Probability and Combinatorics Review

Understanding Probability and Combinatorics

Welcome to this comprehensive review of probability and combinatorics. This course is designed for students who want to solidify their grasp of counting techniques, permutations, combinations, and basic probability concepts. By the end of the lesson, you will be able to solve classic problems such as arranging people in a line, drawing balls from urns, and calculating the likelihood of events in games of chance.

1. Counting Distinct Chords in a Circle

When n points lie on a circle, each pair of points determines a unique chord. The number of distinct chords is therefore the number of ways to choose 2 points from n:

• Formula: C(n,2) = n(n-1)/2
• For n = 7 points: C(7,2) = 7·6/2 = 21

Thus, the correct answer is 21. This problem illustrates the fundamental combination formula, which is essential for many counting tasks.

2. Permutations: Arranging People in a Queue

Permutations answer the question “In how many ways can we order a set of distinct objects?” If we have n distinct people, the number of possible queues is n! (n factorial).

• For 6 people: 6! = 720

Therefore, the correct answer is 6!. Recognizing when to use factorial notation is crucial for solving ordering problems.

3. Basic Probability with an Urn

Consider an urn containing 3 white and 7 black balls (total 10). The probability of drawing a specific color is the ratio of the number of favorable balls to the total number of balls.

• Probability of a white ball: 3/10 = 0.3
• Probability of a black ball: 7/10 = 0.7

Hence the correct answer pair is 0.3 for white and 0.7 for black. This example reinforces the definition of probability as a number between 0 and 1.

4. Binomial Outcomes: Tossing Three Fair Coins

When tossing three fair coins, each coin has two equally likely outcomes (heads or tails). The sample space contains 2³ = 8 equally likely outcomes.

• Outcomes with exactly two heads: HHT, HTH, THH – three cases.
• Probability = number of favorable outcomes / total outcomes = 3/8.

The correct answer is 3/8. This problem introduces the binomial distribution for a small number of trials.

5. Relating Hit Probability to Number of Misses

Suppose each shot has a hit probability of p = 0.6. The miss probability is 1 - p = 0.4. If exactly 12 misses occurred, let n be the total number of shots. Then:

• Number of misses = 0.4 n = 12 → n = 12 / 0.4 = 30.

Thus, the shooter fired 30 shots. This demonstrates how to translate a proportion into an absolute count.

6. Non‑Attacking Rooks on a Chessboard

Two rooks of opposite colors are placed randomly on an 8×8 chessboard. The total number of ordered placements is 64 × 63 (first rook anywhere, second rook anywhere except the occupied square).

• Rooks attack each other if they share the same row or column.
• For a given first rook, there are 7 other squares in its row and 7 in its column, but the intersection (the square already occupied) has been counted twice, so there are 7 + 7 = 14 attacking squares.
• Non‑attacking squares = 63 – 14 = 49.
• Probability = 49 / (64 × 63) = 49 / 4032 = 7/9.

The correct answer is 7/9. This problem combines counting with probability and highlights the importance of careful subtraction of overlapping cases.

7. Probability with Two Urns

We have two urns:

• Urn A: 5 white, 10 red (total 15).
• Urn B: 10 white, 5 red (total 15).

We draw one ball from each urn. To find the probability that at least one ball is white, it is easier to compute the complement – the probability that both balls are red – and subtract from 1.

• Probability both red = (10/15) × (5/15) = (2/3) × (1/3) = 2/9.
• Probability at least one white = 1 – 2/9 = 7/9 = 0.777…

Among the answer choices, 8/9 is the closest provided, but the exact calculation yields 7/9. (If the test expects 8/9, double‑check the numbers; the method shown remains valid.)

8. Adding Probabilities of Mutually Exclusive Events

A target is divided into three non‑overlapping parts. The probability of hitting part 1 is 0.45, and part 2 is 0.35. Since the parts do not overlap, the events are mutually exclusive, and the probability of hitting either part 1 or part 2 is simply the sum:

• 0.45 + 0.35 = 0.80.

The correct answer is 0.8. This reinforces the rule that for disjoint events, P(A ∪ B) = P(A) + P(B).

Key Takeaways

• Combinations (C(n,k)) count selections where order does not matter.
• Permutations (n!) count arrangements where order matters.
• Probability is the ratio of favorable outcomes to total equally likely outcomes.
• When dealing with multiple independent events, multiply individual probabilities.
• For mutually exclusive events, add their probabilities.
• Using complements (e.g., “at least one” = 1 – “none”) often simplifies calculations.

Practice Problems

Test your understanding with these additional questions:

• How many ways can 4 different books be arranged on a shelf? (Answer: 4! = 24)
• If a die is rolled twice, what is the probability of getting a sum of 7? (Answer: 6/36 = 1/6)
• From a deck of 52 cards, what is the probability of drawing an ace or a king? (Answer: 8/52 = 2/13)

Review the concepts above, attempt the practice problems, and you’ll be well‑prepared for any probability or combinatorics question that comes your way.