If two dice are thrown and at least one of the dice shows 5, then the probability that the sum is 10 or more is:

Options

A. `(1)/(6)`
B. `(4)/(11)`
C. `(3)/(11)`
D. `(2)/(11)`

Answer

To solve the problem step by step, we need to find the probability that the sum of two dice is 10 or more given that at least one die shows a 5.

### Step 1: Identify the sample space
When two dice are thrown, and at least one die shows a 5, we need to consider all the possible outcomes that include at least one die showing a 5.

- If the first die is 5, the second die can be any of the numbers from 1 to 6:
- (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

- If the second die is 5 (and the first die is not 5 to avoid double counting), the first die can be any of the numbers from 1 to 6 except 5:
- (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)

Combining these outcomes, the complete sample space (S) is:
- (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
- (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)

Thus, the total number of outcomes in the sample space is:
- Total outcomes = 6 (from first die being 5) + 5 (from second die being 5) = 11 outcomes.

### Step 2: Identify the favorable outcomes
Next, we need to find the outcomes where the sum of the two dice is 10 or more. The possible pairs from our sample space that satisfy this condition are:
- (5, 5) → Sum = 10
- (5, 6) → Sum = 11
- (6, 5) → Sum = 11

Thus, the favorable outcomes (A) are:
- (5, 5), (5, 6), (6, 5)

The number of favorable outcomes is 3.

### Step 3: Calculate the probability
The probability of an event is given by the formula:
\[ P(A) = \frac{N(A)}{N(S)} \]
Where:
- \( N(A) \) = Number of favorable outcomes = 3
- \( N(S) \) = Total number of outcomes in the sample space = 11

So, the probability \( P(A) \) is:
\[ P(A) = \frac{3}{11} \]

### Final Answer
The probability that the sum is 10 or more given that at least one die shows 5 is \( \frac{3}{11} \).

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