The effective rate per annum equivalent to a nominal rate of 8% compounded semi-annually is

The problem involves three students attempting to solve independently with probabilities $\frac{1}{2}$21​, $\frac{1}{3}$31​, and $\frac{1}{4}$41​ respectively.

To find the probability that at least one student solves the problem, calculate the probability that none of them solve it, then subtract from 1.

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The probability each student fails:

Student 1 fails: $1 - \frac{1}{2} = \frac{1}{2}$1−21​=21​

Student 2 fails: $1 - \frac{1}{3} = \frac{2}{3}$1−31​=32​

Student 3 fails: $1 - \frac{1}{4} = \frac{3}{4}$1−41​=43​

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Since the students work independently, the probability all three fail:

$$P(\text{all fail}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}$$P(all fail)=21​×32​×43​

$$P(\text{all fail}) = \frac{6}{24}$$P(all fail)=246​

$$P(\text{all fail}) = \frac{1}{4}$$P(all fail)=41​

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The probability at least one student solves the problem:

$$P(\text{problem solved}) = 1 - P(\text{all fail})$$P(problem solved)=1−P(all fail)

$$P(\text{problem solved}) = 1 - \frac{1}{4}$$P(problem solved)=1−41​

$$P(\text{problem solved}) = \frac{3}{4}$$P(problem solved)=43​

Therefore, the probability that the problem is solved is $\frac{3}{4}$43​.