Key Concepts and Formulas

This problem requires us to calculate the probability of a combined event involving two independent dice, each with non-standard face markings. We will use the fundamental principles of probability:

1. Definition of Probability: For an event $E$, its probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of possible outcomes, assuming all outcomes are equally likely: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$
2. Sample Space for Independent Events: When two independent events occur (like rolling two dice), the total number of possible outcomes is the product of the number of outcomes for each individual event. If Die 1 has $N_1$ faces and Die 2 has $N_2$ faces, the total number of outcomes when both are thrown is $N_1 \times N_2$.
3. Probability of Mutually Exclusive Events: If multiple events, say $A$, $B$, and $C$, cannot occur at the same time (they are mutually exclusive), the probability that any one of them occurs is the sum of their individual probabilities: $P(A \text{ or } B \text{ or } C) = P(A) + P(B) + P(C)$ In this problem, getting a sum of 4, getting a sum of 5, and getting a sum of 6 are mutually exclusive events.
4. Counting Favorable Outcomes for Non-Standard Dice: When dice have repeated numbers on their faces, we must account for the frequency of each number. If Die 1 shows number 'x' in $f_1(x)$ ways and Die 2 shows number 'y' in $f_2(y)$ ways, then the pair $(x, y)$ can occur in $f_1(x) \times f_2(y)$ ways.

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Step-by-Step Solution

Step 1: Understand the Dice and Define the Sample Space

First, let's list the face markings and their frequencies for each die. This will allow us to determine the total number of possible outcomes when both dice are thrown.

• Die 1 ($D_1$):

  • Two faces marked '1' (Frequency $f_1(1) = 2$)
  • Two faces marked '2' (Frequency $f_1(2) = 2$)
  • One face marked '3' (Frequency $f_1(3) = 1$)
  • One face marked '4' (Frequency $f_1(4) = 1$)
  • Total faces on $D_1 = 2+2+1+1 = 6$.

• Die 2 ($D_2$):

  • One face marked '1' (Frequency $f_2(1) = 1$)
  • Two faces marked '2' (Frequency $f_2(2) = 2$)
  • Two faces marked '3' (Frequency $f_2(3) = 2$)
  • One face marked '4' (Frequency $f_2(4) = 1$)
  • Total faces on $D_2 = 1+2+2+1 = 6$.

Since each die has 6 faces and they are thrown independently, the total number of distinct possible outcomes in the sample space is: $\text{Total Outcomes} = (\text{Number of faces on } D_1) \times (\text{Number of faces on } D_2) = 6 \times 6 = 36$ Each of these 36 outcomes (e.g., specific face $1_a$ on $D_1$ and specific face $1_c$ on $D_2$) is equally likely.

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Step 2: Identify Favorable Outcomes for Sum = 4

We need to find all pairs of numbers $(D_1, D_2)$ that sum to 4 and calculate the number of ways each pair can occur using the frequencies from Step 1.

The possible pairs $(D_1, D_2)$ that sum to 4 are:

1. $(1, 3)$:
   • Number of ways to get '1' on $D_1 = f_1(1) = 2$
   • Number of ways to get '3' on $D_2 = f_2(3) = 2$
   • Total ways for $(1, 3) = f_1(1) \times f_2(3) = 2 \times 2 = 4$
2. $(2, 2)$:
   • Number of ways to get '2' on $D_1 = f_1(2) = 2$
   • Number of ways to get '2' on $D_2 = f_2(2) = 2$
   • Total ways for $(2, 2) = f_1(2) \times f_2(2) = 2 \times 2 = 4$
3. $(3, 1)$:
   • Number of ways to get '3' on $D_1 = f_1(3) = 1$
   • Number of ways to get '1' on $D_2 = f_2(1) = 1$
   • Total ways for $(3, 1) = f_1(3) \times f_2(1) = 1 \times 1 = 1$

The total number of favorable outcomes for a sum of 4 is the sum of ways for these pairs: $\text{Favorable Outcomes (Sum=4)} = 4 + 4 + 1 = 9$

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Step 3: Identify Favorable Outcomes for Sum = 5

Next, we find all pairs $(D_1, D_2)$ that sum to 5 and calculate the number of ways each pair can occur.

