Mastering GRE Quantitative Probability & Combinations

GRE Quantitative: Probability and Combinations

The GRE Quantitative section tests your ability to interpret and apply fundamental principles of probability and combinatorics to solve problems, often involving scenarios with multiple conditions or constraints. Examiners assess your precision in identifying the correct counting technique and your understanding of how to calculate the likelihood of events, including conditional probabilities and independent trials.

The Method: A Step-by-Step Approach

Follow these steps consistently to tackle probability and combinatorics problems:

1. Understand the Scenario and Identify Key Elements: Read the problem carefully. What is being selected or arranged? What are the total possible outcomes? What are the favorable outcomes? Are there any restrictions or conditions?
2. Determine if Order Matters (Permutation) or Not (Combination):
   • If the arrangement or sequence of items is important (e.g., forming a password, ranking contestants), use permutations.
   • If the group or selection of items is important, but their order within the group is not (e.g., selecting a committee, choosing cards), use combinations.
3. Identify if Repetition is Allowed:
   • If items can be chosen multiple times (e.g., digits in a PIN, rolling dice), this impacts the counting method.
   • If items cannot be chosen multiple times (e.g., drawing cards without replacement), this also affects the calculation.
4. Apply the Correct Formula or Principle:
   • Permutations:
     • Without repetition: \( P(n, k) = \frac{n!}{(n-k)!} \)
     • With repetition (for choosing \(k\) items from \(n\) types with replacement): \( n^k \)
     • With repetition (for arranging \(n\) items where \(n_1\) are of type 1, \(n_2\) of type 2, etc.): \( \frac{n!}{n_1! n_2! \dots n_k!} \)
   • Combinations:
     • Without repetition: \( C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
     • With repetition (choosing \(k\) items from \(n\) types with replacement, "stars and bars"): \( \binom{n+k-1}{k} \)
   • Probability: \( P(\text{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)
   • Independent Events: \( P(A \text{ and } B) = P(A) \times P(B) \)
   • Mutually Exclusive Events: \( P(A \text{ or } B) = P(A) + P(B) \)
   • Conditional Probability: \( P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \)
5. Calculate Total Possible Outcomes: Use the appropriate counting method from step 4 to determine the size of the sample space.
6. Calculate Favorable Outcomes: Use the appropriate counting method from step 4 to determine the number of outcomes that satisfy the problem's conditions.
7. Compute the Probability (if required): Divide the number of favorable outcomes by the total number of possible outcomes. Simplify the fraction if possible.

Fully Worked Example

A research team consists of 8 senior scientists and 12 junior scientists. A subcommittee of 5 scientists is to be formed. What is the probability that the subcommittee will consist of exactly 3 senior scientists and 2 junior scientists?

1. Understand the Scenario: We are forming a subcommittee (order doesn't matter) from a larger group. We need to find the probability of a specific composition.
2. Order Matters? No, the order in which scientists are chosen for a subcommittee does not matter. This is a combination problem.
3. Repetition Allowed? No, once a scientist is chosen, they cannot be chosen again.
4. Apply Formula: We will use the combination formula \( C(n, k) \).
5. Calculate Total Possible Outcomes:
   • Total scientists: \( 8 \text{ (senior)} + 12 \text{ (junior)} = 20 \text{ scientists} \)
   • Subcommittee size: 5 scientists
   • Total ways to form a subcommittee of 5 from 20:
     $$ C(20, 5) = \binom{20}{5} = \frac{20!}{5!(20-5)!} = \frac{20!}{5!15!} $$
     $$ = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = 19 \times 3 \times 17 \times 16 = 15,504 \text{ ways} $$
6. Calculate Favorable Outcomes:
   • We need exactly 3 senior scientists from 8:
     $$ C(8, 3) = \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \text{ ways} $$
   • We need exactly 2 junior scientists from 12:
     $$ C(12, 2) = \binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12!}{2!10!} = \frac{12 \times 11}{2 \times 1} = 66 \text{ ways} $$
   • Since these selections are independent (choosing seniors doesn't affect choosing juniors), we multiply the ways:
     $$ \text{Favorable outcomes} = C(8, 3) \times C(12, 2) = 56 \times 66 = 3,696 \text{ ways} $$
7. Compute Probability:
   $$ P(\text{3 senior, 2 junior}) = \frac{\text{Favorable outcomes}}{\text{Total possible outcomes}} = \frac{3,696}{15,504} $$
   To simplify, divide by common factors. Both are divisible by 8:
   $$ \frac{3,696 \div 8}{15,504 \div 8} = \frac{462}{1,938} $$
   Both are divisible by 6:
   $$ \frac{462 \div 6}{1,938 \div 6} = \frac{77}{323} $$
   The probability is \( \frac{77}{323} \).

Three Mistakes That Lose Marks

1. Confusing Permutations and Combinations: The most common error is using the wrong counting technique. Always ask yourself: "Does the order of selection matter?" If yes, it's a permutation. If no, it's a combination. A password "123" is different from "321" (permutation); a committee of "Alice, Bob, Carol" is the same as "Bob, Carol, Alice" (combination).
2. Incorrectly Handling "And" vs. "Or" Events: For independent events, "and" usually implies multiplication of probabilities or ways (e.g., \( P(A \text{ and } B) = P(A) \times P(B) \)). For mutually exclusive events, "or" usually implies addition (e.g., \( P(A \text{ or } B) = P(A) + P(B) \)). Misinterpreting these logical connectors leads to fundamental errors in calculation.
3. Forgetting to Account for "Not": Sometimes it's easier to calculate the probability of an event not happening and subtract it from 1. Forgetting this complementary probability rule, or miscalculating the "not" scenario, can lead to complex and error-prone direct calculations. For example, "at least one" is often best solved as \( 1 - P(\text{none}) \).

30-Second Recap

Probability and Combinations on the GRE hinge on correctly identifying if order matters (permutation) or not (combination), and whether repetition is allowed. Calculate total possible outcomes and favorable outcomes using the appropriate formulas, then divide to find probability. Watch out for permutation/combination mix-ups, "and/or" logic, and leveraging complementary probability.