Three students A, B and C can respectively solve 50%, 25% and 20% of the problems in a book. A particular problem is selected at random from the…

Student A can solve 50% of problems, so P(A solves) = $\frac{1}{2}$21​

Student B can solve 25% of problems, so P(B solves) = $\frac{1}{4}$41​

Student C can solve 20% of problems, so P(C solves) = $\frac{1}{5}$51​

To find the probability that at least one of them solves the problem, use the complement approach:

P(at least one solves) = 1 − P(none solve)

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The probability that each student fails to solve the problem:

P(A fails) = $1 - \frac{1}{2}$1−21​

$$= \frac{1}{2}$$=21​

P(B fails) = $1 - \frac{1}{4}$1−41​

$$= \frac{3}{4}$$=43​

P(C fails) = $1 - \frac{1}{5}$1−51​

$$= \frac{4}{5}$$=54​

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Since the students work independently:

P(all fail) = P(A fails) × P(B fails) × P(C fails)

$$= \frac{1}{2} \times \frac{3}{4} \times \frac{4}{5}$$=21​×43​×54​

$$= \frac{12}{40}$$=4012​

$$= \frac{3}{10}$$=103​

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P(at least one solves) = $1 - P(\text{all fail})$1−P(all fail)

$$= 1 - \frac{3}{10}$$=1−103​

$$= \frac{7}{10}$$=107​

Therefore, the probability that at least one of them will solve the problem is $\frac{7}{10}$107​.