Let's analyze the problem presented in the image.

Problem Details:

• The sample space $S = \{0, 1, \dots, 9\}$ consists of integers from 0 to 9.
• The events are given as:
  • $A = \{0, 2, 5, 6\}$
  • $B = \{3, 6\}$

We are tasked with calculating the following probabilities under the assumption that all elementary events are equally likely:

1. $P(A)$ (the probability of event $A$),
2. $P(A \cap B)$ (the probability of the intersection of $A$ and $B$),
3. $P(A \mid B)$ (the conditional probability of $A$ given $B$),
4. $P(B \mid A)$ (the conditional probability of $B$ given $A$).

Solution Steps:

(a) $P(A)$

• The number of favorable outcomes for $A$ is $|A| = 4$ (elements $\{0, 2, 5, 6\}$).
• The total number of outcomes in $S$ is $|S| = 10$. $P(A) = \frac{|A|}{|S|} = \frac{4}{10} = 0.4$

(b) $P(A \cap B)$

• The intersection $A \cap B = \{6\}$ (elements common to $A$ and $B$).
• The number of favorable outcomes is $|A \cap B| = 1$. $P(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{1}{10} = 0.1$

(c) $P(A \mid B)$

• Conditional probability formula: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$
• From (b), $P(A \cap B) = 0.1$.
• The probability $P(B)$ is calculated as:
  • $|B| = 2$ (elements $\{3, 6\}$), $P(B) = \frac{|B|}{|S|} = \frac{2}{10} = 0.2$
• Substituting: $P(A \mid B) = \frac{0.1}{0.2} = 0.5$

(d) $P(B \mid A)$

• Conditional probability formula: $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$
• From (b), $P(A \cap B) = 0.1$.
• From (a), $P(A) = 0.4$.
• Substituting: $P(B \mid A) = \frac{0.1}{0.4} = 0.25$

Final Answers:

• $P(A) = 0.4$
• $P(A \cap B) = 0.1$
• $P(A \mid B) = 0.5$
• $P(B \mid A) = 0.25$

Let me know if you need further details or explanations!

Related Questions:

1. How is the formula for conditional probability derived?
2. Can the probabilities $P(A \mid B)$ and $P(B \mid A)$ ever be equal? Why or why not?
3. How does the assumption of equally likely outcomes affect these calculations?
4. What would $P(A \cup B)$ (union of $A$ and $B$) be for this problem?
5. How would the probabilities change if $S$ contained more elements (e.g., $S = \{0, 1, \dots, 19\}$)?

Tip: Always check if $A \cap B$ is empty to ensure the probabilities are meaningful when solving problems involving intersections.