Q.State whether the following statement is True or False: If $A$, $B$ and $C$ are three independent events such that $P(A) = P(B) = P(C) = p$, then $P(\text{at least two of } A, B, C \text{ occur}) = 3p^2 - 2p^3$.

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

Step 1: "At least two occur" means exactly two occur or all three occur.

Step 2: For exactly two, there are three choices: $A \cap B \cap C^c$, $A \cap B^c \cap C$, $A^c \cap B \cap C$. Each has probability $p \cdot p \cdot (1-p) = p^2(1-p)$. …

We need to check whether the given expression correctly captures the probability that at least two of the three independent events $A, B, C$ happen. Since each event has the same probability $p$ and they are independent, we can compute directly.

Why this approach works:

The phrase “at least two occur” is a union of two disjoint cases:

• Exactly two of the three events occur.
• All three events occur.

Because the events are independent, the probability of any specific combination (like $A$ and $B$ occur, $C$ does not) is simply the product of the individual probabilities. And since all events have the same $p$, many terms will be identical — we just count how many such combinations there are.

Let’s go step by step.

1. Case 1: Exactly two events occur. There are $\binom{3}{2} = 3$ ways to choose which two events occur. For a specific pair, say $A$ and $B$ occur but $C$ does not, the probability is:

The same holds for the other two pairs. So total probability for exactly two is:

2. Case 2: All three events occur. Only one combination: $A$, $B$, and $C$ all happen. Probability:

3. Add the two disjoint cases. …

Method: "At least two of three" independent events

For "at least $k$ of $n$" independent events, break the event into disjoint count-cases (or use the complement) and add.

Steps

Step 1: Decompose by exact count.

"At least two of $A,B,C$" $=$ (exactly two) $\cup$ (all three) — disjoint, so probabilities add.

Step 2: Count and weight each case using independence.

With $P(A)=P(B)=P(C)=p$: each "exactly two" pattern has probability $p^2(1-p)$, and there are $\binom{3}{2}=3$ of them; "all three" has probability $p^3$. …

Common Mistakes

Mistake 1: Leaving out the all-three case.

Why it's wrong: "at least two" includes exactly two and all three; stopping at exactly two gives $3p^2-3p^3$, which is short by $p^3$. Correct approach: add the $p^3$ term to get $3p^2-2p^3$.

Mistake 2: Miscounting the "exactly two" patterns. …