The correct option is C) $1/12$. Let $N=10$ be the total number of distinct balls, numbered 1 to 10. We are choosing $k=4$ balls and arranging them in a linear sequence. Step 1: Calculate the total number of possible outcomes. The total number of ways to choose 4 balls from 10 distinct balls and arrange them in a sequence is given by the permutation formula $P(n,k)$. Total outcomes = $P(10,4) = 10 \times 9 \times 8 \times 7 = 5040$. Step 2: Calculate the number of favorable outcomes satisfying condition (1). Condition (1) states that no two chosen balls have consecutive numbers. This is a classic combinatorics problem. The number of ways to choose $k$ items from $n$ distinct items such that no two are consecutive is given by the formula $\binom{n-k+1}{k}$. For $n=10$ and $k=4$, the number of ways to choose 4 non-consecutive balls is: Number of combinations $ = \binom{10-4+1}{4} = \binom{7}{4}$. $\binom{7}{4} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$. Step 3: Calculate the number of favorable outcomes satisfying condition (2) for each selection. Once 4 non-consecutive balls are chosen, let them be denoted as $x_1, x_2, x_3, x_4$ in ascending order (i.e., $x_1 < x_2 < x_3 < x_4$). Condition (2) states that the ball with the smallest number ($x_1$) must be placed at an end of the sequence. For any set of 4 distinct balls, there are $4!$ ways to arrange them linearly. If $x_1$ must be at an end, it can be either in the first position or the last position. If $x_1$ is in the first position, the remaining 3 balls ($x_2, x_3, x_4$) can be arranged in the other 3 positions in $3!$ ways. Number of arrangements = $1 \times 3! = 6$. If $x_1$ is in the last position, the remaining 3 balls ($x_2, x_3, x_4$) can be arranged in the other 3 positions in $3!$ ways. Number of arrangements = $3! \times 1 = 6$. So, for each set of 4 non-consecutive balls selected, there are $6+6=12$ arrangements that satisfy condition (2). Step 4: Calculate the total number of favorable outcomes for both conditions. Total favorable outcomes = (Number of ways to choose 4 non-consecutive balls) \times (Number of ways to arrange them with the smallest at an end) Total favorable outcomes = $35 \times 12 = 420$. Step 5: Calculate the probability. Probability = $\frac{\text{Total favorable outcomes}}{\text{Total possible outcomes}} = \frac{420}{5040}$. Simplifying the fraction: $\frac{420}{5040} = \frac{42}{504}$. Divide by 2: $\frac{21}{252}$. Divide by 3: $\frac{7}{84}$. Divide by 7: $\frac{1}{12}$. Thus, the probability is $1/12$.