PSTAT 5A Practice Worksheet 3 - SOLUTIONS

Comprehensive Review: Probability, Counting, and Conditional Probability

Author

Narjes Mathlouthi

Published

July 10, 2025

📚 Key Formulas Reference:

Basic Probability:

  • Conditional Probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)

  • Law of Total Probability: \(P(A) = \sum P(A|B_i) \cdot P(B_i)\)

  • Addition Rule: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

  • Multiplication Rule: \(P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)\)

Counting:

  • Multiplication Rule: If a procedure consists of \(k\) steps, with \(n_1\) ways for step 1, \(n_2\) for step 2, …, \(n_k\) for step \(k\), then total ways: \(n_1 \times n_2 \times \cdots \times n_k\)

  • Factorial: \(n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1\)

  • Permutations: \(P(n,r) = \frac{n!}{(n-r)!}\)

  • Combinations: \(C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}\)

Section A: Probability - SOLUTIONS

⏱️ Estimated time: 8 minutes

Problem A1: Probability Distributions - SOLUTION

For a valid probability distribution, two conditions must be met:

  1. All probabilities must be non-negative (≥ 0)

  2. The sum of all probabilities must equal 1

Analysis:

(a) Invalid \[\text{Sum} = 0.3 + 0.3 + 0.3 + 0.2 + 0.1 = 1.2 > 1\] The probabilities sum to more than 1, violating condition 2.

(b) Valid \[\text{Sum} = 0 + 0 + 1 + 0 + 0 = 1\] All probabilities are non-negative and sum to 1. This represents a class where everyone receives a C.

(c) Invalid \[\text{Sum} = 0.3 + 0.3 + 0.3 + 0 + 0 = 0.9 < 1\] The probabilities sum to less than 1, violating condition 2.

(d) Invalid Contains \(P(F) = -0.1 < 0\) Although the sum would equal 1.0, the probability for grade F is negative, violating condition 1.

(e) Valid \[\text{Sum} = 0.2 + 0.4 + 0.2 + 0.1 + 0.1 = 1.0\] All probabilities are non-negative and sum to 1.

(f) Invalid Contains \(P(B) = -0.1 < 0\) Although the sum equals 1.0, the probability for grade B is negative, violating condition 1.

Section B: Permutations and Combinations - SOLUTIONS

⏱️ Estimated time: 15 minutes

Problem B1: Permutations and Combinations - SOLUTION

Part (a): How many 6-character passwords can be formed using 3 specific letters and 3 specific digits if repetitions are not allowed and letters must come before digits?

Solution: Since letters must come before digits, we have a fixed structure: LLL DDD

Step 1: Arrange 3 letters in the first 3 positions

  • This is a permutation: \(P(3,3) = \frac{3!}{(3-3)!} = \frac{3!}{0!} = 3! = 6\) ways

Step 2: Arrange 3 digits in the last 3 positions

  • This is a permutation: \(P(3,3) = \frac{3!}{(3-3)!} = \frac{3!}{0!} = 3! = 6\) ways

Step 3: Apply multiplication principle \[\text{Total passwords} = 6 \times 6 = \boxed{36 \text{ passwords}}\]

Part (b): If the team wants to select 4 people from 12 employees to form a security committee where order doesn’t matter, how many ways can this be done?

Solution: Since order doesn’t matter, this is a combination problem.

\[C(12,4) = \binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12!}{4! \cdot 8!}\]

\[= \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11,880}{24} = \boxed{495 \text{ ways}}\]

Section C: Conditional Probability - SOLUTIONS

⏱️ Estimated time: 12 minutes

Problem C1: Drawing Cards (Without Replacement) - SOLUTION

Given Information:

  • Standard 52-card deck

  • Drawing two cards without replacement

  • \(A = \{\text{"first card is a heart"}\}\)

  • \(B = \{\text{"second card is an ace"}\}\)

Solution:

1. P(A)

There are 13 hearts in a 52-card deck. \[P(A) = \frac{13}{52} = \boxed{\frac{1}{4} = 0.2500}\]

2. P(A and B)

We need both events to occur: first card is a heart AND second card is an ace.

Case 1: First card is the Ace of Hearts - \(P(\text{1st card is Ace of Hearts}) = \frac{1}{52}\)

  • \(P(\text{2nd card is an ace | 1st card is Ace of Hearts}) = \frac{3}{51}\) (3 aces left)

  • \(P(\text{Case 1}) = \frac{1}{52} \times \frac{3}{51} = \frac{3}{2652}\)

Case 2: First card is a non-ace heart - \(P(\text{1st card is non-ace heart}) = \frac{12}{52}\) (12 non-ace hearts)

  • \(P(\text{2nd card is an ace | 1st card is non-ace heart}) = \frac{4}{51}\) (4 aces left)

  • \(P(\text{Case 2}) = \frac{12}{52} \times \frac{4}{51} = \frac{48}{2652}\)

\[P(A \text{ and } B) = \frac{3}{2652} + \frac{48}{2652} = \frac{51}{2652} = \boxed{\frac{1}{52} = 0.0192}\]

3. P(B|A)

Using the definition of conditional probability: \[P(B|A) = \frac{P(A \text{ and } B)}{P(A)} = \frac{\frac{1}{52}}{\frac{1}{4}} = \frac{1}{52} \times \frac{4}{1} = \boxed{\frac{4}{52} = \frac{1}{13} = 0.0769}\]

Alternative approach: Given that the first card is a heart:

