math-eval/TAL-SCQ5K huggingface dataset:
Problem: In a regular hexagonal prism, how many pairs of parallel edges can you find? 
Options:
aoVal: A
content: $$33$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$24$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$36$$ 
Answer: A
Analysis: $\_4C\_2\times3+\_6C\_2=33$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: What is the sum of all the even numbers between $30$ and $40$? 
Options:
aoVal: A
content: $$72$$ 
aoVal: B
content: $$108$$ 
aoVal: C
content: $$140$$ 
aoVal: D
content: $$144$$ 
Answer: C
Analysis: Even number between 30 and 40: 32,34,36,38  $$32+34+36+38=140$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: How many of the following statements are wrongt?  Statement $1$: The probability of a certain event to happen is $$1$$.  Statement $2$: Indefinite events include impossible events.  Statement $3$: The probability of an impossible event to happen is $$0$$.  Statement $4$: The probability of an indefinite event to happen is between $$0$$ and $$1$$. 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$1$$ 
aoVal: C
content: $$2$$ 
aoVal: D
content: $$3$$ 
Answer: B
Analysis: Impossible events are definite events.  $$\text{Statement 2}$$ is wrong.  $$\text{Statement 1}$$, $$\text{Statement 3}$$, and $$\text{Statement 4}$$ are right.  Thus, the answer is $$\text{D}$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: A drawer contains ten identical yellow socks, eight identical blue socks and four identical pink socks. Amrita picks socks from the drawer without looking.  What is the smallest number of socks she must pick to be sure that she has at least two pairs of matching socks? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$13$$ 
Answer: B
Analysis: Two matching pairs of socks could be obtained by choosing four socks, but this is not certain. For instance, two of one colour, one of a second colour and one of a third colour could be drawn. Combinations of five chosen socks would give two matching pairs unless three socks of one colour, one of a second colour and one of a third colour were chosen. When this is the case, drawing a sixth sock would guarantee that there would be two matching pairs as there would now be either four socks of one colour plus two other socks or three socks of one colour, two of a second colour plus one of a third colour. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Pigeonhole Principle->Worst Case in Pigeonhole Principle Problems
Problem: Peter bought a carpet 36dm wide and 60dm long. The figure shows part of this carpet. As seen, the carpet has a small squares contianing either a sun or a moon. You can count that along the width there are nine squares. When the carpet is fully unrolled, how many moons will be seen?  [Insert pic] 
Options:
aoVal: A
content: $$68$$ 
aoVal: B
content: $$67$$ 
aoVal: C
content: $$65$$ 
aoVal: D
content: $$63$$ 
aoVal: E
content: $$60$$ 
Answer: B
Analysis: Each small square is 36/9 = 4 dm on a side. there are 60/4 = 15 squares along the length.  Both dimensions, 15 and 9, are odd. Also the corners are suns, so there is one more sun than there are moons.  Thus, there are a total of (15 x 9 - 1)/2 = 67 moons in the carpet when fully rolled. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures->Classifying and Enumerating->Counting Figures Formed by Grid Points
Source: 其它
Problem: A top hat contains $$3$$ red chips and $$2$$ green chips. Chips are drawn randomly, one at a time without replacement, until all $$3$$ of the reds are drawn or until both green chips are drawn. What is the probability that the $$3$$ reds are drawn? (2016 AMC 8 Problems, Question \#21) 
Options:
aoVal: A
content: $$\frac{3}{10}$$ 
aoVal: B
content: $$\frac{2}{5}$$ 
aoVal: C
content: $$\frac{1}{2}$$ 
aoVal: D
content: $$\frac{3}{5}$$ 
aoVal: E
content: $$\frac{7}{10}$$ 
Answer: B
Analysis: There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out $3$ red chips, $3$ red chips and $1$ green chip, $2$ green chips, $2$ green chips and $1$ red chip, and $2$ green chips and $2$ red chips. Because order is important in this problem, there are $1+4+1+3+6=15$ ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which $15-5=10$. Out of the $10$ ways to end the game, $4$ of them ends with a green chip. The answer is $\frac{4}{10}=\frac{2}{5}$, or $(\mathbf{B}) \frac{2}{5}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability
Source: 其它
Problem: A box contains blue marbles. Another two boxes contain only white marbles.  Label on Box $$\rm A$$: white marbles  Label on Box $$\rm B$$: blue marbles  Label on Box $$\rm C$$: Box $$\rm B$$ contains blue marbles  Which box contains blue marbles if two of the above labels are wrong? 
