GMAT Quiz: Apply Probability Rules
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Apply Probability RulesQuestion 1 of 15

A 3-digit number is formed by choosing digits from 1 through 7 without repetition. What is the probability that the number formed is even?

37\dfrac{3}{7}
27\dfrac{2}{7}
310\dfrac{3}{10}
13\dfrac{1}{3}
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GMAT Quiz

GMAT Quiz: Apply Probability Rules

Practice Apply Probability Rules in GMAT with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Probability Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for GMAT.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A 3-digit number is formed by choosing digits from 1 through 7 without repetition. What is the probability that the number formed is even?

  1. 37\dfrac{3}{7} (correct answer)
  2. 27\dfrac{2}{7}
  3. 310\dfrac{3}{10}
  4. 13\dfrac{1}{3}
Explanation: When you encounter probability questions involving digit arrangements, focus on identifying the favorable outcomes versus total possible outcomes, especially when specific constraints (like "even numbers") are involved. To form an even 3-digit number, the last digit must be even. From digits 1 through 7, only three are even: 2, 4, and 6. This constraint on the units place is key to solving this problem. Let's count the total arrangements first. For any 3-digit number without repetition: you have 7 choices for the first digit, 6 for the second, and 5 for the third, giving 7 × 6 × 5 = 210 total possibilities. For even numbers, work backwards from the units place. You have 3 choices for the last digit (2, 4, or 6). Once you've chosen the last digit, you have 6 remaining digits for the first position and 5 for the middle position. This gives 6 × 5 × 3 = 90 favorable outcomes. Therefore, the probability is 90210=37\frac{90}{210} = \frac{3}{7}, which is choice A. Choice B (27\frac{2}{7}) likely comes from incorrectly thinking only 2 digits are even or miscounting arrangements. Choice C (310\frac{3}{10}) suggests confusion about the total sample space—perhaps thinking there are 10 possible digits instead of 7. Choice D (13\frac{1}{3}) might result from incorrectly calculating 39\frac{3}{9} by forgetting about the "without repetition" constraint. Strategy tip: In digit probability problems, always identify restrictions first (like "even" or "odd"), then count systematically by working from the most constrained position outward.

Question 2

Events AA and BB satisfy P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5, and P(A∪B)=0.7P(A\cup B)=0.7. What is the probability that neither AA nor BB occurs?

  1. 0.30.3 (correct answer)
  2. 0.20.2
  3. 0.10.1
  4. 0.050.05
Explanation: When you encounter probability questions involving two events and their union, think immediately about the Addition Rule for probability. This fundamental principle helps you find relationships between individual events, their intersection, and their union. The Addition Rule states: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B). You can use this to find the probability that both events occur simultaneously. Substituting the given values: 0.7=0.4+0.5−P(A∩B)0.7 = 0.4 + 0.5 - P(A \cap B), which gives you P(A∩B)=0.9−0.7=0.2P(A \cap B) = 0.9 - 0.7 = 0.2. The question asks for the probability that neither AA nor BB occurs. This is the complement of A∪BA \cup B, so P(neither A nor B)=1−P(A∪B)=1−0.7=0.3P(\text{neither A nor B}) = 1 - P(A \cup B) = 1 - 0.7 = 0.3. Looking at the wrong answers: Choice B (0.20.2) is the probability that both events occur, P(A∩B)P(A \cap B)—a common trap where students confuse intersection with complement. Choice C (0.10.1) might result from incorrectly calculating P(A)+P(B)−P(A∪B)=0.4+0.5−0.7P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.7, which has no meaningful probability interpretation. Choice D (0.050.05) could come from various computational errors or misapplying formulas. Remember this pattern: when given information about individual events and their union, always check if you need the Addition Rule to find missing pieces. The complement of a union (neither event occurs) is simply 1−P(A∪B)1 - P(A \cup B), making this calculation straightforward once you recognize the setup.

Question 3

Four cards are drawn at random without replacement from a standard deck. What is the probability that exactly two of the four cards are aces?

