The question asks for the probability that at least one computer has defects, given that 6 out of 20 computers have defects and the University Lab bought 4 computers at random.

To solve this, we can use the complement rule, which is easier in such cases:

1. Step 1: Calculate the probability that none of the computers have defects.

   • There are 14 computers without defects (since 20 - 6 = 14).
   • The probability that all 4 computers selected have no defects is a combination of choosing 4 defect-free computers from the 14 available.

2. Step 2: Use the complement rule.

   • The probability of "at least one defective computer" is equal to 1 minus the probability that all 4 computers are non-defective.

Let's go ahead and calculate this.

We use the hypergeometric probability formula, which is suitable for scenarios where we are drawing without replacement.

$P(\text{no defective computers}) = \frac{\binom{14}{4}}{\binom{20}{4}}$

Where:

• $\binom{14}{4}$ is the number of ways to select 4 non-defective computers from 14.
• $\binom{20}{4}$ is the number of ways to select 4 computers from the total 20.

Finally, the probability that at least one computer has defects is: $P(\text{at least one defective}) = 1 - P(\text{no defective computers})$

I'll now calculate this value.The probability that at least one of the four computers selected has defects is approximately 0.79.

Would you like further clarification or additional details on this solution?

Here are 5 related questions to deepen your understanding:

1. What is the probability that none of the computers selected have defects?
2. How would the result change if only 3 computers were selected?
3. What is the probability of selecting exactly 2 defective computers?
4. How does this problem relate to the binomial distribution?
5. Can you solve this problem using simulation methods?

Tip: The complement rule (1 minus the probability of the opposite event) is very useful when calculating the probability of "at least one" event occurring.