1. Interpret the condition carefully

We are choosing an onto function $g:S\to S$, where $S=\{1,2,3,4,5,6\}.$

Since domain and codomain both have 6 elements, any onto function is automatically one-one as well. Hence every onto function is a bijection.

So the sample space consists of all permutations of $S$: $\text{Total onto functions}=6! = 720.$

2. Apply the condition $g(3)=2g(1)$

Because $g(1),g(3)\in S=\{1,2,3,4,5,6\}$, we need $2g(1)\in S.$

Thus possible values of $g(1)$ are only:

• $g(1)=1 \Rightarrow g(3)=2$
• $g(1)=2 \Rightarrow g(3)=4$
• $g(1)=3 \Rightarrow g(3)=6$

Values $g(1)=4,5,6$ are impossible since then $2g(1)\notin S$.

So there are exactly 3 possible ordered choices for the pair $(g(1),g(3))$: $(1,2),\ (2,4),\ (3,6).$

3. Count bijections for each valid choice

Once $g(1)$ and $g(3)$ are fixed, the remaining 4 elements of the domain $\{2,4,5,6\}$ must be mapped bijectively onto the remaining 4 unused values in $S$.

That can be done in $4! = 24$ ways for each valid pair.

Hence favorable functions: $3\cdot 4! = 3\cdot 24 = 72.$

4. Compute the probability

Therefore, $P(g(3)=2g(1))=\frac{72}{720}=\frac{1}{10}.$

5. Check options

$\frac{1}{10}$ corresponds to Option A.

6. Compare with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.