)
Beth gave all four books to Alice
Christy gave three to Edith remaining one
Therefore the only combinations left for Deborah are:
She gets two books from one girl (Edith or Alice) and one each from two other girls (Christy (compulsorily), and Edith or Alice(depending on who gave her two)). QED

NOTE: Deborah must of necessity be the one who gave two books to one sister and her remaining two books to another sister!!!

lobell:

So, one sister gave all her books to one sister. The second sister gave three and one books to two sisters. The third sister gave two and two books to two sisters. The fourth sister gave two and one and one books to three sisters while the last sister gave one and one and one and one books to four sisters!

EXPLANATION
Beth gave all four books to Alice
Christy gave three to Edith remaining one
Therefore the only combinations left for Deborah are:
She gets two books from one girl and one each from two other girls. QED

please make it clearer....
the question is which sisters gave the four books to Deborah?

Either 3 or 4 picks. But to completely label the 3 bags in any situation d least you must pick is 4. The colours of the marbles picked in the 1st and 2nd attempts will determine whether 3 or 4 picking is required. Follow the analysis below:

(Case 1) Any time you pick 2 marbles of the same colour in d 1st and 2ndattempt, then you must pick 2 more from the next bag to be able to define the contents of the 3 bags. And so in this case you have to pick 4 marbles.
(Case 2) Anytime you pick 2 marbles of different colours in your 1st 1nd 2nd attempt, then you only have to pick one more from the next bag (i.e. 3 marbles) (Eg_ Assuming you picked 1 black m and 1 white m from the 1st bag, you only need to pick one marble from the next bag to define the content of the 3 bags. it is so because the colour of marbles in each of 2nd and 3rd bags will be same).

Therefore, the least no of marbles one needs to sight to be able define d contents of the bags is 3

Answer: Alice and Christie gave 1 book each and Edith gave 2 books to Deborah.

Whooo! I remember this kind of puzzle. They use these puzzles for law school aptitude tests.
You use a chart like this to solve it:

           + - - + - - + - - + - - + - - +
           |  A  |  B  |  C  |  D  |  E  |
+ - - - - -+ - - + - - + - - + - - + - - +
| Alice    |  ×  |  1  |  1  |  1  |  1  |
+ - - - - -+ - - + - - + - - + - - + - - +
| Beth     |  4  |  ×  |  ×  |  ×  |  ×  |
+ - - - - -+ - - + - - + - - + - - + - - +
| Christie |  ×  |  ×  |  ×  |  1  |  3  |
+ - - - - -+ - - + - - + - - + - - + - - +
| Deborah  |  ×  |  2  |  2  |  ×  |  ×  |
+ - - - - -+ - - + - - + - - + - - + - - +
| Edith    |  ×  |  1  |  1  |  2  |  ×  |
+ - - - - -+ - - + - - + - - + - - + - - +

Start with the given information:
• None of them gave books to themselves. (image)
• "Beth gave all her books to Alice", so Beth cannot give any more books to anyone else. Also, Alice cannot receive any more books from anyone else. (image)
• "Christy gave three to Edith" (image)

Now, we reason:
• There are five different ways for the 4 books to be given:
- 4 (already used by Beth)
- 3-1
- 2-2
- 2-1-1
- 1-1-1-1
• Only Alice has enough open spaces to give the 1-1-1-1 combination. Now Edith has reached the four books she was given. This means no one else gave her any more books. (image)
• Deborah only has space to give the following combinations: 4 (already used by Beth), 3-1 (only possible by Christie), and 2-2. So it must be 2-2. This means Christie's remaining book came from Edith. (image)
• Edith can only give the 2-1-1 combination (3-1 is only possible by Christie). So she must give Beth 2 books or 1 book. So Beth's remaining book must come from Edith. This means no one else gave Beth any more books. This means Christie gave her remaining book to Deborah. This means Deborah's remaining 2 books came from Edith. (image)

For us to solve dis we must make assumptions

for me, assuming deborah was d sister tht gave 2 books each to 2 sisters den
Alice = 1betty, 1christy nd 1deborah nd 1Edith
betty= 4 Alice
Christy= 1 Deborah nd 3Edith
Deborah= 2 betty nd 2 christy
Edith= 1 Betty, 1 Christy, 2 Deborah

therefore Alice, christy nd Edith gave books to deborah.

[color=#770077][/color]fewest is 2

the fewest you can pick is 2.. either black or it's white according to your question.the fewest you can pick is 2.. either black or it's white according to your question.the fewest you can pick is 2.. either black or it's white according to your question.

Dgmec: Either 3 or 4 picks. But to completely label the 3 bags in any situation d least you must pick is 4. The colours of the marbles picked in the 1st and 2nd attempts will determine whether 3 or 4 picking is required. Follow the analysis below:

(Case 1) Any time you pick 2 marbles of the same colour in d 1st and 2ndattempt, then you must pick 2 more from the next bag to be able to define the contents of the 3 bags. And so in this case you have to pick 4 marbles.
(Case 2) Anytime you pick 2 marbles of different colours in your 1st 1nd 2nd attempt, then you only have to pick one more from the next bag (i.e. 3 marbles) (Eg_ Assuming you picked 1 black m and 1 white m from the 1st bag, you only need to pick one marble from the next bag to define the content of the 3 bags. it is so because the colour of marbles in each of 2nd and 3rd bags will be same).

Therefore, the least no of marbles one needs to sight to be able define d contents of the bags is 3

Tunexol10: the fewest you can pick is 2.. either black or it's white according to your question.the fewest you can pick is 2.. either black or it's white according to your question.the fewest you can pick is 2.. either black or it's white according to your question.

Already answered. See post #173

To my best knowledge i would say' let the boxes be:
BOX 1 = WW
BOX 2 = BB
BOX 3 = WB,
where
W = white marble
B = black marble. Now info says they were incorrectly labelled, so i put a label on them
LABELS(assuming)
BOX 1 = (HYBRID MARBLES)
BOX 2 = (SNOW MARBLES)
BOX 3 = (CHARCOAL MARBLES).
"The info does nt give a definite way of labelling, it just say they are incorrect, so i cld say i'm right with my assumption". Now, if we are to remove the marbles, so as to determine the content, its definitely 4 marbles, cuz 2 marbles wld eventually be chosen from a box and with that fact you wld be able to determine the content of the rest. I hope i'm correct...