Question 1

A fair six-sided die is rolled three times. What is the probability that the face with number 6 does not appear on all three rolls (that is, it is not the case that all three rolls show 6)?

Accepted Answer

Introduction / Context:
This problem involves repeated rolling of a fair six-sided die and asks for the probability that the outcome is not six on all three rolls simultaneously. In other words, we want the probability that it is not true that each of the three rolls shows a 6. This is a classic use of the complement rule in probability.
Given Data / Assumptions:

A fair six-sided die is rolled three times.
Each roll is independent.
Each face (1 through 6) has probability 1/6 on any single roll.
We are interested in the event that not all three rolls show 6.

Concept / Approach:
It is easier to compute the probability of the complementary event where all three rolls show 6, and then subtract this from 1. The event "all three rolls are 6" is straightforward to evaluate because the rolls are independent. Once that is known, the required probability is 1 minus this value.
Step-by-Step Solution:
Probability that one roll shows 6 is 1/6.
Because the rolls are independent, probability that all three rolls show 6 is (1/6) * (1/6) * (1/6) = 1 / 6^3 = 1 / 216.
Let A be the event that all three rolls are 6.
We need the probability of not A, that is, the event that not all three rolls are 6.
P(not A) = 1 - P(A).
So P(not all three are 6) = 1 - 1/216 = (216/216) - (1/216) = 215/216.

Verification / Alternative check:
We can check that this is very close to 1, which is reasonable, because the event that all three rolls are 6 is extremely rare. The chance 1/216 is small, so its complement 215/216 is large. Also, the result is consistent with the idea that only one outcome out of the 216 total ordered triplets corresponds to (6,6,6), while all other 215 outcomes belong to the desired event.
Why Other Options Are Wrong:
1/216: This is the probability that all three rolls are 6, which is the complementary event, not the required one.
215/256: This denominator 256 is not appropriate for three rolls of a six-sided die, because the correct total number of ordered outcomes is 6^3 = 216.
1/52: This number is unrelated to this experiment and likely comes from thinking about cards, not dice.
None of these: This is incorrect because 215/216 is listed among the options and is correct.
Common Pitfalls:
A usual misunderstanding is to interpret the question as "no 6 appears at all", which would give probability (5/6)^3 instead of the complement of all three being 6. Reading the wording carefully is essential. Another mistake is miscalculating 6^3 or forgetting to use the complement rule when it offers a simpler solution.
Final Answer:
Thus, the probability that it is not the case that all three rolls show 6 is 215/216.

Question 2

In a class there are 15 boys and 10 girls. Three students are selected at random. What is the probability that the selected group consists of exactly 1 girl and 2 boys?

Accepted Answer

Introduction / Context:
This is a selection problem involving two categories, boys and girls. The class has 15 boys and 10 girls, and three students are chosen without replacement. We are asked to find the probability that the group contains exactly 1 girl and 2 boys. This tests competence with combinations and probability based on equally likely subsets.
Given Data / Assumptions:

Number of boys = 15.
Number of girls = 10.
Total students = 25.
Three students are selected at random without replacement.
We want exactly 1 girl and 2 boys.

Concept / Approach:
We use combinations to count the number of ways to choose any 3 students from 25. Then we count the number of favourable selections that contain exactly 1 girl and 2 boys, by choosing 1 girl from the 10 available and 2 boys from the 15 available. The probability is the ratio of favourable selections to total selections.
Step-by-Step Solution:
Total students = 25.
Total number of ways to choose 3 students = C(25,3).
Compute C(25,3) = 25 * 24 * 23 / (3 * 2 * 1) = 2300.
Number of ways to choose exactly 1 girl from 10 = C(10,1) = 10.
Number of ways to choose 2 boys from 15 = C(15,2).
Compute C(15,2) = 15 * 14 / 2 = 105.
Total favourable selections = 10 * 105 = 1050.
Required probability = favourable / total = 1050 / 2300.
Simplify 1050 / 2300 by dividing numerator and denominator by 50 to get 21 / 46.

