Let's go through the problem step-by-step. We are dealing with conditional probability, and we have to calculate the probability for various scenarios involving defective items.

Problem Setup

• Total number of air conditioners = 15
• Number of defective air conditioners = 3
• Number of selections = 2 (selected one after the other without replacement)

Now, let's solve each part.

(a) The probability that the first air conditioner selected is defective.

There are 3 defective air conditioners out of 15, so the probability that the first selected air conditioner is defective is:

$P(\text{first is defective}) = \frac{3}{15} = \frac{1}{5} = 0.2$

(b) The probability that the first air conditioner is defective and the second is not defective.

For this case, we need the probability of selecting a defective air conditioner first, followed by a non-defective air conditioner.

1. Probability that the first air conditioner is defective: $\frac{3}{15} = 0.2$.
2. After selecting a defective one first, we have 2 defective and 12 non-defective air conditioners left out of 14 total.

So, the probability that the second one selected is non-defective given that the first one was defective is:

$P(\text{second is non-defective} | \text{first is defective}) = \frac{12}{14} = \frac{6}{7} \approx 0.857$

Thus, the combined probability is:

$P(\text{first defective and second non-defective}) = \frac{3}{15} \times \frac{12}{14} = \frac{3 \cdot 12}{15 \cdot 14} = \frac{36}{210} = \frac{6}{35} \approx 0.1714$

(c) The probability that both air conditioners selected are defective.

For this scenario, we want the probability that both the first and second air conditioners are defective.

1. Probability that the first air conditioner is defective: $\frac{3}{15} = 0.2$.
2. After selecting a defective air conditioner, 2 defective remain out of 14.

So, the probability that the second one is also defective, given that the first one was defective, is:

$P(\text{second is defective} | \text{first is defective}) = \frac{2}{14} = \frac{1}{7} \approx 0.1429$

Thus, the combined probability is:

$P(\text{both defective}) = \frac{3}{15} \times \frac{2}{14} = \frac{3 \cdot 2}{15 \cdot 14} = \frac{6}{210} = \frac{1}{35} \approx 0.0286$

Summary of Results

• (a) Probability the first is defective: $0.2$
• (b) Probability the first is defective and the second is not defective: $\approx 0.1714$
• (c) Probability both are defective: $\approx 0.0286$

Would you like more details or have any questions?

Related Questions

1. What is the probability that neither of the selected air conditioners is defective?
2. How would the probabilities change if we were selecting with replacement?
3. What is the probability that exactly one of the two selected air conditioners is defective?
4. How would these probabilities change if there were 5 defective air conditioners instead of 3?
5. How can we calculate the probability for selecting three air conditioners in succession without replacement?

Tip

When working with probabilities without replacement, always adjust the total number of items and the count of specific items left after each selection step.