We are given that a computer program is tested by three independent tests, with the probabilities of discovering an error by each test being $P(A_1) = 0.2$, $P(A_2) = 0.3$, and $P(A_3) = 0.5$.

We need to find the probability that at least one test will discover the error.

Step 1: Define Events

Let $A_1$, $A_2$, and $A_3$ represent the events that the first, second, and third tests detect the error, respectively. Since the tests are independent, the probability of none of the tests detecting the error is the product of the probabilities that each individual test fails to detect it.

• The probability that the first test does not detect the error is $P(A_1^c) = 1 - P(A_1) = 1 - 0.2 = 0.8$.
• The probability that the second test does not detect the error is $P(A_2^c) = 1 - P(A_2) = 0.7$.
• The probability that the third test does not detect the error is $P(A_3^c) = 1 - P(A_3) = 0.5$.

Step 2: Probability of No Detection

The probability that none of the tests detect the error is the product of these complementary probabilities: $P(\text{none of the tests detect the error}) = P(A_1^c) \cdot P(A_2^c) \cdot P(A_3^c) = 0.8 \times 0.7 \times 0.5 = 0.28.$

Step 3: Probability of At Least One Detection

The probability that at least one test will detect the error is the complement of the probability that none of the tests detect the error: $P(\text{at least one test detects the error}) = 1 - P(\text{none detect the error}) = 1 - 0.28 = 0.72.$

Thus, the probability that at least one test will detect the error is $\boxed{0.72}$.

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Would you like more details on any part of the solution?

Relative Questions:

1. What is the probability that exactly two tests will detect the error?
2. How would the probability change if the tests were not independent?
3. What is the probability that only the second test detects the error?
4. How would the solution change if there were 4 independent tests?
5. Can we generalize this method for $n$ independent tests with different detection probabilities?

Tip:

For problems involving independent events, it's often easier to calculate the probability of none of the events occurring and subtract that from 1 to find the probability of at least one event happening.