Verification / Alternative check:
We can approximate the value 21/46 numerically as about 0.4565. This is slightly less than one half, which is reasonable given that there are more boys than girls, but the combination requirement restricts the composition. The numerator and denominator have no common factor other than 1, so 21/46 is in simplest form, confirming the correctness of simplification.
Why Other Options Are Wrong:
1/5: This is 0.2, which is too small given that having exactly 1 girl and 2 boys is a fairly likely outcome in this class.
3/25: This is 0.12 and does not correspond to the ratio 1050/2300.
1/50: This is 0.02, which is unrealistically small for this configuration.
None of these: This is incorrect because 21/46 matches the correct calculation and is listed among the options.
Common Pitfalls:
Common mistakes include counting permutations instead of combinations, or misreading the requirement and including cases with 0 or 2 girls. Another error is miscomputing C(25,3) or C(15,2) due to arithmetic slips. Systematic use of the combination formula and good arithmetic practices help to avoid these issues.
Final Answer:
Therefore, the probability that exactly 1 girl and 2 boys are selected is 21/46.

Question 3

An unbiased six-sided die is thrown once. What is the probability of getting a number that is a multiple of 2 or a multiple of 5?

Accepted Answer

Introduction / Context:
This question asks for the probability that the outcome of a single roll of a fair six-sided die is either a multiple of 2 or a multiple of 5. It is a straightforward application of set union in a finite sample space and tests understanding of multiples and basic counting.
Given Data / Assumptions:

The die is fair and has faces numbered 1, 2, 3, 4, 5, 6.
Each face is equally likely, so each outcome has probability 1/6.
We are interested in numbers that are multiples of 2 or multiples of 5.

Concept / Approach:
We list the outcomes that are multiples of 2, the outcomes that are multiples of 5, and then take their union. If there is any overlap, we make sure not to double count. Finally, the probability is the count of favourable outcomes divided by 6.
Step-by-Step Solution:
Possible outcomes on a six-sided die: 1, 2, 3, 4, 5, 6.
Multiples of 2 in this set: 2, 4, 6.
Multiples of 5 in this set: 5.
Union of {multiples of 2} and {multiples of 5} = {2, 4, 5, 6}.
There is no overlap between these sets because no number from 1 to 6 is simultaneously a multiple of both 2 and 5.
Number of favourable outcomes = 4 (namely 2, 4, 5, 6).
Total number of outcomes = 6.
Required probability = favourable / total = 4 / 6.
Simplify 4 / 6 by dividing numerator and denominator by 2 to get 2 / 3.

Verification / Alternative check:
We can check quickly by noticing that the only numbers not in the favourable set are 1 and 3, which are 2 of the 6 possible outcomes. Therefore, the complement probability is 2/6 = 1/3. The required probability is then 1 minus this complement: 1 - 1/3 = 2/3. This matches the earlier direct computation.
Why Other Options Are Wrong:
1/3: This is the probability of the complement event (getting 1 or 3) rather than the event of interest.
1/2: This would imply exactly three favourable outcomes, but there are four.
1/6: This would correspond to only one favourable outcome, which is clearly incorrect.
None of these: This is incorrect because 2/3 is explicitly listed and is the correct answer.
Common Pitfalls:
One common confusion is failing to work with the union of the sets correctly, leading to either missing a favourable outcome or double counting. Another mistake is misidentifying multiples, especially when quickly scanning numbers. Listing the possible outcomes and marking those that satisfy each condition helps avoid such errors.
Final Answer:
Therefore, the probability of getting a multiple of 2 or a multiple of 5 is 2/3.

Question 4

A single card is drawn at random from a well-shuffled standard deck of 52 playing cards. What is the probability that the card is either a heart or a diamond?

Accepted Answer

Introduction / Context:
This is another basic probability question with playing cards. The deck is standard and well shuffled, and we want the probability that the drawn card belongs to one of the two red suits, hearts or diamonds. This reinforces understanding of suit structure and simple probability ratios.
Given Data / Assumptions:

A standard deck of cards has 52 cards.
There are four suits: hearts, diamonds, clubs, and spades.
Hearts and diamonds are red suits; clubs and spades are black suits.
Each suit has 13 cards.
We want the probability that the card is a heart or a diamond.

Concept / Approach:
The set of hearts and diamonds together forms all red cards in the deck. This is exactly half of the deck. We can either reason directly from symmetry or explicitly count the number of hearts and diamonds and divide by 52.
Step-by-Step Solution:
Number of hearts in the deck = 13.
Number of diamonds in the deck = 13.
Total favourable cards (hearts or diamonds) = 13 + 13 = 26.
Total cards in the deck = 52.
Required probability = favourable / total = 26 / 52.
Simplify 26 / 52 by dividing numerator and denominator by 26 to get 1 / 2.

