Key Concepts and Formulas

1. Onto Function (Surjective Function): A function $g: A \to B$ is called onto if every element in the codomain $B$ has at least one corresponding element in the domain $A$. This means the range of the function is equal to its codomain ($g(A) = B$).
2. Onto Function from a Finite Set to Itself: When the domain $A$ and codomain $B$ are the same finite set with the same number of elements (i.e., $g: S \to S$ and $|S|=n$), an onto function is also necessarily a one-to-one (injective) function. Such a function is called a bijection or a permutation. The total number of such functions from a set of $n$ elements to itself is $n!$.
3. Probability Formula: The probability of an event $E$ is the ratio of the number of outcomes favorable to $E$ to the total number of possible outcomes in the sample space: $P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

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Step-by-Step Solution

Let the given set be $S = \{1, 2, 3, 4, 5, 6\}$. We are asked to find the probability that a randomly chosen onto function $g: S \to S$ satisfies the condition $g(3) = 2g(1)$.

Step 1: Determine the Total Number of Possible Outcomes (Sample Space)

• Understanding the Function Type: The problem states that $g$ is an "onto function" from $S$ to $S$. Since the domain and codomain are the same finite set, $S$, with $|S| = 6$ elements, any onto function $g: S \to S$ must also be a one-to-one function. Therefore, $g$ is a bijection (or a permutation) of the elements of $S$.
• Calculation of Total Functions: The total number of bijections from a set of $n$ elements to itself is $n!$. For $S = \{1, 2, 3, 4, 5, 6\}$, the total number of onto functions is $6!$. $\text{Total number of onto functions} = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ This value represents the size of our sample space.

Step 2: Determine the Number of Favorable Outcomes

• Identifying the Condition: We need to find the number of onto functions $g: S \to S$ that satisfy the condition $g(3) = 2g(1)$.
• Finding Valid Pairs for (g(1), g(3)): Since $g$ is a bijection, $g(1)$ and $g(3)$ must be distinct elements from the set $S$. Both $g(1)$ and $g(3)$ must belong to $S = \{1, 2, 3, 4, 5, 6\}$. Let's list the possible values for $g(1)$ and the corresponding $g(3)$ values:
  • If $g(1) = 1$, then $g(3) = 2 \times 1 = 2$. This pair $(1, 2)$ is valid as both $1, 2 \in S$ and they are distinct.
  • If $g(1) = 2$, then $g(3) = 2 \times 2 = 4$. This pair $(2, 4)$ is valid as both $2, 4 \in S$ and they are distinct.
  • If $g(1) = 3$, then $g(3) = 2 \times 3 = 6$. This pair $(3, 6)$ is valid as both $3, 6 \in S$ and they are distinct.
  • If $g(1) = 4$, then $g(3) = 2 \times 4 = 8$. This is not valid because $8 \notin S$.
  • Any other value for $g(1)$ (e.g., 5 or 6) would also result in $g(3)$ being outside of $S$.
• Counting Valid Pairs: There are exactly 3 valid pairs for $(g(1), g(3))$ that satisfy the given condition: $(1, 2)$, $(2, 4)$, and $(3, 6)$.
• Completing the Function for Each Pair: For each of these 3 valid pairs, the values of $g(1)$ and $g(3)$ are fixed.
  • The remaining elements in the domain are $S \setminus \{1, 3\}$, which has $6 - 2 = 4$ elements.
  • The remaining elements in the codomain are $S \setminus \{g(1), g(3)\}$, which also has $6 - 2 = 4$ elements (since $g(1)$ and $g(3)$ are distinct).
  • Since $g$ must be a bijection, the remaining 4 elements from the domain must be mapped bijectively to the remaining 4 elements from the codomain. The number of ways to do this is $4!$. $\text{Number of ways to map remaining elements} = 4! = 4 \times 3 \times 2 \times 1 = 24$
• Total Favorable Outcomes: Multiply the number of valid pairs by the number of ways to complete the function for each pair. $\text{Number of favorable outcomes} = (\text{Number of valid pairs}) \times (4!) = 3 \times 24 = 72$

Step 3: Calculate the Probability

• Now we use the probability formula with the total outcomes from Step 1 and favorable outcomes from Step 2. $P(g(3) = 2g(1)) = \frac{\text{Number of favorable outcomes}}{\text{Total number of onto functions}} = \frac{72}{720}$
• Simplify the fraction: $P(g(3) = 2g(1)) = \frac{72}{720} = \frac{1}{10}$

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Common Mistakes & Tips

• Confusing Onto with One-to-One: Remember that for functions from a finite set to itself, "onto" implies "one-to-one" (and vice versa), making them bijections or permutations. This simplifies counting significantly.
• Incorrectly Identifying Valid Pairs: Be careful to ensure that both $g(1)$ and $g(3)$ are elements of the codomain $S$ and that $g(1) \neq g(3)$ for a bijection.
• Calculation Errors: Factorials can be large; double-check calculations to avoid simple arithmetic mistakes.

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Summary

To find the probability, we first determined the total number of possible onto functions from $S$ to $S$, which are permutations, resulting in $6! = 720$ functions. Next, we identified all pairs $(g(1), g(3))$ that satisfy $g(3) = 2g(1)$ and are valid within the set $S$ and the constraints of a bijection. There were 3 such pairs. For each pair, the remaining 4 domain elements could be mapped to the remaining 4 codomain elements in $4!$ ways. This gave $3 \times 4! = 72$ favorable outcomes. Finally, dividing the favorable outcomes by the total outcomes yielded the probability $\frac{72}{720} = \frac{1}{10}$.

The final answer is $\boxed{\frac{1}{10}}$, which corresponds to option (A).