36. Basic probability (coins, dice, cards) problems

36. Basic probability (coins, dice, cards) problems - step by step solution procedure and detailed explanation for each problem.

Below are 10 basic probability problems involving coins, dice, and cards, each with a step-by-step solution procedure and a detailed explanation. These problems are designed to cover fundamental probability concepts, such as independent events, mutually exclusive events, and conditional probability, while keeping calculations straightforward. Each problem includes the problem statement, solution steps, and an explanation of the reasoning and probability principles involved.

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Problem 1: Flipping a Coin Twice

Problem: A fair coin is flipped twice. What is the probability of getting two heads?

Solution Procedure:

1. Determine the sample space: Each flip has 2 outcomes (H or T), so for 2 flips, the sample space has

   2 \times 2 = 4

   outcomes: {HH, HT, TH, TT}.
2. Identify favorable outcomes: We want two heads, which is only {HH}, so there is 1 favorable outcome.
3. Calculate the probability: Probability =

   \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{4}

   .
4. Alternative method: Each flip is independent, with

   P(\text{Head}) = \frac{1}{2}

   . For two heads:

   P(\text{HH}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

   .
5. State the answer: The probability is

   \frac{1}{4}

   .

Detailed Explanation:

• Why sample space? Listing all possible outcomes ensures we account for every scenario.
• Independent events: The outcome of the first flip does not affect the second, so probabilities multiply.
• Verification: The sample space is small, so we can confirm {HH} is 1 of 4 equally likely outcomes.
• Formula: For independent events,

  P(A \text{ and } B) = P(A) \cdot P(B)

  .

Answer:

\frac{1}{4}

.

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Problem 2: Rolling a Die

Problem: A fair six-sided die is rolled. What is the probability of rolling an even number?

Solution Procedure:

1. Determine the sample space: A die has 6 outcomes: {1, 2, 3, 4, 5, 6}.
2. Identify favorable outcomes: Even numbers are {2, 4, 6}, so there are 3 favorable outcomes.
3. Calculate the probability:

   \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}

   .
4. State the answer: The probability is

   \frac{1}{2}

   .

Detailed Explanation:

• Equally likely outcomes: Each face of a fair die has a probability of

  \frac{1}{6}

  .
• Event definition: The event “rolling an even number” includes specific outcomes, and we count them.
• Simplification: The probability simplifies because the number of even and odd outcomes is equal.
• Formula:

  P(\text{Event}) = \frac{\text{Number of event outcomes}}{\text{Total outcomes}}

  .

Answer:

\frac{1}{2}

.

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Problem 3: Drawing a Card from a Deck

Problem: A card is drawn from a standard 52-card deck. What is the probability it is a heart?

Solution Procedure:

1. Determine the sample space: A standard deck has 52 cards.
2. Identify favorable outcomes: There are 13 hearts in the deck (Ace through King of hearts).
3. Calculate the probability:

   \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{13}{52} = \frac{1}{4}

   .
4. State the answer: The probability is

   \frac{1}{4}

   .

Detailed Explanation:

• Deck composition: A standard deck has 4 suits (hearts, diamonds, clubs, spades), each with 13 ranks, making 52 cards.
• Uniform probability: Each card is equally likely to be drawn (

  \frac{1}{52}

  ).
• Suit probability: Since hearts are one of four suits, the probability is proportional to the number of hearts.
• Formula: Same as for equally likely outcomes.

Answer:

\frac{1}{4}

.

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Problem 4: Flipping a Coin and Rolling a Die

Problem: A fair coin is flipped, and a fair six-sided die is rolled. What is the probability of getting heads and a 3?

Solution Procedure:

1. Determine the sample space: Coin has 2 outcomes (H, T), die has 6 outcomes (1 to 6). Total outcomes =

   2 \times 6 = 12

   .
2. Identify favorable outcomes: We want (H, 3), which is 1 outcome.
3. Calculate the probability:

   \frac{1}{12}

   .
4. Alternative method: Events are independent.
   • P(\text{Heads}) = \frac{1}{2}

     .
   • P(\text{3 on die}) = \frac{1}{6}

     .
   • P(\text{Heads and 3}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

     .
5. State the answer: The probability is

   \frac{1}{12}

   .