Verification / Alternative check:
A symmetry argument provides a quick check. Two of the four suits are hearts and diamonds. Since all suits have equal size and the deck is well shuffled, the chance of drawing a card from any particular pair of suits is exactly 2 out of 4, which equals 1/2. This is consistent with the explicit fraction based calculation.
Why Other Options Are Wrong:
1: This would mean certainty that the card is a heart or diamond, which is not true because it could be a club or spade.
3/4: This would correspond to 39 favourable cards out of 52, which is too many and incorrectly includes black suits.
1/3: This would imply about 17 favourable cards, which does not match the actual count of 26.
None of these: This is incorrect because 1/2 is in the list and is the correct probability.
Common Pitfalls:
Some learners confuse hearts and diamonds with one single suit or incorrectly remember the number of cards in each suit. Others forget that the total deck has 52 cards and miscalculate the fraction. Remembering that there are four suits, each with 13 cards, and that two of them are red, makes the problem straightforward.
Final Answer:
Hence, the probability that the card drawn is either a heart or a diamond is 1/2.

Question 5

From a well shuffled standard pack of 52 playing cards, one card is drawn at random. What is the probability that the card obtained is either a red card or a black card?

Accepted Answer

Introduction / Context:
In this question we work with a standard pack of playing cards and basic probability. The aim is to understand how to calculate the probability of an event that actually covers the entire sample space, namely drawing either a red card or a black card from a standard deck of 52 cards.
Given Data / Assumptions:

There is a standard deck with 52 cards.
There are 26 red cards (hearts and diamonds) and 26 black cards (spades and clubs).
The deck is well shuffled, so every card is equally likely to be drawn.
One card is drawn at random.
We want the probability that the card is red or black.

Concept / Approach:
Probability of an event = (number of favourable outcomes) / (total possible outcomes). Here, the event is "card is red or black". In a standard deck every card is either red or black, so the favourable outcomes are all 52 cards. This is an example of an event that is certain, so its probability is equal to 1.
Step-by-Step Solution:
Total number of cards in the deck = 52
Number of red cards = 26
Number of black cards = 26
Total favourable outcomes (red or black) = 26 + 26 = 52
Probability = favourable / total = 52 / 52 = 1
Thus the required probability is 1, meaning the event is certain to occur.
Verification / Alternative check:
Another way is to note that the sample space consists of all cards. There is no third colour in a standard deck. Therefore every possible draw is favourable to the event. Hence probability must be 1. This matches our calculation above.
Why Other Options Are Wrong:
Option 1/2 suggests only half the deck is favourable, which is incorrect because both colours are included. Option 3/4 would mean only 75 percent of cards are red or black, which is again wrong. Option 1/3 is even smaller and has no connection with the actual card counts.
Common Pitfalls:
A common confusion is to think that red or black is the same as choosing one suit or one colour only. Another mistake is to mix up the idea of "either red or black" with "red or black but not both", which is meaningless here because no card can be both colours at once. Remember that in a standard deck there are no cards that are neither red nor black.
Final Answer:
The required probability that the drawn card is either red or black is 1.

Question 6

A bag contains 3 black balls, 4 white balls and 5 red balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is either white or red?

Accepted Answer

Introduction / Context:
This problem tests your understanding of basic probability with coloured balls in a bag. You need to identify the total number of equally likely outcomes and then count the favourable outcomes where the ball drawn is either white or red.
Given Data / Assumptions:

Number of black balls = 3.
Number of white balls = 4.
Number of red balls = 5.
One ball is drawn at random from the bag.
The event of interest is that the ball is white or red.
All balls are identical apart from their colour and each ball is equally likely to be drawn.