The possible pairs $(D_1, D_2)$ that sum to 5 are:

1. $(1, 4)$:
   • Number of ways to get '1' on $D_1 = f_1(1) = 2$
   • Number of ways to get '4' on $D_2 = f_2(4) = 1$
   • Total ways for $(1, 4) = f_1(1) \times f_2(4) = 2 \times 1 = 2$
2. $(2, 3)$:
   • Number of ways to get '2' on $D_1 = f_1(2) = 2$
   • Number of ways to get '3' on $D_2 = f_2(3) = 2$
   • Total ways for $(2, 3) = f_1(2) \times f_2(3) = 2 \times 2 = 4$
3. $(3, 2)$:
   • Number of ways to get '3' on $D_1 = f_1(3) = 1$
   • Number of ways to get '2' on $D_2 = f_2(2) = 2$
   • Total ways for $(3, 2) = f_1(3) \times f_2(2) = 1 \times 2 = 2$
4. $(4, 1)$:
   • Number of ways to get '4' on $D_1 = f_1(4) = 1$
   • Number of ways to get '1' on $D_2 = f_2(1) = 1$
   • Total ways for $(4, 1) = f_1(4) \times f_2(1) = 1 \times 1 = 1$

The total number of favorable outcomes for a sum of 5 is the sum of ways for these pairs: $\text{Favorable Outcomes (Sum=5)} = 2 + 4 + 2 + 1 = 9$

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Step 4: Identify Favorable Outcomes for Sum = 6

To ensure we cover all outcomes relevant to the required probability, we also calculate the favorable outcomes for a sum of 6.

The possible pairs $(D_1, D_2)$ that sum to 6 are:

1. $(2, 4)$:
   • Number of ways to get '2' on $D_1 = f_1(2) = 2$
   • Number of ways to get '4' on $D_2 = f_2(4) = 1$
   • Total ways for $(2, 4) = f_1(2) \times f_2(4) = 2 \times 1 = 2$
2. $(3, 3)$:
   • Number of ways to get '3' on $D_1 = f_1(3) = 1$
   • Number of ways to get '3' on $D_2 = f_2(3) = 2$
   • Total ways for $(3, 3) = f_1(3) \times f_2(3) = 1 \times 2 = 2$
3. $(4, 2)$:
   • Number of ways to get '4' on $D_1 = f_1(4) = 1$
   • Number of ways to get '2' on $D_2 = f_2(2) = 2$
   • Total ways for $(4, 2) = f_1(4) \times f_2(2) = 1 \times 2 = 2$

The total number of favorable outcomes for a sum of 6 is the sum of ways for these pairs: $\text{Favorable Outcomes (Sum=6)} = 2 + 2 + 2 = 6$

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Step 5: Calculate the Total Favorable Outcomes for the Event

Since getting a sum of 4, a sum of 5, and a sum of 6 are mutually exclusive events (they cannot happen at the same time), the total number of favorable outcomes for the event is the sum of the favorable outcomes for each case: $\text{Total Favorable Outcomes} = \text{Favorable Outcomes (Sum=4)} + \text{Favorable Outcomes (Sum=5)} + \text{Favorable Outcomes (Sum=6)} = 9 + 9 + 6 = 24$

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Step 6: Calculate the Final Probability

Now, we can calculate the probability of the event by dividing the total favorable outcomes by the total possible outcomes: $P(\text{Sum=4 or Sum=5}) = \frac{\text{Total Favorable Outcomes}}{\text{Total Outcomes}} = \frac{24}{36} = \frac{2}{3}$

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Common Mistakes & Tips

• Non-Standard Dice: Always carefully list the frequencies of each number for non-standard dice. Do not assume standard 1/6 probability for each number.
• Total Outcomes: Ensure you correctly calculate the total number of possible outcomes. For two dice with $N_1$ and $N_2$ faces, it's $N_1 \times N_2$.
• Counting Favorable Outcomes: Systematically list all combinations that lead to the desired sum. For each combination $(x,y)$, multiply their individual frequencies $f_1(x) \times f_2(y)$.
• Mutually Exclusive Events: Remember that for "OR" events that are mutually exclusive, you add the number of favorable outcomes (or probabilities).
• Double-Checking: It's a good practice to sum up the frequencies of all possible outcomes to ensure it matches the total sample space.

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Summary and Key Takeaway

By systematically listing the frequencies of each face value on both non-standard dice and then calculating the number of ways to achieve a sum of 4, 5, or 6, we found that there are 9 ways to get a sum of 4, 9 ways to get a sum of 5, and 6 ways to get a sum of 6. Combining these mutually exclusive outcomes, there are $9 + 9 + 6 = 24$ total favorable outcomes. With a total of 36 possible outcomes in the sample space, the probability of the event is $\frac{24}{36} = \frac{2}{3}$.

The final answer is $\boxed{\text{A}}$.