Detailed Explanation:

• Independent events: The coin flip and die roll do not influence each other.
• Sample space: We can list outcomes like (H, 1), (H, 2), ..., (T, 6) to confirm 12 total.
• Multiplication rule: For independent events, multiply individual probabilities.
• Verification: The single outcome (H, 3) aligns with the calculated probability.

Answer:

\frac{1}{12}

.

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Problem 5: Rolling Two Dice (Sum of 7)

Problem: Two fair six-sided dice are rolled. What is the probability that the sum of the numbers is 7?

Solution Procedure:

1. Determine the sample space: Each die has 6 outcomes, so total outcomes =

   6 \times 6 = 36

   .
2. Identify favorable outcomes: List pairs where the sum is 7:
   • (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).
   • There are 6 favorable outcomes.
3. Calculate the probability:

   \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6}

   .
4. State the answer: The probability is

   \frac{1}{6}

   .

Detailed Explanation:

• Sample space: Each die is independent, and all 36 outcomes are equally likely.
• Sum calculation: We systematically check pairs to find those summing to 7.
• Why list pairs? The order matters (e.g., (1, 6) is different from (6, 1)), as each die is distinct.
• Formula: Probability is the ratio of favorable outcomes to total outcomes.

Answer:

\frac{1}{6}

.

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Problem 6: Drawing Two Cards Without Replacement

Problem: Two cards are drawn from a 52-card deck without replacement. What is the probability that both are aces?

Solution Procedure:

1. Determine the sample space for the first draw: There are 52 cards.
2. Favorable outcomes for the first draw: There are 4 aces, so

   P(\text{First ace}) = \frac{4}{52}

   .
3. Update for the second draw: After drawing one ace, 51 cards remain, with 3 aces.

   P(\text{Second ace} | \text{First ace}) = \frac{3}{51}

   .
4. Calculate the joint probability: Since the draws are dependent:
   • P(\text{Both aces}) = P(\text{First ace}) \cdot P(\text{Second ace} | \text{First ace}) = \frac{4}{52} \cdot \frac{3}{51}

     .
5. Simplify:

   \frac{4}{52} = \frac{1}{13}

   ,

   \frac{3}{51} = \frac{1}{17}

   , so

   \frac{1}{13} \cdot \frac{1}{17} = \frac{1}{221}

   .
6. Alternative method: Use combinations:
   • Ways to choose 2 aces:

     C(4, 2) = \frac{4!}{2!2!} = 6

     .
   • Total ways to choose 2 cards:

     C(52, 2) = \frac{52!}{2!50!} = \frac{52 \cdot 51}{2} = 1326

     .
   • Probability:

     \frac{C(4, 2)}{C(52, 2)} = \frac{6}{1326} = \frac{1}{221}

     .
7. State the answer: The probability is

   \frac{1}{221}

   .

Detailed Explanation:

• Without replacement: The deck changes after the first draw, making the events dependent.
• Conditional probability: The probability of the second ace depends on the first being an ace.
• Combinations approach: Useful for unordered selections, it confirms the sequential calculation.
• Formula: For dependent events,

  P(A \text{ and } B) = P(A) \cdot P(B|A)

  .

Answer:

\frac{1}{221}

.

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Problem 7: Flipping Three Coins (At Least One Head)

Problem: Three fair coins are flipped. What is the probability of getting at least one head?

Solution Procedure:

1. Determine the sample space: Each coin has 2 outcomes, so total outcomes =

   2^3 = 8

   : {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
2. Identify favorable outcomes: “At least one head” means excluding the outcome with no heads (TTT). Favorable outcomes: {HHH, HHT, HTH, HTT, THH, THT, TTH} = 7 outcomes.
3. Calculate the probability:

   \frac{7}{8}

   .
4. Alternative method: Calculate the complement (no heads):
   • P(\text{No heads}) = P(\text{TTT}) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}

     .
   • P(\text{At least one head}) = 1 - P(\text{No heads}) = 1 - \frac{1}{8} = \frac{7}{8}

     .
5. State the answer: The probability is

   \frac{7}{8}

   .