Concept / Approach:
Use the classical definition of probability:
Probability = number of favourable outcomes / total number of outcomes.
Here, favourable outcomes are balls that are white or red. We first compute the total number of balls, then the number that are white or red, and finally their ratio. We then express the answer as a decimal to match the options.
Step-by-Step Solution:
Total balls = 3 + 4 + 5 = 12
Number of white balls = 4
Number of red balls = 5
Favourable balls (white or red) = 4 + 5 = 9
Probability = favourable / total = 9 / 12
Simplify 9 / 12 = 3 / 4 = 0.75
Therefore the required probability is 0.75.
Verification / Alternative check:
We can also view the probability as one minus the probability of drawing a black ball. The probability of a black ball is 3 / 12 = 1 / 4 = 0.25. So the probability of not drawing black (that is, drawing white or red) is 1 - 0.25 = 0.75. This matches our main calculation.
Why Other Options Are Wrong:
Option 0.5 corresponds to 6 favourable balls out of 12, which does not match the actual 9 favourable balls. Option 0.25 is the probability of drawing a black ball, the complement of what we need. Option 0.8 would require 9.6 favourable outcomes out of 12, which is impossible for a discrete count of balls.
Common Pitfalls:
Students sometimes forget to include both colours in the favourable count or mistakenly divide by the number of colours instead of the total number of balls. Another mistake is to misread the question and include black balls as well. Always recheck the total number of balls and the exact event described in the wording.
Final Answer:
The probability that the ball drawn is either white or red is 0.75.

Question 7

The odds in favour of a person hitting a target are 3 to 5. The odds against another person hitting the same target are 3 to 2. Each of them fires once at the target independently. What is the probability that both of them hit the target?

Accepted Answer

Introduction / Context:
This question checks your understanding of odds, conversion of odds into probabilities, and the multiplication rule for independent events. Two people shoot at the same target under different odds conditions, and you must find the probability that both shots are successful.
Given Data / Assumptions:

Odds in favour of the first shooter hitting the target are 3 to 5.
Odds against the second shooter hitting the target are 3 to 2.
Each shooter fires exactly once.
The events of their hits are independent.
We want the probability that both shooters hit the target.

Concept / Approach:
For odds in favour a : b, the probability of success is a / (a + b). For odds against a : b, the probability of success is b / (a + b), because a corresponds to failure and b corresponds to success. Once we find the individual probabilities of hitting, we use the multiplication rule:
P(both hit) = P(first hits) * P(second hits)
Step-by-Step Solution:
For the first shooter, odds in favour are 3 : 5.
Probability first shooter hits = 3 / (3 + 5) = 3 / 8.
For the second shooter, odds against are 3 : 2.
So probability second shooter hits = 2 / (3 + 2) = 2 / 5.
Now, P(both hit) = (3 / 8) * (2 / 5) = 6 / 40.
Simplify 6 / 40 = 3 / 20.
Thus, the probability that both shooters hit the target is 3/20.
Verification / Alternative check:
We can verify by converting to decimals. Probability for the first shooter is 0.375, for the second shooter is 0.4. Multiplying gives 0.375 * 0.4 = 0.15. Converting 3/20 to decimal gives 0.15, which is consistent. This confirms that our fraction is correct.
Why Other Options Are Wrong:
Option 1/5 equals 0.2, which is larger than the correct probability of 0.15. Option 1/20 is 0.05 and would require either much smaller odds or dependent events. Option 3/40 is 0.075 and corresponds to halving the correct probability accidentally.
Common Pitfalls:
Typical mistakes include misinterpreting odds in favour or against, reversing success and failure, or adding probabilities instead of multiplying for independent events. Another frequent error is simplifying odds directly as probabilities without dividing by the total of both parts. Always convert odds to probabilities carefully before applying multiplication rules.
Final Answer:
The probability that both people hit the target is 3/20.

Question 8

A group consists of 3 men, 2 women and 4 children. Four persons are chosen at random from the group. What is the probability that exactly two of the selected persons are women?

Accepted Answer

Introduction / Context:
This problem involves combinations and basic probability. From a mixed group of men, women and children, we choose four persons and want the probability that exactly two of the chosen persons are women. Because there are only two women in the entire group, this means both women must be selected.
Given Data / Assumptions:

Number of men = 3.
Number of women = 2.
Number of children = 4.
Total persons in the group = 3 + 2 + 4 = 9.
We select 4 persons at random, all selections equally likely.
The required event is: exactly two women are selected.