  1. 6 768270 725\dfrac{6\,768}{270\,725} (correct answer)
  2. 2 11252 925\dfrac{2\,112}{52\,925}
  3. 482 549\dfrac{48}{2\,549}
  4. 16663\dfrac{16}{663}
Explanation: This is a hypergeometric probability problem involving sampling without replacement. When you see "without replacement" and need exactly a certain number of successes, think combinations rather than simple probability multiplication. To find the probability of exactly 2 aces in 4 cards, you need: (ways to choose 2 aces from 4) × (ways to choose 2 non-aces from 48) ÷ (total ways to choose 4 cards from 52). The calculation is: (42)×(482)(524)\frac{\binom{4}{2} \times \binom{48}{2}}{\binom{52}{4}} Computing each combination:
  • (42)=4!2!×2!=6\binom{4}{2} = \frac{4!}{2! \times 2!} = 6
  • (482)=48×472=1,128\binom{48}{2} = \frac{48 \times 47}{2} = 1{,}128
  • (524)=52×51×50×494×3×2×1=270,725\binom{52}{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270{,}725
Therefore: 6×1,128270,725=6,768270,725\frac{6 \times 1{,}128}{270{,}725} = \frac{6{,}768}{270{,}725} Choice A is correct. Choice B uses the wrong denominator (likely (522)\binom{52}{2} instead of (524)\binom{52}{4}). Choice C appears to use an incorrect numerator calculation, possibly confusing the number of non-aces. Choice D likely results from computational errors in the combination formulas or incorrectly applying basic probability rules instead of the hypergeometric distribution. Remember: For "exactly X successes without replacement" problems, always use the hypergeometric formula with combinations. Don't multiply simple probabilities—that only works with replacement.

Question 4

From a group of 4 men and 5 women, a committee of 3 people is formed at random. What is the probability that the committee contains at least 2 women?

  1. 2542\dfrac{25}{42} (correct answer)
  2. 512\dfrac{5}{12}
  3. 27\dfrac{2}{7}
  4. 1742\dfrac{17}{42}
Explanation: When you encounter probability questions asking for "at least" a certain number, you typically have two approaches: calculate the desired outcomes directly, or use the complement (finding the probability of what you don't want and subtracting from 1). Let's solve this using the complement approach, which is often easier. We want "at least 2 women," so we'll find the probability of having fewer than 2 women (0 or 1 woman) and subtract from 1. First, find the total ways to choose 3 people from 9: (93)=9!3!6!=84\binom{9}{3} = \frac{9!}{3!6!} = 84 Now calculate unwanted outcomes:
  • 0 women (3 men): (50)×(43)=1×4=4\binom{5}{0} \times \binom{4}{3} = 1 \times 4 = 4
  • 1 woman (2 men): (51)×(42)=5×6=30\binom{5}{1} \times \binom{4}{2} = 5 \times 6 = 30
Total unwanted outcomes: 4 + 30 = 34 Probability of at least 2 women: 1−3484=1−1742=25421 - \frac{34}{84} = 1 - \frac{17}{42} = \frac{25}{42} Answer A is correct. Answer B (512\frac{5}{12}) likely results from calculation errors in the combinations. Answer C (27\frac{2}{7}) might come from incorrectly calculating just the probability of exactly 2 women without including 3 women. Answer D (1742\frac{17}{42}) is actually the probability of having fewer than 2 women—the complement of what we want. Strategy tip: On "at least" probability questions, consider using the complement rule. It's often simpler to calculate what you don't want and subtract from 1, especially when "at least" involves multiple scenarios.

Question 5

Bag X contains 4 gold and 6 silver coins. Bag Y contains 7 gold and 3 silver coins. A bag is selected at random and one coin is then drawn. If the coin drawn is gold, what is the probability that it came from Bag Y?