Options:
aoVal: A
content: Box A 
aoVal: B
content: Box B 
aoVal: C
content: Box C 
Answer: C
Analysis: If Box $$\rm A$$ has the wrong label, it contains blue marbles. This indicates that Box $$\rm B$$ has the wrong label too, because only one box has blue marbles.  It also means that Box $$\rm C$$ has the wrong label. We will then have a scenario of three wrong labels, so Box $$\rm A$$ cannot be the one having the wrong label.  If Box $$\rm A$$ has the right label, Box $$\rm B$$ and Box $$\rm C$$ will have the wrong labels.  Box $$\rm A\rightarrow$$ white marbles  Box $$\rm B\rightarrow$$ white marbles  Box $$\rm C\rightarrow$$ blue marbles  Box $$\rm C$$ contains blue marbles. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning using Hypothesis
Problem: Each of the 5 sides and the 5 diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color? 
Options:
aoVal: A
content: $\frac{2}{3}$ 
aoVal: B
content: $\frac{105}{128}$ 
aoVal: C
content: $\frac{125}{128}$ 
aoVal: D
content: $\frac{253}{256}$ 
aoVal: E
content: $$1$$ 
Answer: D
Analysis: Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear. After drawing the regular pentagon out, note the topmost vertex; it has $4$ sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.  Case 1: all $4$ are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection.  Case $2$: $3$ are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.  Case $3$: $2$ are one color and $2$ are of the other color. Using the same logic as previously, we can color the pentagon $2$ different ways by inspection to satisfy the requirements. There are $\left(\begin{array}{l}4 \textbackslash\textbackslash{} 2\end{array}\right)$ ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of $6 \cdot 2=12$ ways to color the pentagon so that no such triangle has the same color for all of its sides. These are all the cases, and there are a total of $2^{10}$ ways to color the pentagon. Therefore the answer is $1-\frac{12}{1024}=1-\frac{3}{256}=\frac{253}{256}=D$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: In my hometown, city streets are numbered with odd numbers in increasing order from south to north and with even numbers in increasing order from west to east. In what direction must I travel if I want to go directly from $$241\text{st}$$ Street to $$225\text{th}$$ Street? 
Options:
aoVal: A
content: $$$$north 
aoVal: B
content: $$$$south 
aoVal: C
content: $$$$east 
aoVal: D
content: $$$$west 
Answer: B
Analysis: In my hometown, city streets are numbered with odd numbers in increasing order from south to north. I must travel from north to south to go directly from $$241\text{st}$$ Street to $$225\text{th}$$ Street.~ When I travel from north to south, I travel south. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Directions and Coordinates->Directions
Problem: Anna went to travel by bike. She started at $8$ am. But she had to stop for $15$ minutes on the way and arrived at $12$ am. How long did Anna ride her bicycle? (adapted from 2011 Math kangaroo Problems, Level 3-4 , Question \#4) 
Options:
aoVal: A
content: $3$ hours $$45$$ min 
aoVal: B
content: $3$ hours $$35$$ min 
aoVal: C
content: $2$ hours $$45$$ min 
aoVal: D
content: $3$ hours $$25$$ min 
aoVal: E
content: $3$ hours $$45$$ min 
Answer: A
Analysis: $$12:00-8:00=4$$ h  $4$h- $15$min =$3$ h $45$min 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Problem: Peter rolls two standard six-sided dice at the same time. What is the probability that the sum of both rolls is a prime number less than $6$? 