Concept / Approach:
Because there are only two women in the entire group, having exactly two women in the selected four means both women must be included. The remaining two positions must be filled from the seven non-women. We will use combinations:
Number of ways to choose r items from n = nCr.
Probability is then the ratio of favourable combinations to total combinations.
Step-by-Step Solution:
Total persons = 9.
Total ways to choose any 4 persons = 9C4.
Compute 9C4 = 9! / (4! * 5!) = 126.
To have exactly two women, we must select both women: 2C2 = 1 way.
We then choose 2 more persons from the remaining 7 (3 men + 4 children): 7C2.
Compute 7C2 = 7! / (2! * 5!) = 21.
Favourable ways = 1 * 21 = 21.
Probability = favourable / total = 21 / 126 = 1 / 6.
So the required probability is 1/6.
Verification / Alternative check:
We can verify numerically. If we expand 9C4 as 126 and observe that there is no way to get exactly one woman or more than two women because only two exist, then 21 favourable combinations is reasonable. Dividing 21 by 126 gives 0.1666..., which is exactly 1/6.
Why Other Options Are Wrong:
Option 1/5 corresponds to 25.2 favourable selections, which is not an integer. Option 1/7 is about 0.1429, which would require 18 favourable selections. Option 1/9 is about 0.1111 and does not match the computed value of 1/6.
Common Pitfalls:
Many learners mistakenly use permutations instead of combinations, or they forget that the two women must both be chosen, not just one. Another common error is to choose four persons and then adjust by conditional reasoning, which is more complex and prone to mistakes. It is much simpler to count combinations carefully as done here.
Final Answer:
The probability that exactly two of the four selected persons are women is 1/6.

Question 9

In a lottery there are 10 prize tickets and 25 blank tickets, making 35 tickets in total. One ticket is drawn at random. What is the probability that the ticket drawn is a prize ticket?

Accepted Answer

Introduction / Context:
This question is a straightforward probability problem involving a simple lottery. You need to calculate the probability of drawing a prize ticket when the total number of prize and blank tickets is known.
Given Data / Assumptions:

Number of prize tickets = 10.
Number of blank tickets = 25.
Total number of tickets = 10 + 25 = 35.
One ticket is drawn at random.
All tickets are equally likely to be drawn.
The event of interest is drawing a prize ticket.

Concept / Approach:
We apply the basic probability formula:
Probability = number of favourable outcomes / total number of outcomes.
Here, the favourable outcomes are the prize tickets, and the total outcomes are all tickets in the lottery. Because each ticket is equally likely, we just take the ratio of counts.
Step-by-Step Solution:
Total tickets = 35.
Prize tickets = 10.
Probability of a prize = 10 / 35.
Simplify 10 / 35 by dividing numerator and denominator by 5.
10 / 35 = 2 / 7.
So the probability of drawing a prize ticket is 2/7.
Verification / Alternative check:
We can also look at the probability of drawing a blank ticket which is 25 / 35 = 5 / 7. The probabilities of prize and blank must sum to 1. Indeed, 2/7 + 5/7 = 1. This consistency check confirms our calculation.
Why Other Options Are Wrong:
Option 1/7 would be correct if there were only 5 prizes out of 35. Option 3/7 would require 15 prize tickets, which we do not have. Option 2/5 corresponds to 14 prize tickets out of 35, which is again incorrect.
Common Pitfalls:
A typical mistake is to forget to include both prize and blank tickets when counting the total. Another error is to mis-simplify fractions, for example reducing 10/35 incorrectly. Always double check that the numerator reflects favourable outcomes and that the denominator is the total number of possible outcomes.
Final Answer:
The probability of getting a prize ticket is 2/7.

Question 10

A bag contains 3 black balls, 4 white balls and 5 red balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is either black or red?

Accepted Answer

Introduction / Context:
Here we again consider a simple bag of coloured balls. The task is to find the probability that a randomly drawn ball is either black or red. This is a basic example of adding probabilities for mutually exclusive events.
Given Data / Assumptions:

Black balls = 3.
White balls = 4.
Red balls = 5.
Total balls = 3 + 4 + 5 = 12.
One ball is selected at random.
We require the probability that the ball is black or red.