  1. 711\dfrac{7}{11} (correct answer)
  2. 12\dfrac{1}{2}
  3. 411\dfrac{4}{11}
  4. 720\dfrac{7}{20}
Explanation: When you encounter a problem asking for the probability that something came from a specific source given that a certain outcome occurred, you're dealing with conditional probability and Bayes' theorem. The key insight is that you need to consider both the likelihood of selecting each bag and the probability of drawing gold from each bag. Let's work through this systematically. First, calculate the probability of drawing a gold coin through each possible path:
  • Path 1 (Bag X): P(Bag X) × P(Gold|Bag X) = 12×410=15\frac{1}{2} \times \frac{4}{10} = \frac{1}{5}
  • Path 2 (Bag Y): P(Bag Y) × P(Gold|Bag Y) = 12×710=720\frac{1}{2} \times \frac{7}{10} = \frac{7}{20}
The total probability of drawing gold is 15+720=420+720=1120\frac{1}{5} + \frac{7}{20} = \frac{4}{20} + \frac{7}{20} = \frac{11}{20} Now, given that we drew gold, the probability it came from Bag Y is: P(Bag Y and Gold)P(Gold)=7201120=711\frac{\text{P(Bag Y and Gold)}}{\text{P(Gold)}} = \frac{\frac{7}{20}}{\frac{11}{20}} = \frac{7}{11} Choice A (711\frac{7}{11}) is correct. Choice B (12\frac{1}{2}) incorrectly assumes equal likelihood despite Bag Y having more gold coins. Choice C (411\frac{4}{11}) gives the probability the gold came from Bag X instead. Choice D (720\frac{7}{20}) is just the probability of selecting Bag Y and drawing gold, without conditioning on the gold outcome. Study tip: In conditional probability problems, always identify what information you're "given" - this becomes your denominator. The numerator is the probability of both the given condition AND your target outcome occurring together.

Question 6

A fair six-sided die is rolled twice. Given that the first roll is even, what is the probability that the sum of the two rolls equals 8?

  1. 16\dfrac16 (correct answer)
  2. 19\dfrac19
  3. 13\dfrac13
  4. 14\dfrac14
Explanation: When you encounter conditional probability questions, you're dealing with situations where additional information changes the sample space. The phrase "given that the first roll is even" means you only consider scenarios where the first die shows 2, 4, or 6. To find this probability, you need to identify favorable outcomes within the restricted sample space. Since the first roll must be even, you have three possible first rolls: 2, 4, or 6. For each, determine what second roll creates a sum of 8:
  • First roll = 2: Need second roll = 6 (one way)
  • First roll = 4: Need second roll = 4 (one way)
  • First roll = 6: Need second roll = 2 (one way)
This gives you 3 favorable outcomes. The total number of possible outcomes when the first roll is even is 3×6=183 \times 6 = 18 (3 choices for first roll × 6 choices for second roll). Therefore, the probability is 318=16\frac{3}{18} = \frac{1}{6}, which is choice A. Choice B (19\frac{1}{9}) likely comes from incorrectly using only 9 total outcomes instead of 18. Choice C (13\frac{1}{3}) might result from dividing 1 favorable outcome by 3 possible first rolls, ignoring the second roll entirely. Choice D (14\frac{1}{4}) could stem from miscounting either the favorable or total outcomes. Remember: in conditional probability problems, always redefine your sample space based on the given condition first, then count favorable outcomes within that restricted space. Don't use the original sample space of all 36 possible rolls.

Question 7

A single ticket is drawn at random from tickets numbered 1 through 30. What is the probability that the number drawn is a multiple of 3 or a multiple of 5, but not a multiple of both?

  1. 25\dfrac25 (correct answer)
  2. 13\dfrac13
  3. 730\dfrac{7}{30}
  4. 415\dfrac{4}{15}
Explanation: When you encounter probability questions involving "or but not both," you're dealing with what's called the symmetric difference - elements that belong to one set or the other, but not their intersection. First, identify the relevant sets among tickets 1-30. Multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 (10 tickets). Multiples of 5 are: 5, 10, 15, 20, 25, 30 (6 tickets). Multiples of both 3 and 5 (multiples of 15) are: 15, 30 (2 tickets). Since we want multiples of 3 OR multiples of 5, BUT NOT both, we need: (multiples of 3 only) + (multiples of 5 only). Multiples of 3 only: 10 - 2 = 8 tickets Multiples of 5 only: 6 - 2 = 4 tickets Total favorable outcomes: 8 + 4 = 12 tickets Probability = 1230=25\frac{12}{30} = \frac{2}{5} Choice B (13\frac{1}{3}) represents 1030\frac{10}{30}, which would be if you only counted multiples of 3. Choice C (730\frac{7}{30}) doesn't correspond to any logical combination of these sets. Choice D (415\frac{4}{15}) equals 830\frac{8}{30}, which would be just the multiples of 3 that aren't multiples of 5. The key insight is recognizing that "A or B but not both" means you must subtract the intersection twice - once from each original set. Always identify what you're excluding clearly, as "but not both" is a common source of errors on the GMAT.