Detailed Explanation:

• Complement rule: It’s often easier to calculate the probability of the opposite event and subtract from 1:

  P(\text{Event}) = 1 - P(\text{Not Event})

  .
• Why easier? “No heads” is a single outcome, while “at least one head” includes multiple outcomes.
• Verification: Listing outcomes confirms 7 of 8 have at least one head.

Answer:

\frac{7}{8}

.

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Problem 8: Rolling a Die (Number Greater Than 4)

Problem: A fair six-sided die is rolled. What is the probability of rolling a number greater than 4?

Solution Procedure:

1. Determine the sample space: Outcomes are {1, 2, 3, 4, 5, 6}, so 6 outcomes.
2. Identify favorable outcomes: Numbers greater than 4 are {5, 6}, so 2 outcomes.
3. Calculate the probability:

   \frac{2}{6} = \frac{1}{3}

   .
4. State the answer: The probability is

   \frac{1}{3}

   .

Detailed Explanation:

• Simple event: The problem involves a single roll with a clear condition.
• Equally likely: Each outcome has a probability of

  \frac{1}{6}

  .
• Counting outcomes: We count outcomes satisfying the condition (>4).
• Formula: Basic probability ratio for equally likely outcomes.

Answer:

\frac{1}{3}

.

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Problem 9: Drawing a Card (Ace or King)

Problem: A card is drawn from a 52-card deck. What is the probability it is an ace or a king?

Solution Procedure:

1. Determine the sample space: 52 cards.
2. Identify favorable outcomes:
   • Aces: 4 (one per suit).
   • Kings: 4 (one per suit).
   • Total:

     4 + 4 = 8

     , since aces and kings are mutually exclusive (no card is both).
3. Calculate the probability:

   \frac{8}{52} = \frac{2}{13}

   .
4. Alternative method: Use the addition rule for mutually exclusive events:
   • P(\text{Ace}) = \frac{4}{52}

     ,

     P(\text{King}) = \frac{4}{52}

     .
   • P(\text{Ace or King}) = P(\text{Ace}) + P(\text{King}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}

     .
5. State the answer: The probability is

   \frac{2}{13}

   .

Detailed Explanation:

• Mutually exclusive: A card cannot be both an ace and a king, so we add probabilities.
• Addition rule: For mutually exclusive events,

  P(A \text{ or } B) = P(A) + P(B)

  .
• Verification: The 8 favorable cards (4 aces + 4 kings) align with the calculation.

Answer:

\frac{2}{13}

.

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Problem 10: Rolling Two Dice (Same Number)

Problem: Two fair six-sided dice are rolled. What is the probability that both dice show the same number?

Solution Procedure:

1. Determine the sample space: Total outcomes =

   6 \times 6 = 36

   .
2. Identify favorable outcomes: Same number means (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). There are 6 favorable outcomes.
3. Calculate the probability:

   \frac{6}{36} = \frac{1}{6}

   .
4. Alternative method: For each number (1 to 6), the probability of both dice showing it is:
   • P(\text{Both show } k) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}

     .
   • Sum for all 6 numbers:

     6 \cdot \frac{1}{36} = \frac{6}{36} = \frac{1}{6}

     .
5. State the answer: The probability is

   \frac{1}{6}

   .

Detailed Explanation:

• Independent rolls: Each die’s outcome is independent.
• Favorable pairs: We count pairs where the numbers match.
• Alternative approach: Summing probabilities for each matching pair confirms the result.
• Formula: Probability is based on counting or summing probabilities of specific outcomes.

Answer:

\frac{1}{6}

.

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Summary of Key Concepts

• Sample space: All possible outcomes of an experiment (e.g., 4 for two coin flips, 36 for two dice).
• Probability formula:

  P(\text{Event}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}}

  .
• Independent events: Multiply probabilities (

  P(A \text{ and } B) = P(A) \cdot P(B)

  ).
• Dependent events: Use conditional probability (

  P(A \text{ and } B) = P(A) \cdot P(B|A)

  ).
• Mutually exclusive events: Add probabilities (

  P(A \text{ or } B) = P(A) + P(B)

  ).
• Complement rule:

  P(\text{Event}) = 1 - P(\text{Not Event})

  .
• Combinations: Useful for problems involving selections without regard to order.

These problems introduce foundational probability concepts using familiar objects like coins, dice, and cards.