Concept / Approach:
The events "ball is black" and "ball is red" are mutually exclusive. Therefore, the probability that the ball is black or red is the sum of their individual probabilities. Using the formula:
Probability = favourable outcomes / total outcomes.
we can count the favourable balls and then divide by the total number of balls.
Step-by-Step Solution:
Total balls = 12.
Black balls = 3.
Red balls = 5.
Favourable balls (black or red) = 3 + 5 = 8.
Probability = 8 / 12.
Simplify 8 / 12 = 2 / 3.
Therefore, the required probability is 2/3.
Verification / Alternative check:
An alternative perspective is to find the probability that the ball is white and subtract from 1. Probability of a white ball is 4 / 12 = 1 / 3. Thus, probability of black or red is 1 - 1 / 3 = 2 / 3. This matches the main method, giving extra confidence in the result.
Why Other Options Are Wrong:
Option 1/4 equals 3 / 12, the probability of drawing a black ball only. Option 5/12 corresponds to just the red balls. Option 1/2 would imply 6 favourable outcomes, which does not match the total of 8 black or red balls.
Common Pitfalls:
Learners often forget to include both colours in the favourable count or mistakenly divide by the number of colours instead of the number of balls. Another issue is not simplifying the fraction fully, which can make it harder to compare with the given options.
Final Answer:
The probability that the ball drawn is black or red is 2/3.

Question 11

A leap year has 366 days. If a leap year is selected at random, what is the probability that it contains exactly 53 Sundays?

Accepted Answer

Introduction / Context:
This question involves calendar-based probability. You need to know how the days of the week distribute in a leap year and then determine the chance that there are exactly 53 Sundays. Understanding how extra days beyond full weeks work is the key idea.
Given Data / Assumptions:

A leap year has 366 days.
366 days equal 52 full weeks plus 2 extra days.
Each year can start on any day of the week with equal likelihood.
We want the probability that there are 53 Sundays in the leap year.

Concept / Approach:
In any year, each day of the week occurs at least 52 times. In a leap year there are 2 extra days beyond the 52 full weeks. These 2 extra days determine which days of the week occur 53 times. There are 7 possible ordered pairs of consecutive extra days, depending on the starting day of the year.
Step-by-Step Solution:
Total days in a leap year = 366 = 52 weeks + 2 days.
Every day of the week appears at least 52 times.
The 2 extra days are consecutive days of the week.
Possible ordered pairs of extra days: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun).
There are 7 equally likely possibilities.
A leap year will have 53 Sundays if Sunday is one of the extra two days. This happens when the pair is (Sun, Mon) or (Sat, Sun), so there are 2 favourable cases.
Verification / Alternative check:
We can consider the starting day. If the year starts on Sunday, the extra days are Sunday and Monday, giving 53 Sundays. If it starts on Saturday, the extra days are Saturday and Sunday, again giving 53 Sundays. For any other starting day, Sunday is not among the extra two, so there are only 52 Sundays. Therefore the probability is 2 favourable starting days out of 7.
Why Other Options Are Wrong:
Option 1/7 corresponds to only one favourable case, which is not correct. Option 1/3 is approximately 0.333 but the correct probability is about 0.2857. Option 4/7 is greater than 0.5 and would require four favourable starting days, which does not match the actual possibilities.
Common Pitfalls:
A frequent error is to miscount the number of extra days in a leap year or to assume each day appears the same number of times. Another mistake is to think that any year has 53 Sundays with probability 1/7, which is not accurate for leap years with two extra days. Always consider how many extra days the year has and how they are distributed.
Final Answer:
The probability that a leap year has exactly 53 Sundays is 2/7.

Question 12

Directions: Read the following passage and answer the given questions.  There are five pouches (A, B, C, D, and E) which contain candies of four flavors (Orange, Pineapple, Caramel, and Coffee). Below table shows the number of candies of four differ…

Accepted Answer

Concept & Probability Logic
To solve this Data Interpretation problem, we first determine the missing values in the table by forming equations from the given probabilities.
P(	ext{Event}) = \frac{	ext{Favorable Outcomes}}{	ext{Total Outcomes}}

