We are given a sample space $S = \{a, b, c, d\}$ with the following probabilities:

• $P(a) = 0.2$
• $P(b) = 0.3$
• $P(c) = 0.4$
• $P(d) = 0.1$

Let:

• Event $A = \{a, b\}$
• Event $B = \{b, c, d\}$

Now, let’s solve the given probabilities step by step:

(a) $P(A)$

The probability of event $A$, which is $\{a, b\}$, is the sum of the probabilities of the elements in $A$: $P(A) = P(a) + P(b) = 0.2 + 0.3 = 0.5$

(b) $P(B)$

The probability of event $B$, which is $\{b, c, d\}$, is the sum of the probabilities of the elements in $B$: $P(B) = P(b) + P(c) + P(d) = 0.3 + 0.4 + 0.1 = 0.8$

(c) $P(A^c)$

The complement of event $A$, denoted $A^c$, consists of all elements in the sample space $S$ that are not in $A$. Thus, $A^c = \{c, d\}$.

$P(A^c) = P(c) + P(d) = 0.4 + 0.1 = 0.5$

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

The union of $A$ and $B$, denoted $A \cup B$, consists of all elements that are in either $A$ or $B$. So, $A \cup B = \{a, b, c, d\} = S$ Since the union covers the entire sample space, the probability is: $P(A \cup B) = P(S) = 1$

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

The intersection of $A$ and $B$, denoted $A \cap B$, consists of all elements that are in both $A$ and $B$. So, $A \cap B = \{b\}$ The probability of this intersection is simply the probability of $b$: $P(A \cap B) = P(b) = 0.3$

Summary of Results:

• (a) $P(A) = 0.5$
• (b) $P(B) = 0.8$
• (c) $P(A^c) = 0.5$
• (d) $P(A \cup B) = 1$
• (e) $P(A \cap B) = 0.3$

Would you like further clarification on any of the steps?

Here are 5 additional questions that build on this topic:

1. How would you calculate $P(A^c \cap B)$?
2. If $P(B^c)$ were required, how would you approach it?
3. Can you derive the formula for $P(A \cup B^c)$?
4. What if event $A$ and $B$ were mutually exclusive, how would $P(A \cup B)$ change?
5. How do we interpret $P(A^c \cup B^c)$ in terms of this experiment?

Tip: The complement rule $P(A^c) = 1 - P(A)$ is a useful shortcut to quickly find the probability of the complement of an event.