To solve this problem, let's go through it step-by-step.

Problem Analysis

• A manufacturer shipped 50 cameras, out of which 8 are defective.
• The electronics store sold 15 cameras without knowing which are defective.
• We are asked to find the probability that at least one defective camera was sold.

Steps to Solution

To determine this probability, we can first calculate the probability that no defective cameras were sold and then subtract this from 1 to get the probability that at least one defective camera was sold.

1. Total Ways to Choose 15 Cameras (from 50): This is given by $C(50, 15)$, where $C(n, k)$ represents a combination or "n choose k".

2. Ways to Choose 15 Cameras That Are All Non-Defective: Since there are 8 defective cameras, there are $50 - 8 = 42$ non-defective cameras. The number of ways to select 15 non-defective cameras from these 42 is $C(42, 15)$.

3. Probability of No Defective Cameras Being Sold: This probability is the ratio of the ways to choose only non-defective cameras to the total ways of choosing 15 cameras: $P(\text{no defective cameras sold}) = \frac{C(42, 15)}{C(50, 15)}$

4. Probability of At Least One Defective Camera Being Sold: To find this, subtract the probability of no defective cameras from 1: $P(\text{at least one defective camera sold}) = 1 - \frac{C(42, 15)}{C(50, 15)}$

Answer

Using the formula above will give the probability that at least one defective camera was sold.

Would you like the calculations for the exact values of $C(42, 15)$ and $C(50, 15)$, or do you have any other questions?

Further Questions

1. How do combinations work, and how are they calculated?
2. What would change if we had more or fewer defective cameras?
3. Can you explain why we subtract from 1 to find the probability of "at least one defective"?
4. What is the general formula for solving similar probability problems involving "at least one"?
5. How does this method change if replacements were allowed when choosing items?

Tip

In probability problems involving "at least one," it’s often easier to calculate the probability of the complement event ("none") and subtract it from 1.