Step-by-Step Solution
Step 1: Complete the Candy Distribution Table
• Pouch E: P(	ext{pineapple}) = 1/3 \implies 10/T_E = 1/3 \implies T_E = 30. Since P(	ext{orange}) = 1/6, O_E = 5. Given O_E = Ca_E, Ca_E = 5. Co_E = 30-20 = 10.
• Pouch D: Ca_D = 10. P(	ext{coffee}) = 1/5 \implies Co_D/T_D = 1/5. Testing multiples of 5 with P(	ext{none is orange}) = 17/35 yields T_D = 50, O_D = 15. Thus, P_D = 15, Co_D = 10.
• Pouch C: Ca_C = 16. P(	ext{not caramel}) = 4/5 \implies T_C = 80. P(	ext{neither Ca nor Co}) = 1/2 \implies O_C + P_C = 40. Since P_C = P_D = 15, O_C = 25. Co_C = 24.
• Pouch A: Ca_D = Ca_A + 2 \implies Ca_A = 8. P(	ext{coffee}) = 1/3 \implies Co_A/T_A = 1/3. Using P(	ext{two pineapple}) = 1/22 gives T_A = 45, P_A = 10, Co_A = 15. 
• Pouch B: Total Caramel = 69 \implies Ca_B = 30. P(	ext{orange or caramel}) = 25/42 \implies T_B = 84, leaving O_B = 20.

Step 2: Evaluate New Quantities
• Shift operations: 5 Orange moved from A to C. 2 Caramel moved from D to A.
• New Totals:
 - Pouch A: T_A = 42 (Orange=7, Pineapple=10, Caramel=10, Coffee=15)
 - Pouch C: T_C = 85 (Orange=30, Pineapple=15, Caramel=16, Coffee=24)
 - Pouch D: T_D = 48 (Orange=15, Pineapple=15, Caramel=8, Coffee=10)

Step 3: Calculate Given Probabilities
• Quantity I: Pick 2 candies from D. P(	ext{both Caramel}) = \frac{^8C_2}{^{48}C_2} = \frac{28}{1128} = \frac{7}{282} \approx 0.024
• Quantity II: Pick 3 candies from A. P(	ext{exactly 2 Pineapple}) = \frac{^{10}C_2 	imes ^{32}C_1}{^{42}C_3} = \frac{45 	imes 32}{11480} = \frac{36}{287} \approx 0.125
• Quantity III: Pick 1 candy from C. P(	ext{Orange}) = \frac{30}{85} = \frac{6}{17} \approx 0.353

Comparing the values, Quantity I < Quantity II < Quantity III.

Exam Strategy & Shortcut
Whenever tackling quantity comparison questions with probability, it is not always necessary to solve to the last decimal point. Since 7/282 is very small, 36/287 is around 1/8, and 6/17 is over 1/3, the inequality is distinct and can be quickly judged visually once the fractions are simplified.

Common Pitfall
Forgetting to adjust the total number of candies in each pouch after shifting specific flavors between them, causing incorrect denominator calculations.

Final Answer
Therefore, the correct answer is Quantity I < Quantity II < Quantity III.

Question 13

Directions: Read the following passage and answer the given questions.  There are five pouches (A, B, C, D, and E) which contain candies of four flavors (Orange, Pineapple, Caramel, and Coffee). Below table shows the number of candies of four differ…

Accepted Answer

Concept & Probability Logic
To solve this Data Interpretation problem, we first determine the missing values in the table by forming equations from the given probabilities.
P(	ext{Event}) = \frac{	ext{Favorable Outcomes}}{	ext{Total Outcomes}}

Step-by-Step Solution
Step 1: Complete the Candy Distribution Table
• Pouch E: P(	ext{pineapple}) = 1/3 \implies 10/T_E = 1/3 \implies T_E = 30. Since P(	ext{orange}) = 1/6, O_E = 5. Given O_E = Ca_E, Ca_E = 5. Co_E = 30-20 = 10.
• Pouch D: Ca_D = 10. P(	ext{coffee}) = 1/5 \implies Co_D/T_D = 1/5. Testing multiples of 5 with P(	ext{none is orange}) = 17/35 yields T_D = 50, O_D = 15. Thus, P_D = 15, Co_D = 10.
• Pouch C: Ca_C = 16. P(	ext{not caramel}) = 4/5 \implies T_C = 80. P(	ext{neither Ca nor Co}) = 1/2 \implies O_C + P_C = 40. Since P_C = P_D = 15, O_C = 25. Co_C = 24.
• Pouch A: Ca_D = Ca_A + 2 \implies Ca_A = 8. P(	ext{coffee}) = 1/3 \implies Co_A/T_A = 1/3. Using P(	ext{two pineapple}) = 1/22 gives T_A = 45, P_A = 10, Co_A = 15. 
• Pouch B: Total Caramel = 69 \implies Ca_B = 30. P(	ext{orange or caramel}) = 25/42 \implies T_B = 84, leaving O_B = 20.

