There are 16 balls and six empty boxes with the natural numbers 1 through 6 written on them. A trial is performed using one die. When the die is rolled and the result is $k$: If $k$ is odd, place 1 ball each in the boxes labeled $1, 3, 5$, and if $k$ is even, place 1 ball each in the boxes labeled with the divisors of $k$. After repeating this trial 4 times, given that the sum of all balls in the six boxes is odd, what is the probability that the number of balls in the box labeled 3 is 1 more than the number of balls in the box labeled 2? [4 points] (1) $\frac { 1 } { 8 }$ (2) $\frac { 3 } { 16 }$ (3) $\frac { 1 } { 4 }$ (4) $\frac { 5 } { 16 }$ (5) $\frac { 3 } { 8 }$
There are 16 balls and six empty boxes with the natural numbers 1 through 6 written on them. A trial is performed using one die.
When the die is rolled and the result is $k$:
If $k$ is odd, place 1 ball each in the boxes labeled $1, 3, 5$, and
if $k$ is even, place 1 ball each in the boxes labeled with the divisors of $k$.
After repeating this trial 4 times, given that the sum of all balls in the six boxes is odd, what is the probability that the number of balls in the box labeled 3 is 1 more than the number of balls in the box labeled 2? [4 points]\\
(1) $\frac { 1 } { 8 }$\\
(2) $\frac { 3 } { 16 }$\\
(3) $\frac { 1 } { 4 }$\\
(4) $\frac { 5 } { 16 }$\\
(5) $\frac { 3 } { 8 }$