I see you've uploaded a worksheet containing probability problems. Let's break down each question and work through the solutions:

Problem 1:

Two independent events $A$ and $B$ are given with probabilities:

• $P(A \cap B') = 0.09$
• $P(A' \cap B) = 0.49$

Let $x = P(A \cap B)$.

(a) Express $P(A)$ in terms of $x$.

Since $P(A \cap B') + P(A \cap B) = P(A)$, we know:

$P(A) = P(A \cap B') + P(A \cap B) = 0.09 + x$

(b) Express $P(B)$ in terms of $x$.

Using the fact that $P(A' \cap B) + P(A \cap B) = P(B)$:

$P(B) = P(A' \cap B) + P(A \cap B) = 0.49 + x$

(c) Find the value of $x$.

The total probability must sum to 1, as it covers all possible outcomes:

$P(A \cap B') + P(A' \cap B) + P(A \cap B) + P(A' \cap B') = 1$

We already know:

$P(A \cap B') = 0.09,\ P(A' \cap B) = 0.49$

Also, $P(A' \cap B') = 1 - P(A \cup B)$.

We can calculate:

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = (0.09 + x) + (0.49 + x) - x$

Simplifying:

$P(A \cup B) = 0.58 + x$

So,

$P(A' \cap B') = 1 - (0.58 + x) = 0.42 - x$

Since all probabilities must sum to 1:

$0.09 + 0.49 + x + 0.42 - x = 1$

This equation holds, so no contradictions arise.

(d) Find $P(B | A)$.

Using the definition of conditional probability:

$P(B | A) = \frac{P(A \cap B)}{P(A)} = \frac{x}{0.09 + x}$

Problem 2:

Given the cycling scenario, we're provided with the following probabilities:

• $P(\text{rain tomorrow} | \text{dry today}) = 0.25$
• $P(\text{rain tomorrow} | \text{rainy today}) = 0.85$
• $P(\text{dry on 1st June}) = 0.8$

(a) Draw a tree diagram for the first three days of June.

The tree diagram will have branches for each day's weather condition (rainy or dry) and the associated conditional probabilities for the next day's weather.

• First branch: 1st June (Dry $P = 0.8$, Rainy $P = 0.2$)
• From each of these branches, you have the 2nd June with the conditional probabilities.
• Similarly, branches for 3rd June come from the outcomes of 2nd June.

(b) Find the probability that the 3rd of June is rainy.

To solve this, consider all paths leading to rain on the 3rd of June:

• Dry on 1st $\rightarrow$ Dry on 2nd $\rightarrow$ Rain on 3rd
• Dry on 1st $\rightarrow$ Rain on 2nd $\rightarrow$ Rain on 3rd
• Rain on 1st $\rightarrow$ Dry on 2nd $\rightarrow$ Rain on 3rd
• Rain on 1st $\rightarrow$ Rain on 2nd $\rightarrow$ Rain on 3rd

Use the probabilities provided to calculate the total.

(c) Find the probability that the 1st of June was dry given that the 3rd of June is dry.

Use Bayes' Theorem:

$P(\text{Dry on 1st} | \text{Dry on 3rd}) = \frac{P(\text{Dry on 1st and Dry on 3rd})}{P(\text{Dry on 3rd})}$

The numerator is the probability of being dry on both the 1st and 3rd, and the denominator is the total probability of the 3rd being dry.

Would you like more detailed calculations or help with a specific part of the solution?

Here are 5 relative questions:

1. How can we use the complement rule in probability to simplify calculations?
2. How does independence between events affect their joint probability?
3. Can conditional probability be used to predict long-term trends in weather or other real-world applications?
4. How does Bayes' Theorem adjust probabilities based on new information?
5. How are tree diagrams useful in visualizing sequential probability problems?

Tip: Always check if events are independent or conditional when determining the appropriate formula for joint probabilities!