Problem 146

First, let's consider all possible combinations for selecting 3 questions out of 10. The number of combinations is given by the binomial coefficient $C(10,3)$. This can be computed as:

$$C(10,3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{6} = 120$$

The student will pass if they answer correctly to exactly 2 questions or all 3 of the selected questions.

The number of ways to select 1 question that the student cannot answer (from the 4 that they cannot answer correctly) and 2 questions they can answer (from the 6 that they can) is given by the product $C(4,1) \cdot C(6,2)$. This can be computed as:

$$C(4,1) \cdot C(6,2) = 4 \times \frac{6 \times 5}{2} = 4 \times 15 = 60$$

The number of ways to select 3 questions they can answer is given by $C(6,3)$. This can be computed as:

$$C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{6} = 20$$

Adding these two numbers up gives us the total number of successful outcomes, which is $60 + 20 = 80$.

Hence, the probability of the student passing is the ratio of the number of successful outcomes to the total number of possible outcomes, which is:

$$\frac{80}{120} = \frac{2}{3}$$

So, the probability that the student passes the test is $\boxed{\frac{2}{3}}$.