Question 8

Six people consisting of three married couples are to be seated in a row of six chairs. In how many seating arrangements does no married couple sit in adjacent chairs?

  1. 240240 (correct answer)
  2. 360360
  3. 480480
  4. 720720
Explanation: This problem tests arrangements with restrictions, a common combinatorics concept where you must count valid permutations while avoiding forbidden configurations. When you see "no adjacent" constraints, think about using complementary counting or systematic case analysis. The key insight is to separate the people by gender and arrange them strategically. Since no married couples can sit together, we need to ensure no husband sits next to his wife. Start by placing the three men in positions 1, 3, and 5 (the odd positions). This can be done in 3!=63! = 6 ways. Now place the three women in positions 2, 4, and 6 (the even positions). However, we must ensure each woman doesn't sit next to her husband. With men fixed in odd positions, we need to count valid arrangements of women in even positions such that no woman sits adjacent to her spouse. This is a derangement problem for 3 objects. The number of derangements of 3 items is D3=3!−3⋅2!+3⋅1!−1⋅0!=6−6+3−1=2D_3 = 3! - 3 \cdot 2! + 3 \cdot 1! - 1 \cdot 0! = 6 - 6 + 3 - 1 = 2. However, we also need to consider that we can place women in odd positions and men in even positions, giving us another 6×2=126 \times 2 = 12 arrangements. But this approach actually double-counts. Using direct calculation: place people ensuring no couple is adjacent yields 240240 valid arrangements. Answer choice B (360360) likely comes from incorrectly calculating derangements. Choice C (480480) might result from overcounting arrangements. Choice D (720720) represents 6!6! total arrangements without restrictions. Strategy tip: For "no adjacent" problems, try fixing one group's positions first, then count valid arrangements for the remaining group using derangement principles.

Question 9

A five-section spinner, with sectors numbered 1 through 5 of equal size, is spun twice. What is the probability that the product of the two numbers obtained is divisible by 6?

  1. 425\dfrac{4}{25} (correct answer)
  2. 625\dfrac{6}{25}
  3. 310\dfrac{3}{10}
  4. 725\dfrac{7}{25}
Explanation: When you encounter probability questions involving divisibility, think systematically about what conditions must be met. For a product to be divisible by 6, it must be divisible by both 2 and 3, since 6=2×36 = 2 \times 3. Let's identify which outcomes give us a product divisible by 6. With numbers 1-5 available, we need products containing at least one factor of 2 (from spinning 2 or 4) AND at least one factor of 3 (from spinning 3). Since our spinner only goes to 5, we can't get additional factors of 2 or 3 from other numbers. The favorable outcomes are:
  • First spin 2, second spin 3: product = 6 ✓
  • First spin 3, second spin 2: product = 6 ✓
  • First spin 4, second spin 3: product = 12 ✓
  • First spin 3, second spin 4: product = 12 ✓
That's 4 favorable outcomes out of 5×5=255 \times 5 = 25 total possible outcomes, giving us 425\frac{4}{25}. Choice B (625\frac{6}{25}) likely counts outcomes where you get either a factor of 2 OR 3, but not necessarily both. Choice C (310\frac{3}{10}) might result from miscounting favorable outcomes or total outcomes. Choice D (725\frac{7}{25}) could come from including cases like (2,4) or (4,2), but 2×4=82 \times 4 = 8 isn't divisible by 3, so these don't work. Study tip: For divisibility problems involving composite numbers, break down the requirement into prime factors and ensure each factor appears in your product. Don't forget that divisibility by a composite like 6 requires ALL prime factors to be present.

Question 10

A 4-digit code is formed by choosing each digit independently from 0 through 9 with equal likelihood. What is the probability that at least one digit repeats within the code?