Options:
aoVal: A
content: $\dfrac{7}{36}$. 
aoVal: B
content: $\dfrac{1}{6}$. 
aoVal: C
content: $\dfrac{5}{36}$. 
aoVal: D
content: $\dfrac{1}{9}$. 
aoVal: E
content: $\dfrac{5}{12}$. 
Answer: A
Analysis: There are $7$ out of $36$ outcomes which are prime numbers under $6$: $1+1$, $1+2$, $1+4$, $2+1$, $2+3$, $3+2$, $4+1$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Abe has $1$ green and $1$ red jelly bean in his hand. Bob has $1$ green and $2$ yellow jelly beans in his hand. Each randomly picks a jelly bean to show to the other. What is the probability that the colours match? ． 
Options:
aoVal: A
content: $\dfrac{1}{2}$ 
aoVal: B
content: $\dfrac{1}{3}$ 
aoVal: C
content: $\dfrac{3}{4}$ 
aoVal: D
content: $\dfrac{1}{6}$ 
aoVal: E
content: $\frac{1}{9}$ 
Answer: D
Analysis: The probability that both show a green bean is $\dfrac{1}{2}\cdot \dfrac{1}{3}=\dfrac{1}{6}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: Mary, Nancy, and Peter are playing archery. They have $$7$$ identical arrows in total. Each of them must shoot at least once. How many different ways can they shoot the arrows? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$15$$ 
Answer: D
Analysis: There are 3 different people here, so order does matter. Each of them must shoot at least once:  $$7=1+1+5$$  $$7=1+2+4$$  $$7=1+3+3$$  $$7=1+4+2$$  $$7=1+5+1$$  $\textasciitilde$  $$7=2+1+4$$  $$7=2+2+3$$  $$7=2+3+2$$  $$7=2+4+1$$  $\cdots$  And so on. If we keep listing, total will be:  $$5+4+3+2+1=15$$ ways 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: There are $2$ black balls and $3$ white balls in an opaque bag. They are of identical shape, size and quality except for color. Under the condition that the balls cannot be seen, take out $3$ balls from the bag at a time randomly. Among the following events, the certain one is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: Among the $3$ balls taken out, at least $1$ is white. 
aoVal: B
content: Among the $3$ balls taken out, at least $1$ is black. 
aoVal: C
content: Among the $3$ balls taken out, at least $2$ are black. 
aoVal: D
content: Among the $3$ balls taken out, at least $2$ are white. 
Answer: A
Analysis: $$\text{A}$$ is a certain event.  $$\text{B}$$ is a random event, so it\textquotesingle s wrong.  $$\text{C}$$ is a random event, so it\textquotesingle s wrong.  $$\text{D}$$ is a random event, so it\textquotesingle s wrong. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: During Black Friday, a store earned a revenue of $80000$ dollars in one week. How much money can the store earn in a year, approximately, considering there are $52$ weeks in a year? 
Options:
aoVal: A
content: 4.16 million dollars 
aoVal: B
content: 4.17 million dollars 
aoVal: C
content: 4.18 million dollars 
aoVal: D
content: unable to determine 
Answer: D
Analysis: The sample is biased becuase Black Friday is a promotion which allowed the store to earn way much more than they usually earn. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Four friends, Edwin, Fred, Gary and Howard, were playing together  when one of them broke a vase.  The teacher asked: "Who is the culprit?"  Both Edwin and Howard said, "Not me."  Fred said, "Howard broke the vase."  Gary said, "Fred is the culprit."  If only one of four boys was lying,~\uline{~~~~~~~~~~}~broke the vase. 
Options:
aoVal: A
content: Edwin 
aoVal: B
content: Fred 
aoVal: C
content: Gary 
aoVal: D
content: Howard 
Answer: B
Analysis: Either Fred or Howard must be lying since what they said did not tally.  Since only one person was lying, Gary was telling the truth i.e, Fred broke the vase. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: SASMO 2015 P2 Q2  For the number sentence below, what is the answer?  $$15+15+15+15+15+15+15+15+15+15+15+15+15+15+15$$ 
Options:
aoVal: A
content: $$210$$ 
aoVal: B
content: $$215$$ 
aoVal: C
content: $$220$$ 
aoVal: D
content: $$225$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: 7 x 30 + 15 = 210 + 15 = 225 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: $$2$$ beds and $$4$$ chairs have a total mass of $$240$$$$\text{kg}$$. $$1$$ bed weighs as much as $$6$$ chairs. Find the mass of $$1$$ chair. 
