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Re: 12 Easy Pieces (or not?)
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23 Oct 2013, 13:22
Bunuel wrote: 7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were nonnegative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to nonnegative numbers be 2 to 1? A. 11/14 B. 13/18 C. 4/7 D. 3/7 E. 3/14
If choose variable for set A there will be too many fractions to manipulate with, so pick some smart #: let set A contain 18 numbers.
"2/9 of the numbers in a data set A were observed" > 4 observed and 184=14 numbers left to observe; "3/4 of those numbers were nonnegative" > 3 nonnegative and 1 negative; Ratio of negative numbers to nonnegative numbers to be 2 to 1 there should be total of 18*2/3=12 negative numbers, so in not yet observed part there should be 121=11 negative numbers. Thus 11/14 of the remaining numbers in set A must be negative.
Answer: A. Hey Bunuel, Can you please advice how you chose 18 as the smart number? Thanks,



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Re: 12 Easy Pieces (or not?)
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24 Oct 2013, 02:22
prsnt11 wrote: Bunuel wrote: 7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were nonnegative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to nonnegative numbers be 2 to 1? A. 11/14 B. 13/18 C. 4/7 D. 3/7 E. 3/14
If choose variable for set A there will be too many fractions to manipulate with, so pick some smart #: let set A contain 18 numbers.
"2/9 of the numbers in a data set A were observed" > 4 observed and 184=14 numbers left to observe; "3/4 of those numbers were nonnegative" > 3 nonnegative and 1 negative; Ratio of negative numbers to nonnegative numbers to be 2 to 1 there should be total of 18*2/3=12 negative numbers, so in not yet observed part there should be 121=11 negative numbers. Thus 11/14 of the remaining numbers in set A must be negative.
Answer: A. Hey Bunuel, Can you please advice how you chose 18 as the smart number? Thanks, Choose a number which is a multiple of 9 (since we are told that "2/9 of the numbers in a data set..."). Try to choose a multiple so that you get another number ("3/4 of those numbers...") an integer too.
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Re: 12 Easy Pieces (or not?)
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06 Nov 2013, 21:01
Bunuel wrote: SOLUTIONS:
Notice that most of the problems have short, easy and elegant solutions, since you've noticed a trick/trap hidden in the questions.
1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color? A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1
No formula is need to answer this one. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.
Answer: E. Hi Bunnel, Isn't this is the probability of getting atleast 1 pair of socks? This will include cases where we have 2 pairs or all socks of same color. Question seems to be asking the probability of exactly 1 pair.



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Re: 12 Easy Pieces (or not?)
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07 Nov 2013, 02:29
cumulonimbus wrote: Bunuel wrote: SOLUTIONS:
Notice that most of the problems have short, easy and elegant solutions, since you've noticed a trick/trap hidden in the questions.
1. There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color? A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1
No formula is need to answer this one. The trick here is that we have only 3 different color socks but we pick 4 socks, which ensures that in ANY case we'll have at least one pair of the same color (if 3 socks we pick are of the different color, then the 4th sock must match with either of previously picked one). P=1.
Answer: E. Hi Bunnel, Isn't this is the probability of getting atleast 1 pair of socks? This will include cases where we have 2 pairs or all socks of same color. Question seems to be asking the probability of exactly 1 pair. In that case it would be asking about the probability of EXACTLY two socks of the same color.
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Re: 12 Easy Pieces (or not?)
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26 Nov 2013, 07:04
Hi Bunuel,
Thank you for the very high quality questions. Can you please explain your thinking behind the solution in question 8?
My confusion stems from the word "guarantee". Ideally the least number of chosen chips, that could result in 2 different colours, is indeed 3. However, the probability of this event is certainly not 1.



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Re: 12 Easy Pieces (or not?)
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Re: 12 Easy Pieces (or not?)
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21 Jan 2014, 13:33
Bunuel wrote: 6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colours. How many different arrangements are possible? A. 30 B. 60 C. 120 D. 240 E. 480
Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.
Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.
Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.
Answer: B. what about R B G R Y R G R B R ?