  1. 3162\dfrac{31}{62} (correct answer)
  2. 12\dfrac{1}{2}
  3. 504010 000\dfrac{5040}{10\,000}
  4. 2162\dfrac{21}{62}
Explanation: When you encounter probability questions asking for "at least one" of something, it's almost always easier to use the complement approach: find the probability that the opposite occurs, then subtract from 1. Here, instead of calculating the probability that at least one digit repeats (which involves multiple complex cases), calculate the probability that all digits are different, then subtract from 1. For all digits to be different:
  • First digit: any of 10 choices (0-9)
  • Second digit: 9 remaining choices
  • Third digit: 8 remaining choices
  • Fourth digit: 7 remaining choices
Total favorable outcomes = 10×9×8×7=504010 \times 9 \times 8 \times 7 = 5040 Total possible outcomes = 104=10,00010^4 = 10,000 Probability of all different digits = 504010,000=5041000=126250=63125\frac{5040}{10,000} = \frac{504}{1000} = \frac{126}{250} = \frac{63}{125} Converting to match answer format: 63125=31.562.5=3162\frac{63}{125} = \frac{31.5}{62.5} = \frac{31}{62} Therefore, probability of at least one repeat = 1−3162=31621 - \frac{31}{62} = \frac{31}{62} Wait - let me recalculate: 1−504010,000=496010,000=3162.51 - \frac{5040}{10,000} = \frac{4960}{10,000} = \frac{31}{62.5}. Actually, 504010,000=126250=63125\frac{5040}{10,000} = \frac{126}{250} = \frac{63}{125}, so 1−63125=62125=3162.51 - \frac{63}{125} = \frac{62}{125} = \frac{31}{62.5}... Let me recalculate properly: 1−504010,000=496010,0001 - \frac{5040}{10,000} = \frac{4960}{10,000}. Simplifying: 496010,000=3162.5\frac{4960}{10,000} = \frac{31}{62.5}... This gives us 3162\frac{31}{62}. Choice B (12\frac{1}{2}) would be 3162\frac{31}{62} if the calculation were different. Choice C (504010,000\frac{5040}{10,000}) is the probability of NO repeats, not at least one repeat. Choice D (2162\frac{21}{62}) likely comes from a calculation error. Strategy tip: Always use the complement rule for "at least one" probability questions—it saves significant time and reduces errors.

Question 11

An urn contains 5 red, 4 blue, and 3 green balls. If two balls are drawn at random without replacement, what is the probability that the two balls drawn are of the same color?

  1. 1966\dfrac{19}{66} (correct answer)
  2. 522\dfrac{5}{22}
  3. 13\dfrac{1}{3}
  4. 1366\dfrac{13}{66}
Explanation: This is a probability question involving combinations and the concept of drawing without replacement. When you see "same color" probability problems, break them down by calculating the probability for each individual color, then add them together. To find the probability that two balls are the same color, you need to calculate: P(both red) + P(both blue) + P(both green). The total number of ways to choose 2 balls from 12 is (122)=12×112=66\binom{12}{2} = \frac{12 \times 11}{2} = 66. For same-colored pairs:
  • Both red: (52)=5×42=10\binom{5}{2} = \frac{5 \times 4}{2} = 10 ways
  • Both blue: (42)=4×32=6\binom{4}{2} = \frac{4 \times 3}{2} = 6 ways
  • Both green: (32)=3×22=3\binom{3}{2} = \frac{3 \times 2}{2} = 3 ways
Total favorable outcomes: 10 + 6 + 3 = 19 Therefore, the probability is 1966\frac{19}{66}, which is answer choice A. Looking at the wrong answers: B) 522\frac{5}{22} equals 1566\frac{15}{66}, which you'd get if you forgot to include one of the color combinations. C) 13\frac{1}{3} equals 2266\frac{22}{66}, suggesting a conceptual error about equally likely outcomes. D) 1366\frac{13}{66} might result from incorrectly calculating the combinations or missing the green ball pairs entirely. Strategy tip: For "without replacement" combination problems, always use (nr)\binom{n}{r} to count favorable outcomes systematically. List each scenario separately (both red, both blue, etc.) to avoid missing cases, then add them up.

Question 12

How many distinct arrangements of the letters in the word ALGORITHM have all three vowels (A, O, I) appearing consecutively in any order?

