Topic 4.6 Independent Events and Unions of Events: determine whether events are independent, apply the multiplication rule for independent events, and combine the addition and multiplication rules to find probabilities of unions and intersections.

What this topic is asking

The College Board (Topic 4.6) wants you to determine whether events are independent, apply the multiplication rule for independent events, and combine the addition and multiplication rules to find probabilities of unions ("or") and intersections ("and").

Independence

You can check independence two ways: see whether the conditional probability equals the unconditional one ($P(A \mid B) = P(A)$), or see whether the joint probability factors ($P(A \text{ and } B) = P(A)P(B)$). Separate physical trials, two coin flips, two dice, draws with replacement, are independent because nothing carries over. Draws without replacement are dependent, because removing one item changes the makeup for the next.

The multiplication rule for independent events

This simple product is the engine behind the binomial distribution (Topic 4.10), where $n$ independent trials each contribute a factor. Whenever a problem describes repeated independent events all happening, you multiply; whenever you need "at least one," you multiply the complements and subtract from $1$.

Combining the rules

Most exam problems mix "and" and "or," so you combine the multiplication and addition rules. For an intersection of independent events, multiply. For a union, use the general addition rule, computing the intersection by multiplication if the events are independent. The single most useful combination is "at least one," which is best handled by the complement: the opposite of "at least one occurs" is "none occur," and for independent events $P(\text{none}) = P(A_1^c)\,P(A_2^c)\cdots$, so $P(\text{at least one}) = 1 - P(\text{none})$. This turns a messy sum of many cases into one tidy product subtracted from $1$. In the factory example, "at least one machine fails" is found as $1 - (0.95)(0.97)$, far easier than adding the "only A," "only B," and "both" cases. Recognizing the "at least one" cue and reaching for the complement-times-product is a high-value habit that recurs through the binomial and geometric topics.

Independence versus mutually exclusive, again

Because Topic 4.4 and Topic 4.6 sit side by side, the exam keeps probing whether you can separate independent from mutually exclusive. They answer different questions. Mutually exclusive: can the events happen together? (If not, $P(A \text{ and } B) = 0$.) Independent: does one affect the other's probability? (If not, $P(A \mid B) = P(A)$.) Independent events generally do co-occur; their joint probability is the product, which is non-zero whenever both are possible. Mutually exclusive events have zero joint probability, so far from being independent, they are strongly dependent (one happening forces the other not to). A clean way to remember it: mutually exclusive is about overlap (use it with the addition rule); independent is about influence (use it with the multiplication rule). Keeping these in separate mental boxes prevents the single most common probability error on the AP exam.

Try this

Q1. Events $A$ and $B$ are independent with $P(A) = 0.5$, $P(B) = 0.2$. Find $P(A \text{ and } B)$ and $P(A \text{ or } B)$. [2 points]

• Cue. $P(A \text{ and } B) = 0.5 \times 0.2 = 0.1$; $P(A \text{ or } B) = 0.5 + 0.2 - 0.1 = 0.6$.

Q2. Are draws without replacement from a deck independent? Explain. [1 point]

• Cue. No; removing the first card changes the composition of the deck, so the second draw's probabilities depend on the first, making them dependent.