Step 2: Calculate Rotten and Good Candies in Pouch B
• Total candies in Pouch B = 84.
• Rotten candies = 25% of 84 = 21.
• Good candies = 84 - 21 = 63.
• Rotten Caramel candies = 1/3 of 21 = 7.

Step 3: Find the Required Probability
• We need 1 rotten caramel candy and 1 good candy.
• Ways to choose 1 rotten caramel = ^7C_1 = 7.
• Ways to choose 1 good candy = ^{63}C_1 = 63.
• Total ways to choose 2 candies = ^{84}C_2 = \frac{84 	imes 83}{2} = 3486.
• Probability = \frac{7 	imes 63}{3486} = \frac{441}{3486} = \frac{21}{166}.

Exam Strategy & Shortcut
Cancel out common factors before multiplying fully. Notice that 3486 = 42 	imes 83. So, \frac{7 	imes 63}{42 	imes 83} = \frac{63}{6 	imes 83} = \frac{21}{2 	imes 83} = \frac{21}{166}. Keeping terms factored simplifies the math greatly.

Common Pitfall
A common error is confusing the total pool of candies to select from; ensure you clearly separate "rotten caramel" and the overall "good" candies, ignoring the other flavors of rotten candies.

Final Answer
Therefore, the correct answer is 21/166.

Question 14

Directions: Read the following passage and answer the given questions.  There are five pouches (A, B, C, D, and E) which contain candies of four flavors (Orange, Pineapple, Caramel, and Coffee). Below table shows the number of candies of four differ…

Accepted Answer

Concept & Probability Logic
To solve this Data Interpretation problem, we first determine the missing values in the table by forming equations from the given probabilities.
P(	ext{Event}) = \frac{	ext{Favorable Outcomes}}{	ext{Total Outcomes}}

Step-by-Step Solution
Step 1: Complete the Candy Distribution Table
• Pouch E: P(	ext{pineapple}) = 1/3 \implies 10/T_E = 1/3 \implies T_E = 30. Since P(	ext{orange}) = 1/6, O_E = 5. Given O_E = Ca_E, Ca_E = 5. Co_E = 30-20 = 10.
• Pouch D: Ca_D = 10. P(	ext{coffee}) = 1/5 \implies Co_D/T_D = 1/5. Testing multiples of 5 with P(	ext{none is orange}) = 17/35 yields T_D = 50, O_D = 15. Thus, P_D = 15, Co_D = 10.
• Pouch C: Ca_C = 16. P(	ext{not caramel}) = 4/5 \implies T_C = 80. P(	ext{neither Ca nor Co}) = 1/2 \implies O_C + P_C = 40. Since P_C = P_D = 15, O_C = 25. Co_C = 24.
• Pouch A: Ca_D = Ca_A + 2 \implies Ca_A = 8. P(	ext{coffee}) = 1/3 \implies Co_A/T_A = 1/3. Using P(	ext{two pineapple}) = 1/22 gives T_A = 45, P_A = 10, Co_A = 15. 
• Pouch B: Total Caramel = 69 \implies Ca_B = 30. P(	ext{orange or caramel}) = 25/42 \implies T_B = 84, leaving O_B = 20.

Step 2: Apply the Transfer Operations
• Pouch D initially has 50 candies (O=15, P=15, Ca=10, Co=10).
• Move K pineapple to E: Pouch D loses K candies.
• Move 2K caramel and K coffee from E to D: Pouch D gains 3K candies.
• Net change in Pouch D = +2K. 
• New Total in Pouch D = 50 + 2K. 
• The number of Orange candies remains unchanged at 15.

Step 3: Solve for K
• New Probability of getting an Orange candy = 1/4.
• \frac{15}{50 + 2K} = \frac{1}{4}
• 60 = 50 + 2K
• 2K = 10 \implies K = 5.

Exam Strategy & Shortcut
You only need to track the total number of candies in Pouch D and the number of orange candies in Pouch D. Since no orange candies were moved, the numerator stays exactly 15, drastically simplifying the algebra to a single fast equation.

Common Pitfall
A common mistake is meticulously tracking every flavor dynamically when only the total count and the specific target flavor (orange) are relevant to the final question.

Final Answer
Therefore, the correct answer is 5.

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