  • If it’s the Ace of Hearts: 3 aces remain out of 51 cards

  • If it’s a non-ace heart: 4 aces remain out of 51 cards

  • \(P(B|A) = \frac{1}{13} \times \frac{3}{51} + \frac{12}{13} \times \frac{4}{51} = \frac{3 + 48}{13 \times 51} = \frac{51}{663} = \frac{4}{52} = \frac{1}{13}\)

4. P(B)

Using the Law of Total Probability. Let \(A^c\) = “first card is not a heart”

\[P(B) = P(B|A) \cdot P(A) + P(B|A^c) \cdot P(A^c)\]

We know:

  • \(P(A) = \frac{1}{4}\) and \(P(A^c) = \frac{3}{4}\)

  • \(P(B|A) = \frac{1}{13}\) (from part 3)

  • \(P(B|A^c) = \frac{4}{51}\) (if first card isn’t a heart, all 4 aces remain)

\[P(B) = \frac{1}{13} \times \frac{1}{4} + \frac{4}{51} \times \frac{3}{4} = \frac{1}{52} + \frac{12}{204} = \frac{1}{52} + \frac{3}{51}\]

\[= \frac{51 + 156}{52 \times 51} = \frac{207}{2652} = \boxed{\frac{4}{51} = 0.0784}\]

5. Comparison of P(B|A) vs P(B)

\[P(B|A) = \frac{1}{13} = 0.0769\] \[P(B) = \frac{4}{51} = 0.0784\]

Analysis: \(P(B|A) < P(B)\)

Explanation: The probability of getting an ace on the second draw is slightly lower when we know the first card is a heart compared to when we don’t know anything about the first card. This happens because:

  • When the first card is a heart, there’s a \(\frac{1}{13}\) chance it’s the Ace of Hearts, removing one ace from the deck

  • This makes it slightly less likely to draw an ace on the second draw

  • This demonstrates dependence - the events are not independent because drawing without replacement creates dependence between successive draws

Section D: Advanced Counting with Restrictions - SOLUTIONS

⏱️ Estimated time: 15 minutes

Problem D1: Advanced Counting with Restrictions - SOLUTION

Given:

  • 6 appetizer options (including 1 seafood)

  • 8 main course options (including 1 vegetarian, and 3 that are beef or chicken)

  • 5 dessert options (including 1 chocolate)

Restrictions:

  1. Seafood appetizer → cannot choose vegetarian main course

  2. Chocolate dessert → must choose beef or chicken main course (3 specific options)

Part (a): How many valid meal combinations are possible?

Solution using cases:

Case 1: Seafood appetizer is chosen

  • 1 appetizer choice (seafood)

  • 7 main course choices (8 total minus 1 vegetarian)

  • 5 dessert choices (no restrictions)

  • Total: \(1 \times 7 \times 5 = 35\) combinations

Case 2: Non-seafood appetizer + chocolate dessert

  • 5 appetizer choices (6 total minus 1 seafood)

  • 3 main course choices (only beef or chicken allowed with chocolate)

  • 1 dessert choice (chocolate)

  • Total: \(5 \times 3 \times 1 = 15\) combinations

Case 3: Non-seafood appetizer + non-chocolate dessert

  • 5 appetizer choices (6 total minus 1 seafood)

  • 8 main course choices (no restrictions since no seafood appetizer)

  • 4 dessert choices (5 total minus 1 chocolate)

  • Total: \(5 \times 8 \times 4 = 160\) combinations

Total valid combinations: \[35 + 15 + 160 = \boxed{210 \text{ combinations}}\]

Verification using complementary counting:

  • Total unrestricted combinations: \(6 \times 8 \times 5 = 240\)

  • Invalid combinations to subtract:

    • Seafood + vegetarian + any dessert: \(1 \times 1 \times 5 = 5\)

    • Non-seafood + chocolate + non-beef/chicken: \(5 \times 5 \times 1 = 25\)

  • Valid combinations: \(240 - 5 - 25 = 210\) ✓

Part (b): If customers choose randomly among valid combinations, what is the probability someone chooses the chocolate dessert?

Solution: From our case analysis, combinations with chocolate dessert come only from Case 2:

  • Combinations with chocolate dessert: 15

  • Total valid combinations: 210

\[P(\text{chocolate dessert}) = \frac{15}{210} = \frac{1}{14} = \boxed{0.0714}\]

Alternative verification:

We can also calculate this directly:

  • Non-seafood appetizers: 5 choices

  • With chocolate dessert, must choose from 3 main courses

  • Valid chocolate combinations: \(5 \times 3 = 15\)

  • Probability: \(\frac{15}{210} = \frac{1}{14} = 0.0714\) ✓

Summary of Key Concepts

Probability Distributions

  • Valid distributions require: all probabilities \(\geq 0\) and sum \(= 1\)

  • Check both conditions systematically

Permutations vs Combinations

  • Permutations: Order matters, use \(P(n,r) = \frac{n!}{(n-r)!}\)

  • Combinations: Order doesn’t matter, use \(C(n,r) = \frac{n!}{r!(n-r)!}\)

  • Multiplication principle: Combine independent choices

Conditional Probability

  • Without replacement: Creates dependence between events

  • Use definition: \(P(B|A) = \frac{P(A \cap B)}{P(A)}\)

  • Law of Total Probability: For calculating marginal probabilities

Advanced Counting

  • Case analysis: Break complex problems into manageable parts

  • Handle restrictions: Consider what’s allowed vs. not allowed

  • Verification: Use complementary counting or direct calculation