Probability Calculation for Two Dice

When rolling two fair cubical dice, we first need to determine the total number of possible outcomes. Each die has 6 faces, numbered 1 through 6. Since the outcome of one die does not affect the outcome of the other, the total number of outcomes is the product of the number of outcomes for each die.

• Total outcomes for one die = 6
• Total outcomes for two dice = \(6 \times 6 = 36\)

These 36 outcomes can be represented as pairs \((a, b)\), where \(a\) is the result of the first die and \(b\) is the result of the second die, and both \(a, b \in \{1, 2, 3, 4, 5, 6\}\).

Finding Favorable Outcomes: At Least One 4

The question asks for the probability that at least one of the dice shows a 4. This means we are interested in outcomes where the first die is a 4, OR the second die is a 4, OR both dice are 4.

We can list the outcomes where at least one die shows a 4:

• Outcomes where the first die is 4: \((4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\)
• Outcomes where the second die is 4: \((1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)\)

If we simply add these lists, the outcome \((4, 4)\) is included in both lists. To count the unique outcomes where at least one 4 appears, we can list them out:

• \((4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)\)
• \((1, 4), (2, 4), (3, 4), (5, 4), (6, 4)\) <-- (4,4) is already listed

Counting these unique outcomes, we find there are \(6 + 5 = 11\) favorable outcomes.

Calculating the Probability

The probability of an event is calculated as:

Probability = \(\frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}\)

In this case:

• Number of Favorable Outcomes (at least one 4) = 11
• Total Number of Possible Outcomes = 36

So, the probability that at least one of the dice shows 4 is \(\frac{11}{36}\).

Alternative Method: Using the Complement

Sometimes, calculating the probability of "at least one" is easier by calculating the probability of the opposite event (the complement) and subtracting it from 1. The complement of "at least one shows 4" is "neither die shows 4".

• For a single die, there are 5 outcomes that are not 4 (i.e., 1, 2, 3, 5, 6).
• For two dice, the number of outcomes where neither shows 4 is \(5 \times 5 = 25\).

The probability that neither die shows 4 is \(\frac{25}{36}\).

The probability that at least one die shows 4 is:

\(P(\text{at least one 4}) = 1 - P(\text{neither shows 4})\)

\(P(\text{at least one 4}) = 1 - \frac{25}{36}\)

\(P(\text{at least one 4}) = \frac{36}{36} - \frac{25}{36}\)

\(P(\text{at least one 4}) = \frac{11}{36}\)

Both methods give the same result.