Options:
aoVal: A
content: $$12$$ $$\text{kg}$$ 
aoVal: B
content: $$15$$ $$\text{kg}$$ 
aoVal: C
content: $$24$$ $$\text{kg}$$ 
aoVal: D
content: $$30$$ $$\text{kg}$$ 
Answer: B
Analysis: $$240\div16=15\text{kg}$$ 
Knowledge points: Overseas In-curriculum->Knowledge Point->Algebra-> Numbers, Letters and Equations->Equivalent Substitution->Direct Substitution
Overseas Competition->Knowledge Point->Counting Modules
Problem: A box contains seven cards, each with a different integer from $$1$$ to $$7$$ written on it. Avani takes three cards from the box and then Niamh takes two cards, leaving two cards in the box. Avani looks at her cards and then tells Niamh "I know the sum of the numbers on your cards is even." What is the sum of the numbers on Avani\textquotesingle s cards? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$11$$ 
aoVal: E
content: $$12$$ 
Answer: E
Analysis: To obtain an even number when adding two integers, both integers must be even or both integers must be odd. Therefore the four integers remaining once Avani has removed her three integers must all be odd or all be even or there would be a possibility that the sum of Niamh\textquotesingle s two integers could be odd. Since there were four odd integers and three even integers on the cards in the box initially, the integers on the cards remaining once Avani has removed her cards are all odd. Therefore the cards Avani removed had the three even integers $$2$$, $$4$$ and $$6$$ written on them which have sum $$12$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Nine chairs are in a straight line and numbered $$1$$ to $$9$$ from left to right. Five girls and four boys sit in the chairs so that no girl is next to another girl. A boy could be sitting in the chair with which number? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$5$$ 
Answer: C
Analysis: The girls must be in the odd-numbered seats; otherwise, two girls would be seated next to each other. This leaves only the even-numbered seats for the boys, so a boy is seated in seat $$4$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Fill in the blank:~\uline{~~~~~~~~~~}~is 2 tens 8 ones less than 5 tens 5 ones. 
Options:
aoVal: A
content: $$27$$ 
aoVal: B
content: $$37$$ 
aoVal: C
content: $$73$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: 55 - 28 = 27 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: In Tim\textquotesingle s class, there are $$20$$ students who can swim, $$25$$ students who can play basketball, and $$10$$ students who can do both. If everyone in the class plays at least one sport, how many students are there in Tim\textquotesingle s class?~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$32$$ 
aoVal: B
content: $$33$$ 
aoVal: C
content: $$34$$ 
aoVal: D
content: $$35$$ 
Answer: D
Analysis: $$20+25-10=35$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Alice\textquotesingle s average score on the first four assignments is $$89$$. She got $$94$$ on the fifth assignment. What is Alice\textquotesingle s average score on all five assignments? 
Options:
aoVal: A
content: $$90$$ 
aoVal: B
content: $$91$$ 
aoVal: C
content: $$92$$ 
aoVal: D
content: $$93$$ 
Answer: A
Analysis: $$89\times4=356$$  $$356+94=450$$  $$450\div5=90$$ 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)
Problem: Jack, Sarah, and Jimmy participated in a Maths competition.  \textbf{Jack says: "I won the competition."}  \textbf{Sarah says: "I didn\textquotesingle t win the competition."}  \textbf{Jimmy says: "Jack didn\textquotesingle t win the competition."}  Only one of them told the truth. Who won the Maths competition? 