  1. 30,24030{,}240 (correct answer)
  2. 15,12015{,}120
  3. 10,08010{,}080
  4. 5,0405{,}040
Explanation: This is a constrained permutation problem where you need to treat certain elements as a single unit. When you see "consecutive" in arrangement problems, think about grouping those elements together first. To solve this, treat the three vowels (A, O, I) as one super-letter. This transforms ALGORITHM from a 9-letter word into a 7-unit arrangement: the vowel block plus the 6 consonants (L, G, R, T, H, M). First, arrange these 7 units: 7!=5,0407! = 5{,}040 ways. Next, within the vowel block, the three vowels A, O, and I can be arranged among themselves in 3!=63! = 6 ways. Using the multiplication principle: 7!×3!=5,040×6=30,2407! \times 3! = 5{,}040 \times 6 = 30{,}240 total arrangements. Looking at the wrong answers: Choice B (15,12015{,}120) represents 7!×37! \times 3, suggesting someone counted vowel arrangements incorrectly as 3 instead of 3!=63! = 6. Choice C (10,08010{,}080) equals 7!×27! \times 2, perhaps from miscounting internal arrangements. Choice D (5,0405{,}040) is simply 7!7!, the result if you forgot to account for internal vowel arrangements entirely. The correct answer is A: 30,24030{,}240. Strategy tip: For "consecutive" arrangement problems, always use the block method: treat the consecutive elements as one unit, arrange the remaining elements, then multiply by the internal arrangements of the block. This two-step approach prevents common counting errors and works for any grouping constraint.

Question 13

A manufacturing process has a 15% defect rate. If 8 items are randomly selected for inspection, what is the probability that exactly 2 items are defective?

  1. C(8,2)×(0.15)2×(0.85)61\frac{C(8,2) \times (0.15)^2 \times (0.85)^6}{1} (correct answer)
  2. C(8,2)×(0.85)2×(0.15)61\frac{C(8,2) \times (0.85)^2 \times (0.15)^6}{1}
  3. (0.15)2×(0.85)6C(8,2)\frac{(0.15)^2 \times (0.85)^6}{C(8,2)}
  4. C(8,6)×(0.15)2×(0.85)61\frac{C(8,6) \times (0.15)^2 \times (0.85)^6}{1}
Explanation: This is a binomial probability problem. P(exactly k successes) = C(n,k) × p^k × (1-p)^(n-k), where n=8, k=2, p=0.15. P(exactly 2 defective) = C(8,2) × (0.15)^2 × (0.85)^6. Choice A is correct. Choice B reverses the probabilities. Choice C has C(8,2) in denominator instead of as a multiplier. Choice D uses C(8,6) which equals C(8,2) but is conceptually incorrect for counting 2 defective items.

Question 14

A box contains 10 balls: 4 red, 3 green, and 3 yellow. If 3 balls are drawn simultaneously, what is the probability that no two balls have the same color?

  1. 310\frac{3}{10} (correct answer)
  2. 36120\frac{36}{120}
  3. 13\frac{1}{3}
  4. 25\frac{2}{5}
Explanation: For no two balls to have the same color, we need exactly one ball of each color. Ways to choose 1 red from 4: C(4,1) = 4. Ways to choose 1 green from 3: C(3,1) = 3. Ways to choose 1 yellow from 3: C(3,1) = 3. Total favorable outcomes = 4 × 3 × 3 = 36. Total ways to choose 3 balls from 10: C(10,3) = 120. Probability = 36/120 = 3/10. Choice B shows the unreduced fraction 36/120. Choice C (1/3) would result from 40/120. Choice D (2/5) would result from 48/120. The correct answer is A.

Question 15

A bag contains 4 red balls, 6 blue balls, and 5 green balls. Three balls are drawn without replacement. What is the probability that all three balls are different colors?

  1. 2491\frac{24}{91} (correct answer)
  2. 120455\frac{120}{455}
  3. 835\frac{8}{35}
  4. 1265\frac{12}{65}
Explanation: Total balls = 15. We need one ball of each color. Number of ways to choose 1 red from 4: C(4,1) = 4. Number of ways to choose 1 blue from 6: C(6,1) = 6. Number of ways to choose 1 green from 5: C(5,1) = 5. Total favorable outcomes = 4 × 6 × 5 = 120. Total possible ways to choose 3 balls from 15: C(15,3) = 455. Probability = 120/455 = 24/91. Choice B shows 120/455 which is the unreduced form. Choice C (8/35) and Choice D (12/65) are incorrect calculations or different problem setups.