If you select two cards from a pile of cards numbered 1 to 10, what is

Question

If you select two cards from a pile of cards numbered 1 to 10, what is the probability that the sum of the numbers is less than the average of the pile?

(A) 1/100
(B) 2/45
(C) 2/25
(D) 4/45
(E) 1/10

sudh · Accepted Answer

Average of the pile  \frac{1+10}{2}  = 5.5

Pairs whose sum  < 5.5
(1,2), (1,3), (1,4) ---> 3
(2,1), (2,3) --------- -> 2
(3,1), (3,2) --------- -> 2
(4,1) --- --------> 1; Total 3+2+2+1 = 8

Total number of outcomes 
(1,2), (1,3) ........... (1,10) --> 9
(2,1), (2,3) ........... (2,10) --> 9
.
.
.
(10,1), (10,2) ...... (10,9) --> 9; Total 10*9 = 90

p =  \frac{favorable outcome}{possible outcome} 
 =  \frac{8}{90} 
 =  \frac{4}{45}

Answer D

Lucky2783 · Answer

brute force approach :

total # ways =  ^1^0C_2=45 
favorable cases : 
3 (1,2)
4 (1,3)
5 (3,2) (1,4) 
4 cases

answer  \frac{4}{45}  D

SherLocked2018 · Answer

I solved this Q in 2 ways, and got 2 different answers.
#1 
Total no of ways : 10C2 = 45. 
(Sum<5) ways : {1,2},{1,3},{1,4},{2,3}. 4 cases.
Thus 4/45. D

But I got a bit confused with a similar Question of Rolling dice. 
Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?
Expected value of a roll of one dice is 1/6(1+2+3+4+5+6)=3.5.
Expected value of three dices is 3*3.5=10.5.
Mary scored 10 so the probability to have more then 10, or more then average is the same as to have less than average=1/2.
P=1/2.

I dont understand the differences in these 2 questions. I thought that maybe we can use the same approach here as well, but this would give different answers. 
And moreover, im not sure if we should use same approach. 
Kindly explain. Thanks!

Zhenek · Answer

The only reason this approach worked for the given example, imo, is because we got perfect values, if the case was abit different (say, Mary scored 9 and not 10), then this approach wouldn't work at all.
To explain it in abit more detail, say we got that 10,5 point average, ye, the odds that you score above it are same as if you score below it, right? Coz its an average that is not part of the set of numbers that you can roll and coz its average, you got same amount of numbers "to the right" as you got "to the left" from it, so that means that the odds that you score above 10,5 (11, 12, 13 ,14,15,16,17,18) are same as the odds that you score below 10,5 (10, 9, 8, 7,6 ,5, 4, 3). The question was " what are the odds that we roll above 10" (not below! We had to outscore her, keep in mind if the question was "what are the odds that we score LESS than her", it wouldn't be 1/2 at all), this perfectly matches our scenario about the odds of scoring above 10,5, which is nothing else but 1/2.
In our question if you try to interpret it as rolling 2 1-10 dices, your average will be 11 and the score you have to roll lower than is 6 (5 4 3 2 1). Not only you can't really relate averages to this "roll below 6"condition but also its actually impossible to set the same example for this case. The reason is simple:
Odds of rolling (2 3 4 5 6 7 8 9 10) and odds of rolling 12 13 14 15 16 17 18 19 20 (above and below average) are same but they are  not 1/2  because you can also roll 11. (But you can't roll 10,5 in joe&mary's case).

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

The first hurdle here is interpreting the question. To paraphrase, if I were to choose two random cards, would their sum be less than a certain other number. This is essentially a probability question, as evidenced by the answer choices as fractions. However there are a couple of elements to keep in mind. The first task is to determine the average of the pile.

Given 10 numbers, we could simply sum them up and divide by 10, but it’s probably much faster to recognize that the mean of an evenly spaced set is equal to the median of the set. A set with 10 numbers has a median that’s the average of the 5th and 6th elements (Not the Bruce Willis movie). Conveniently, the 5th element is 5 and the 6th element is 6, yielding an average of 5.5. Since we’re dealing with integers, we must now determine the number of possibilities that give a sum of 5 or less.

The options are limited enough that we can just reason out the choices. A good strategy is just to assume that the first card is a 1, and figure out what numbers work for the second number. If we pick 1, the next smallest card is 2. Thus the possibility (1,2) works. Similarly, we can see that (1,3) and (1,4) will work. (1,5) is too big, so we can stop there as any other option would only be bigger than this benchmark. It’s worth noting that the question is set up so that there’s no repetition, thus the option (1,1) cannot be considered. If the first card picked is a 1, there are three options that will keep the average below 5.5 (like a Russian judge at the Winter Olympics).

Next, supposing that the first card were a 2, there would be the separate option of (2,1). Since the order matters, (2,1) is not the same as the aforementioned (1,2). This is another valid choice. (2,2) is eliminated because of duplication, leaving us only with (2,3) that will also work if the first card is a 2. Since (2,4) is too big, we don’t need to examine any further. That’s two more options to add to our running tally.

Continuing, if the first card were a 3, then (3,1) and (3,2) would work. (3,3) is above the average, and it is a duplicate, so it can be eliminated for either reason. That gives us two more options for our running tally. The final option is to start with a 4, giving (4,1). Anything bigger is above the average. Similarly, anything starting with 5, 6, 7, 8, 9 or 10 will be above the average. Only eight options work out of all the possibilities.

The question is almost over, but there is one final trap we need to avoid before locking in our answer. The stimulus purported 10 different cards to select. If we were to compile all the possibilities, a natural total to think of would be 100 (10×10). However, since there is no replacement, we’re first selecting from 10 choices, and then from 9 choices. Exactly as a permutation of two selections out of 10, this gives us a total of 90 possible choices. If there are eight options that satisfy the conditions out of 90 choices, then the correct answer must be 8/90, which simplifies to 4/45.

Answer choice D.

Examining the answer choices, we can see some of the more obvious traps. Compiling eight options out of 100 choices would give us the erroneous 2/25 fraction in answer choice C. Overlooking the lack of replacement would give us 10 total choices (the same eight plus (1,1) and (2,2) out of 100 possibilities, or answer choice E. The exam is designed to ask tricky questions, which means that the answer choices will often be answers you can get if you make a single calculation error or unfounded assumption. Be vigilant until the end of the question, as you don’t want to spend a full two minutes on a complicated question just to falter at the finish line. Questions can have many aspects to consider and many steps to execute, but by continuously thinking in a logical manner, you can solve any GMAT question. Remember that even the longest journey begins with a single step.

mvictor · Answer

very tricky one...
i counted both options
1,2 and 2,1...and I guess this is why I made the mistake.

nevertheless, we can rule out A and E right away.

a true 700 lvl question.

dlabarb · Answer

Wonderful question! Please post more

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