To solve the problem, we need to find the probability that \( g(3) = 2g(1) \) for an onto function \( g \) defined from the set \( \{1, 2, 3, 4, 5, 6\} \) to itself.
### Step-by-step Solution:
1. **Understanding Onto Functions**:
An onto function (or surjective function) from a set \( A \) to a set \( B \) means that every element in \( B \) is mapped to by at least one element in \( A \). In our case, both sets are \( \{1, 2, 3, 4, 5, 6\} \).
2. **Total Number of Onto Functions**:
The total number of onto functions from a set of \( n \) elements to itself is given by \( n! \). Here, \( n = 6 \), so the total number of onto functions is:
\[
6! = 720
\]
3. **Finding Favorable Conditions**:
We need to find the number of favorable conditions where \( g(3) = 2g(1) \).
- **Case 1**: If \( g(1) = 1 \), then \( g(3) = 2 \).
- Remaining values: \( 3, 4, 5, 6 \)
- The number of ways to assign these values to \( g(2), g(4), g(5), g(6) \) is \( 4! = 24 \).
- **Case 2**: If \( g(1) = 2 \), then \( g(3) = 4 \).
- Remaining values: \( 1, 3, 5, 6 \)
- The number of ways to assign these values is \( 4! = 24 \).
- **Case 3**: If \( g(1) = 3 \), then \( g(3) = 6 \).
- Remaining values: \( 1, 2, 4, 5 \)
- The number of ways to assign these values is \( 4! = 24 \).
- **Case 4**: If \( g(1) = 4 \), then \( g(3) = 8 \) (not possible since 8 is not in the set).
- **Case 5**: If \( g(1) = 5 \), then \( g(3) = 10 \) (not possible).
- **Case 6**: If \( g(1) = 6 \), then \( g(3) = 12 \) (not possible).
Thus, the only valid cases are Case 1, Case 2, and Case 3. The total number of favorable conditions is:
\[
3 \times 4! = 3 \times 24 = 72
\]
4. **Calculating Probability**:
The probability \( P \) that \( g(3) = 2g(1) \) is given by the ratio of favorable conditions to the total number of onto functions:
\[
P = \frac{\text{Favorable Conditions}}{\text{Total Onto Functions}} = \frac{72}{720} = \frac{1}{10}
\]
### Final Answer:
The probability that \( g(3) = 2g(1) \) is \( \frac{1}{10} \).