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Re: 12 Easy Pieces (or not?)
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23 Jan 2014, 03:53
mariofelix wrote: Bunuel wrote: 6. Anna has 10 marbles: 5 red, 2 blue, 2 green and 1 yellow. She wants to arrange all of them in a row so that no two adjacent marbles are of the same color and the first and the last marbles are of different colours. How many different arrangements are possible? A. 30 B. 60 C. 120 D. 240 E. 480
Seems tough and complicated but if we read the stem carefully we find that the only way both conditions to be met for 5 red marbles, which are half of total marbles, they can be arranged only in two ways: R*R*R*R*R* or *R*R*R*R*R.
Here comes the next good news, in these cases BOTH conditions are met for all other marbles as well: no two adjacent marbles will be of the same color and the first and the last marbles will be of different colors.
Now, it's easy: 2 blue, 2 green and 1 yellow can be arranged in 5 empty slots in 5!/(2!*2!)=30 ways (permutation of 5 letters BBGGY out of which 2 B's and 2 G' are identical). Finally as there are two cases (R*R*R*R*R* and *R*R*R*R*R. ) then total # of arrangement is 30*2=60.
Answer: B. what about R B G R Y R G R B R ? This arrangement violates the condition that the first and the last marbles must be of different colours.
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Re: 12 Easy Pieces (or not?)
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30 Jan 2014, 00:23
Bunuel wrote: 7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were nonnegative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to nonnegative numbers be 2 to 1? A. 11/14 B. 13/18 C. 4/7 D. 3/7 E. 3/14
If choose variable for set A there will be too many fractions to manipulate with, so pick some smart #: let set A contain 18 numbers.
"2/9 of the numbers in a data set A were observed" > 4 observed and 184=14 numbers left to observe; "3/4 of those numbers were nonnegative" > 3 nonnegative and 1 negative; Ratio of negative numbers to nonnegative numbers to be 2 to 1 there should be total of 18*2/3=12 negative numbers, so in not yet observed part there should be 121=11 negative numbers. Thus 11/14 of the remaining numbers in set A must be negative.
Answer: A. Hi Bunuel, Thanks for this explanation. I tried real hard to solve this ques, when I first attempted, algebraically but couldnt do it... I know its far too easy to solve it this way .. but I always go for algebraic ... please help... If possible please to provide an algebraic solution for this.... thanks a ton for all the help... Cheers!



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Re: 12 Easy Pieces (or not?)
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30 Jan 2014, 01:11
rawjetraw wrote: Bunuel wrote: 7. After 2/9 of the numbers in a data set A were observed, it turned out that 3/4 of those numbers were nonnegative. What fraction of the remaining numbers in set A must be negative so that the total ratio of negative numbers to nonnegative numbers be 2 to 1? A. 11/14 B. 13/18 C. 4/7 D. 3/7 E. 3/14
If choose variable for set A there will be too many fractions to manipulate with, so pick some smart #: let set A contain 18 numbers.
"2/9 of the numbers in a data set A were observed" > 4 observed and 184=14 numbers left to observe; "3/4 of those numbers were nonnegative" > 3 nonnegative and 1 negative; Ratio of negative numbers to nonnegative numbers to be 2 to 1 there should be total of 18*2/3=12 negative numbers, so in not yet observed part there should be 121=11 negative numbers. Thus 11/14 of the remaining numbers in set A must be negative.
Answer: A. Hi Bunuel, Thanks for this explanation. I tried real hard to solve this ques, when I first attempted, algebraically but couldnt do it... I know its far too easy to solve it this way .. but I always go for algebraic ... please help... If possible please to provide an algebraic solution for this.... thanks a ton for all the help... Cheers! Algebraic solution is a trap here and a waste of time... But anyway: Say there are x numbers in set A. "2/9 of the numbers in a data set A were observed" > \(\frac{2x}{9}\) observed and \(x\frac{2x}{9}=\frac{7x}{9}\) numbers left to observe; "3/4 of those numbers were nonnegative" > \(\frac{3}{4}*\frac{2x}{9}=\frac{x}{6}\) nonnegative and \(\frac{1}{4}*\frac{2x}{9}=\frac{x}{18}\) negative; Ratio of negative numbers to nonnegative numbers to be 2 to 1 there should be total of \(x*\frac{2}{3}\) negative numbers, so in not yet observed part there should be \(\frac{2x}{3}\frac{x}{18}=\frac{11x}{18}\) negative numbers. Thus \(\frac{(\frac{11x}{18})}{(\frac{7x}{9})}=\frac{11}{14}\) of the remaining numbers in set A must be negative. Hope it helps.
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Re: 12 Easy Pieces (or not?)
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05 May 2014, 06:07
because they are in parentheses means they are a set and no number can repeat? as in we can have only 3 7s, one 3, one 6 and one 1?
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Re: 12 Easy Pieces (or not?)
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Re: 12 Easy Pieces (or not?)
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18 Sep 2014, 23:54
Bunuel wrote: 8. There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color? A. 3 B. 5 C. 6 D. 16 E. 19
Worst case scenario would be if the first two chips we pick will be of the different colors. But the next chip must match with either of two, so 3 is the answer.
Answer: A. Bunuel i m not able to get the essence of this one..esp the "least" term here. Also, while attempting such worst case scenario problems what will be deciding factor int he question stem??
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Re: 12 Easy Pieces (or not?)
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19 Sep 2014, 02:24
Vinitkhicha1111 wrote: Bunuel wrote: 8. There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color? A. 3 B. 5 C. 6 D. 16 E. 19
Worst case scenario would be if the first two chips we pick will be of the different colors. But the next chip must match with either of two, so 3 is the answer.
Answer: A. Bunuel i m not able to get the essence of this one..esp the "least" term here. Also, while attempting such worst case scenario problems what will be deciding factor int he question stem?? The question asks about the minimum number of chips we should pick to guarantee that we have 2 chips of the same color.
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Re: 12 Easy Pieces (or not?)
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15 Jan 2015, 13:29
Question 11,
I guess it should have stated that we can only unse each one of these 3 nunnbers only once, right? Because, I used the same approach as Bunuel, but without knowing that you can only use these nubers once the process an be very long... and I am not even sure if it would end up in the same result  i guess not.
Is there sth in the way that the question is phrased that should have warned me that these numbers were only to be used once?



