To solve the problem, we need to find the probability that either the first roll was a 4 or the sum of the two rolls was 6.
Let’s denote:
- Event A: The first roll was a 4.
- Event B: The sum of the two rolls was 6.
Step 1: Calculate the probabilities of each event
Total outcomes for two rolls of a number cube: Each roll can result in any of the numbers 1 through 6. Thus, there are a total of \(6 \times 6 = 36\) possible outcomes when rolling two number cubes.
Event A (First roll is a 4): If the first roll is a 4, the second roll can be any number from 1 to 6. The outcomes that satisfy this event are:
- (4, 1)
- (4, 2)
- (4, 3)
- (4, 4)
- (4, 5)
- (4, 6)
This gives us 6 favorable outcomes for event A.
Event B (Sum of the two rolls = 6): The pairs of rolls that produce a sum of 6 are as follows:
- (1, 5)
- (2, 4)
- (3, 3)
- (4, 2)
- (5, 1)
This gives us 5 favorable outcomes for event B.
Step 2: Calculate the intersection of events A and B (A ∩ B)
Next, we need to find the outcomes that are common to both events A and B—the scenarios where the first roll is a 4 and the total sum is 6. The only outcome that meets these criteria is:
- (4, 2)
Thus, \( |A \cap B| = 1\).
Step 3: Use the inclusion-exclusion principle to find the union of A and B
The probability of A or B is given by: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Calculating each probability:
- \( P(A) = \frac{6}{36} = \frac{1}{6} \)
- \( P(B) = \frac{5}{36} \)
- \( P(A \cap B) = \frac{1}{36} \)
Now, we can substitute the values into the formula: \[ P(A \cup B) = \frac{1}{6} + \frac{5}{36} - \frac{1}{36} \]
First, convert \(\frac{1}{6}\) to have a common denominator of 36: \[ \frac{1}{6} = \frac{6}{36} \]
Now, substitute: \[ P(A \cup B) = \frac{6}{36} + \frac{5}{36} - \frac{1}{36} = \frac{6 + 5 - 1}{36} = \frac{10}{36} \]
Step 4: Reduce the fraction
To reduce \(\frac{10}{36}\), divide both the numerator and the denominator by their greatest common divisor, which is 2: \[ \frac{10 \div 2}{36 \div 2} = \frac{5}{18} \]
Conclusion
The probability that the first roll was a 4 or the sum of the two rolls was 6 is: \[ \boxed{\frac{5}{18}} \]
5 answers