Options:
aoVal: A
content: $$$$Jack 
aoVal: B
content: Sarah 
aoVal: C
content: Jimmy 
Answer: B
Analysis: We can spot that Jack\textquotesingle s statement and Jimmy\textquotesingle s statement contradict each other, so one of them is telling the truth. Therefore, Sarah tells a lie. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Maria had $$28$$ dreams last month. If $$16$$ of them involved monkeys, $$15$$ involved squirrels, and $$4$$ involved no animals, then at least how many dreams involved both monkeys and squirrels? 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$11$$ 
Answer: B
Analysis: Maria had $$28$$ dreams last month, $$24$$ of which involved animals. Since $$16+ 15 =31$$ involved moneys or squirrels, then at least $$31 - 24 = 7$$ dreams involved both monkeys and squirrels. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Source: 其它
Problem: Joann and Sana have $20$ dresses in total. Joann gives half of her dresses to Claire, and then she and Sana have $14$ dresses in total. How many dresses does Sana have? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$8$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$14$$ 
Answer: B
Analysis: Joann gives $20-14=6$ dresses to Claire.  Joann has $6+6=12$ dresses originally.  Sana has $20-12=8$ dresses. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Vansen left for school at $6:30$ am. He took half an hour to walk from his house to school. Ron reached school $1$ hour later than Vansen. At what time did Ron reach school? (adapted from 2009 Math Kangaroo Problem, Level 3-4, Question \#8) 
Options:
aoVal: A
content: $7:00$ am. 
aoVal: B
content: $8:00$ am. 
aoVal: C
content: $8:45$ am. 
aoVal: D
content: $9:00$ am. 
aoVal: E
content: $10:00$ am. 
Answer: B
Analysis: Half an hour after $6:30$ is $7:00$  $7+1=8$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Source: 其它
Problem: In a certain country, a part of the residents can speak English only, a part can speak French only and the rest can speak both languages. It is known that $$85\textbackslash\%$$ residents can speak English and $$75\textbackslash\%$$ can speak French. What percent of the residents of this country can speak both English and French? ($$2002$$ Math kangaroo Problem, Level $$7-8$$, Question \#$$17$$) 
Options:
aoVal: A
content: $$50\textbackslash\%$$ 
aoVal: B
content: $$57\textbackslash\%$$ 
aoVal: C
content: $$25\textbackslash\%$$ 
aoVal: D
content: $$60\textbackslash\%$$ 
aoVal: E
content: $$40\textbackslash\%$$ 
Answer: D
Analysis: $85\textbackslash\%+75\textbackslash\%-100\textbackslash\%=60\textbackslash\%$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: Tim, an adventurer, found three treasure chests in the cave. Only one of the chests contains the treasure. The following clues are written on the three treasure chests respectively:  \textbf{Treasure chest A: "The treasure is not in Treasure chest C!"}  \textbf{Treasure chest B: "The treasure is not here!"}  \textbf{Treasure chest C: "The treasure is here!"}  Given that only one sentence of the above three is true, which chest should Tim open to get the treasure? 
Options:
aoVal: A
content: A 
aoVal: B
content: B 
aoVal: C
content: C 
Answer: B
Analysis: The clues on treasure chest A and C are conflicting, so either A or C is telling the truth. Thus, the clue on treasure chest B is a lie. Therefore, the treasure is in chest B. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning using Hypothesis
Problem: Eddie, Avril and Pip want to share 10 cakes. Eddie only wants to get 3 cakes. Avril and Pip want to get at least 1 cake. How many ways are there to share the cakes? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: C
Analysis: $$Omitted.$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
Source: 其它
Problem: The correct information James can get from the statement "there is a $11\textbackslash\%$ chance that it will rain tomorrow in Los Angeles"~is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: It will rain $11\textbackslash\%$ of the time tomorrow in Los Angeles. 
aoVal: B
content: It will rain in $11\textbackslash\%$ of the regions in Los Angeles tomorrow. 
aoVal: C
content: It will definitely rain tomorrow in Los Angeles. 
aoVal: D
content: The probability of raining in Los Angeles is high. 
aoVal: E
content: The probability of raining in Los Angeles is low. 
Answer: E
Analysis: It is less likely to rain tomorrow in Los Angeles. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它