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Re: 12 Easy Pieces (or not?)
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16 Jan 2015, 02:45
pacifist85 wrote: Question 11,
I guess it should have stated that we can only unse each one of these 3 nunnbers only once, right? Because, I used the same approach as Bunuel, but without knowing that you can only use these nubers once the process an be very long... and I am not even sure if it would end up in the same result  i guess not.
Is there sth in the way that the question is phrased that should have warned me that these numbers were only to be used once? We are given a set: {1, 3, 6, 7, 7, 7}. I think it should be clear that 1, 3, and 6 should be used once, and 7 thrice.
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Re: 12 Easy Pieces (or not?)
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08 Dec 2015, 12:25
There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color? A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1
I totally understand that we don't need to do the calculations here because the answer is obvious. However, I would like to know how we do calculate the answer to get to the official answer.



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Re: 12 Easy Pieces (or not?)
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08 Dec 2015, 20:32
Peltina wrote: There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four socks are picked at random what is the probability of getting two socks of the same color? A. 1/5 B. 2/5 C. 3/4 D. 4/5 E. 1
I totally understand that we don't need to do the calculations here because the answer is obvious. However, I would like to know how we do calculate the answer to get to the official answer. Hi, in such questions, we look at the worst scenario.. here it would be getting different coloured socks , every time we pick socks.. but there are only three different coloured socks, so as a worst case , the first three picked up are different colours.. the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour.. hence prob is 1, that is it is sure to have two socks of atleast one colour..
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Re: 12 Easy Pieces (or not?)
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09 Dec 2015, 14:34
chetan2u wrote: Hi,
in such questions, we look at the worst scenario.. here it would be getting different coloured socks , every time we pick socks..
but there are only three different coloured socks, so as a worst case , the first three picked up are different colours.. the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour.. hence prob is 1, that is it is sure to have two socks of atleast one colour.. Thank you! I was looking more for the formula that we should use to get to the answer P=1 in case we fail to notice that the answer is obvious and needs no calculation.



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Re: 12 Easy Pieces (or not?)
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09 Dec 2015, 22:39
Peltina wrote: chetan2u wrote: Hi,
in such questions, we look at the worst scenario.. here it would be getting different coloured socks , every time we pick socks..
but there are only three different coloured socks, so as a worst case , the first three picked up are different colours.. the fourth which you pick up has to be one of these three colours, thus ensuring that we have two socks of same colour.. hence prob is 1, that is it is sure to have two socks of atleast one colour.. 5 pairs of white, 3 pairs of black and 2 pairs of grey socks Thank you! I was looking more for the formula that we should use to get to the answer P=1 in case we fail to notice that the answer is obvious and needs no calculation. Hi, there cant be direct formulas for most of the questions but they are just a step to correct answer.. so if you want the answer the formula way, then it would be something like this... there are three different coloured socks.. let us first find the scenario where we do not get two of one kind... 1)the first can be picked up, say white, prob=5/10.. 2)the second can be picked up, say black, prob=3/9.. 3)the third can be picked up, the remaining grey, prob=2/8.. 4)the fourth picking up is 0 as no other colour is left prob=0/7.. now these four can be arranged in 4!/2! ways.. prob of two colours of one kind = 1 5/10 *3/9 * 2/8* 0/7 * 4!/2!=10=1
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Re: 12 Easy Pieces (or not?) &nbs
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