Let's analyze the two questions in the image:

Question #8

The problem defines two events:

• $R$: A student drives a car to school.
• $L$: A student arrives late to school.

Given probabilities:

• $P(R \cap L) = 0.03$
• $P(L | R) = 0.0625$

To find the correct answer, let’s work through the probabilities.

1. Conditional Probability: We know $P(L | R) = \frac{P(R \cap L)}{P(R)}$. Given $P(L | R) = 0.0625$ and $P(R \cap L) = 0.03$, we can rearrange this equation to find $P(R)$: $P(R) = \frac{P(R \cap L)}{P(L | R)} = \frac{0.03}{0.0625} = 0.48$

So, $P(R) = 0.48$.

We can now compare this calculation to the answer choices to determine which option is correct.

Question #9

The question provides:

• Total cans: 8
• Cans with beans: 4
• Cans with sliced pineapples: 4

If the chef opens 4 randomly selected cans, we want the probability that all 4 contain sliced pineapples.

1. Calculate Probability: Since the chef wants all 4 cans to be sliced pineapples, we need to calculate the probability of drawing only the sliced pineapple cans in four successive selections without replacement.

The probability that the chef picks all 4 sliced pineapple cans from 8 with 4 of each type can be calculated based on combinations.

If you would like detailed solutions or have any questions about these steps, please let me know.

Here are five related questions that expand on this information:

1. How do you calculate conditional probability using joint and marginal probabilities?
2. What is the difference between conditional probability and independent events?
3. How would the answer change in Question #8 if $P(R \cap L)$ or $P(L | R)$ had different values?
4. What are some common mistakes when calculating probabilities with replacement versus without replacement?
5. How would the answer change in Question #9 if there were more cans with beans or sliced pineapples?

Tip: When solving probability questions, remember to carefully identify whether selections are made with or without replacement, as it significantly affects the calculations.