Concept: Set Operations — the key is to match each event condition with its correct set-theoretic description.

Step 1 — (a) Mutually exclusive events

Two events are mutually exclusive if they cannot occur together. In set terms:

$E_1 \cap E_2 = \phi$ → matches (iv).

Step 2 — (b) Mutually exclusive and exhaustive events

Exhaustive means their union is the whole sample space $S$. So:

$E_1 \cap E_2 = \phi$ and $E_1 \cup E_2 = S$ → matches (iii).

Step 3 — (c) Events with common outcomes …

The entire language of probability is built on set operations. When we talk about events, we're really talking about subsets of a sample space $S$. Mutually exclusive means the sets don't overlap — their intersection is empty. Exhaustive means their union covers the whole sample space. Subset means one event is completely contained inside another. And the expression $(E_1 - E_2) \cup (E_1 \cap E_2)$ is just a fancy way of writing $E_1$ itself, since every element of $E_1$ is either in $E_2$ or not in $E_2$.

Let's go through each pair one by one.

1. (a) If $E_1$ and $E_2$ are two mutually exclusive events

   Mutually exclusive means the two events cannot happen at the same time. In set terms, they have no common outcomes.

   That is exactly $E_1 \cap E_2 = \phi$, which is option (iv).

   Tip

   The word "exclusive" is the clue — they exclude each other, so no overlap.

2. (b) If $E_1$ and $E_2$ are mutually exclusive and exhaustive events

   "Mutually exclusive" again gives $E_1 \cap E_2 = \phi$.

   "Exhaustive" means together they cover every possible outcome in the sample space $S$, so $E_1 \cup E_2 = S$.

   This matches option (iii): $E_1 \cap E_2 = \phi$, $E_1 \cup E_2 = S$.

3. (c) If $E_1$ and $E_2$ have common outcomes

   This means they are not mutually exclusive — their intersection is non-empty.

   Now look at option (ii): $(E_1 - E_2) \cup (E_1 \cap E_2) = E_1$. …