If two dice are thrown and at least one the dice show 5, then the probability that the sum is 10 or more is
3/11
Calculating Probability with Two Dice
This problem involves finding a conditional probability. We are given that at least one of the two dice shows a 5, and we need to find the probability that the sum of the two dice is 10 or more under this condition.
Let's define the events:
- Event A: The sum of the two dice is 10 or more.
- Event B: At least one of the two dice shows a 5.
We want to find the probability of event A occurring given that event B has occurred, which is denoted as \(P(A|B)\). The formula for conditional probability is:
\(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
Alternatively, we can use the reduced sample space approach, considering only the outcomes where event B occurs.
Sample Space of Two Dice
When two fair dice are thrown, there are \(6 \times 6 = 36\) possible outcomes. Each outcome is an ordered pair \((d_1, d_2)\), where \(d_1\) is the result of the first die and \(d_2\) is the result of the second die, and \(d_1, d_2 \in \{1, 2, 3, 4, 5, 6\}\).
Outcomes for Event B (At Least One Die Shows 5)
Event B occurs if the first die is a 5 OR the second die is a 5 (or both). Let's list these outcomes:
- First die is 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
- Second die is 5: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5)
The outcome (5,5) is listed in both sets, so we count it only once. The unique outcomes for event B are:
(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6)
There are 11 outcomes where at least one die shows 5. The probability of event B is \(P(B) = \frac{11}{36}\).
Outcomes for Event A (Sum is 10 or More)
Event A occurs if the sum of the two dice is 10, 11, or 12. Let's list these outcomes:
- Sum is 10: (4,6), (5,5), (6,4)
- Sum is 11: (5,6), (6,5)
- Sum is 12: (6,6)
The outcomes for event A are: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6). There are 6 such outcomes.
Outcomes for Event A \(\cap\) B (Sum is 10 or More AND At Least One Die Shows 5)
We need the outcomes that are in both event A and event B. These are the outcomes where the sum is 10 or more AND at least one die is a 5.
Let's look at the outcomes in event B and check their sum:
| Outcome | Sum | Sum \(\ge\) 10? |
|---|---|---|
| (1,5) | 6 | No |
| (2,5) | 7 | No |
| (3,5) | 8 | No |
| (4,5) | 9 | No |
| (5,5) | 10 | Yes |
| (6,5) | 11 | Yes |
| (5,1) | 6 | No |
| (5,2) | 7 | No |
| (5,3) | 8 | No |
| (5,4) | 9 | No |
| (5,6) | 11 | Yes |
The outcomes that are in both A and B are (5,5), (6,5), and (5,6). There are 3 such outcomes.
The probability of event A \(\cap\) B is \(P(A \cap B) = \frac{3}{36}\).
Calculating the Conditional Probability
Now we can calculate \(P(A|B)\) using the formula:
\(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{3/36}{11/36}\)
To simplify, we can multiply the numerator and denominator by 36:
\(P(A|B) = \frac{3}{11}\)
Using the Reduced Sample Space
Alternatively, we know that event B (at least one die shows 5) has 11 outcomes. These 11 outcomes form our new sample space for the conditional probability. We just need to count how many of these 11 outcomes result in a sum of 10 or more.
The outcomes in event B are: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6).
From the table above, the outcomes from this list with a sum of 10 or more are: (5,5), (6,5), (5,6).
There are 3 such outcomes within the reduced sample space of 11 outcomes.
So, the probability is \(\frac{\text{Number of outcomes in B with sum} \ge 10}{\text{Total number of outcomes in B}} = \frac{3}{11}\).
Conclusion
The probability that the sum is 10 or more, given that at least one of the dice shows 5, is \(\frac{3}{11}\).
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