Understanding the Probability Problem

This problem involves calculating the probability of a specific event (Event E) when a fair six-sided die is rolled three times. Event E occurs if either all three outcomes are the same number or if none of the outcomes is the number 4. We need to find the probability of this combined event.

Defining the Sample Space

A standard fair six-sided die has possible outcomes: {1, 2, 3, 4, 5, 6}. Since the die is tossed three times, each toss is independent. The total number of possible sequences of outcomes is calculated by multiplying the number of outcomes for each toss:

Total Possible Outcomes = $6 \times 6 \times 6 = 6^3 = 216$

Breaking Down Event E

Event E is the union of two separate events:

• Event A: All three numbers rolled are equal.
• Event B: None of the three numbers rolled is a 4.

We are looking for the probability of Event A happening OR Event B happening, which is represented as P(A ∪ B). The formula to calculate this is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Here, P(A ∩ B) represents the probability that *both* Event A and Event B occur simultaneously.

Calculating Probability of Event A (All Numbers Equal)

Event A includes sequences where all three tosses show the same number. These are:

(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6)

There are 6 outcomes that satisfy Event A.

The probability of Event A is:

$P(A) = \frac{\text{Number of outcomes in A}}{\text{Total Outcomes}} = \frac{6}{216}$

Calculating Probability of Event B (No 4s Rolled)

Event B occurs when none of the three tosses is a 4. This means for each toss, the outcome must be from the set {1, 2, 3, 5, 6}. There are 5 possible outcomes for each toss that satisfy this condition.

The total number of outcomes for Event B is:

Number of outcomes in B = $5 \times 5 \times 5 = 5^3 = 125$

The probability of Event B is:

$P(B) = \frac{\text{Number of outcomes in B}}{\text{Total Outcomes}} = \frac{125}{216}$

Calculating Probability of the Intersection (A ∩ B)

The intersection (A ∩ B) represents the outcomes where *both* Event A (all numbers equal) and Event B (no 4s) are true. We need to find the sequences from Event A that do not contain the number 4.

Looking at the outcomes for Event A: (1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6).

The outcomes that satisfy both conditions (all equal and no 4) are:

(1,1,1), (2,2,2), (3,3,3), (5,5,5), (6,6,6)

There are 5 outcomes in the intersection (A ∩ B).

The probability of the intersection is:

$P(A \cap B) = \frac{\text{Number of outcomes in (A ∩ B)}}{\text{Total Outcomes}} = \frac{5}{216}$

Calculating the Final Probability P(E)

Now we apply the inclusion-exclusion formula using the probabilities calculated above:

$P(E) = P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(E) = \frac{6}{216} + \frac{125}{216} - \frac{5}{216}$

Combine the numerators over the common denominator:

$P(E) = \frac{6 + 125 - 5}{216}$

$P(E) = \frac{126}{216}$

Simplifying the Result

The final step is to simplify the fraction $\frac{126}{216}$. We can divide both the numerator and the denominator by their greatest common divisor.

Let's simplify in steps. Both numbers are divisible by 6:

$\frac{126 \div 6}{216 \div 6} = \frac{21}{36}$

Now, both 21 and 36 are divisible by 3:

$\frac{21 \div 3}{36 \div 3} = \frac{7}{12}$

Thus, the probability of Event E is $\frac{7}{12}$.

Summary Table