Q.Let $E_1$ and $E_2$ be two independent events such that $P(E_1) = p_1$ and $P(E_2) = p_2$. Describe in words of the events whose probabilities are:

Concept: Event Independence — For independent events, the probability of their intersection is the product of their individual probabilities.

Reasoning:

1. $p_1 p_2 = P(E_1) \cdot P(E_2)$. Since $E_1$ and $E_2$ are independent, this equals $P(E_1 \cap E_2)$ — the event that both $E_1$ and $E_2$ occur.
2. $(1-p_1)\,p_2 = P(E_1^c) \cdot P(E_2)$. By independence, this is $P(E_1^c \cap E_2)$ — the event that $E_1$ does not occur and $E_2$ occurs.
3. $1-(1-p_1)(1-p_2) = 1 - P(E_1^c \cap E_2^c) = P(E_1 \cup E_2)$ — the event that at least one of $E_1$ or $E_2$ occurs. …

We have two independent events $E_1$ and $E_2$ with $P(E_1) = p_1$ and $P(E_2) = p_2$. Independence means that the occurrence of one does not affect the probability of the other — mathematically, $P(E_1 \cap E_2) = P(E_1) \cdot P(E_2) = p_1 p_2$.

Let’s interpret each expression step by step.

1. Expression (i): $p_1 p_2$

   Since $P(E_1 \cap E_2) = p_1 p_2$ for independent events, this is simply the probability that both $E_1$ and $E_2$ occur. In words: “$E_1$ and $E_2$ both happen.”

2. Expression (ii): $(1-p_1)\,p_2$

   Here $1-p_1 = P(\text{not } E_1)$, the complement of $E_1$. Because $E_1$ and $E_2$ are independent, so are $\text{not } E_1$ and $E_2$. Thus $(1-p_1)p_2 = P(\text{not } E_1 \cap E_2)$. This is the probability that $E_1$ does not occur but $E_2$ does. In words: “Only $E_2$ occurs.”

3. Expression (iii): $1-(1-p_1)(1-p_2)$

   The product $(1-p_1)(1-p_2)$ is $P(\text{not } E_1 \cap \text{not } E_2)$, the probability that neither event occurs. Subtracting this from 1 gives the probability that at least one of the events occurs. In words: “$E_1$ or $E_2$ (or both) occur.” This is the union $P(E_1 \cup E_2)$.

   Tip

   For independent events, the probability of the union is not simply $p_1 + p_2$ — that would double-count the intersection. The correct formula is $P(E_1 \cup E_2) = p_1 + p_2 - p_1 p_2$, which is exactly $1 - (1-p_1)(1-p_2)$ after expanding. The complement form is often quicker.

4. Expression (iv): $p_1 + p_2 - 2p_1 p_2$

   Let’s break this down. The probability that exactly one event occurs is the sum of two mutually exclusive cases: …

Method: Translating Probability Expressions into Event Descriptions

Use this reasoning pattern to read an algebraic expression in $p_1,p_2$ and say, in words, which compound event it measures (given $E_1,E_2$ independent).

Steps

Step 1: Read a product as an intersection.

For independent events a product of probabilities is the probability that all those events occur together: $p_1p_2=P(E_1\cap E_2)$.

Step 2: Read $(1-p)$ as a complement.

$1-p_1=P(\overline{E_1})$; independence is preserved for complements, so $(1-p_1)p_2=P(\overline{E_1}\cap E_2)$ — "only $E_2$".

Step 3: Read $1-(\text{neither})$ as "at least one". …

Common Mistakes

Mistake 1: Confusing "exactly one" with "at least one".

Why it's wrong: $p_1+p_2-2p_1p_2$ subtracts the "both" term twice, giving exactly one; "at least one" is $p_1+p_2-p_1p_2$, larger by $p_1p_2$. Correct approach: check the coefficient of $p_1p_2$ to tell them apart.

Mistake 2: Reading $(1-p_1)p_2$ as "$E_1$ or $E_2$". …