6.2 Basic and Compound Probability

Probability is a measure of the likelihood that a specific event will occur. On the SHSAT, probability questions range from basic single-stage events to complex multi-stage compound scenarios. Mastery of probability requires understanding the rules of combining independent and dependent events, as well as recognizing the distinction between events with and without replacement.

Probability of a Single Event

The probability of an event $E$, denoted as $P(E)$, is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes in a sample space: $P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

All probabilities are expressed as fractions, decimals, or percentages between 0 and 1, inclusive:

• A probability of 0 means the event is impossible.
• A probability of 1 means the event is certain to occur.
• The sum of the probabilities of all possible outcomes in a sample space always equals 1.

The complement of an event $A$, written as $P(\text{not } A)$ or $P(A')$, represents the probability that event $A$ does not occur. The formula is: $P(\text{not } A) = 1 - P(A)$

For example, if the probability of selecting a red marble is $\frac{3}{8}$, the probability of selecting a non-red marble is $1 - \frac{3}{8} = \frac{5}{8}$.

The Addition Rule for Mutually Exclusive Events

When finding the probability of event A OR event B occurring, you must look at whether the two events can happen at the same time. If they cannot occur simultaneously, they are called mutually exclusive (or disjoint) events. In this case, you add their individual probabilities: $P(A \text{ or } B) = P(A) + P(B)$ For example, if you roll a fair six-sided die, the event of rolling a 2 and the event of rolling a 5 are mutually exclusive. The probability of rolling a 2 or a 5 is $\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

However, if the events are not mutually exclusive—meaning they can happen at the same time—you must subtract the overlapping probability to avoid double-counting. The general addition rule is: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ For example, if you draw a single card from a standard deck of 52 cards, the probability of drawing a King (4 cards) or a Heart (13 cards) is calculated by adding the individual probabilities and subtracting the King of Hearts (1 card) since it is counted in both groups: $P(\text{King or Heart}) = P(\text{King}) + P(\text{Heart}) - P(\text{King of Hearts}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$ Understanding this overlap prevents common calculation mistakes on the exam.

Independent vs. Dependent Events

When dealing with more than one event, you must determine whether the outcome of the first event affects the outcome of the subsequent events.

• Independent Events: Two events are independent if the occurrence of the first event does not change the probability of the second event. For example, tossing a coin and rolling a die are independent. The formula for the probability of both independent events $A$ and $B$ occurring is: $P(A \text{ and } B) = P(A) \times P(B)$
• Dependent Events: Two events are dependent if the occurrence of the first event alters the probability of the second event. For example, drawing a card from a deck and then drawing a second card without returning the first card to the deck. The probability of both dependent events occurring is: $P(A \text{ and } B) = P(A) \times P(B \text{ after } A)$

Compound Probability: With and Without Replacement

Many SHSAT problems involve selecting objects from a container. The wording of these problems dictates whether the events are independent or dependent.

Walkthrough 1: With Replacement

Suppose a bag contains 5 red marbles and 7 blue marbles (12 total). You draw one marble, record its color, return it to the bag, and then draw a second marble. What is the probability that both marbles are red?

1. First Draw: The probability of drawing a red marble is $\frac{5}{12}$.
2. Replacement: Since the marble is replaced, the bag still contains 5 red and 12 total marbles.
3. Second Draw: The probability of drawing a red marble is still $\frac{5}{12}$.
4. Compound Probability: Multiply the probabilities. $P(\text{Red and Red}) = \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}$

Walkthrough 2: Without Replacement

Using the same bag of 5 red and 7 blue marbles, you draw two marbles sequentially without replacing the first marble. What is the probability that both marbles are red?

1. First Draw: The probability of drawing a red marble is $\frac{5}{12}$.
2. No Replacement: The drawn red marble is kept out. The bag now contains 4 red marbles and 7 blue marbles, making a new total of 11 marbles.
3. Second Draw: The probability of drawing a red marble is now $\frac{4}{11}$.
4. Compound Probability: Multiply the probabilities. $P(\text{Red and Red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$

Notice that the probability is lower when there is no replacement because the removal of the first favorable outcome reduces the ratio of remaining favorable outcomes.

The "At Least One" Shortcut

When asked to find the probability that an event occurs "at least once" in multiple trials, calculating all the successful combinations directly can be extremely tedious. Instead, use the complement rule: $P(\text{at least one successful outcome}) = 1 - P(\text{no successful outcomes})$

Example: If you roll a standard six-sided die three times, what is the probability of rolling at least one 6?

1. Find the probability of NOT rolling a 6 on a single roll: The probability is $\frac{5}{6}$.
2. Find the probability of rolling NO 6s on all three rolls: Since the rolls are independent, multiply the probabilities: $P(\text{no 6s}) = \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{216}$
3. Use the complement rule to find the probability of at least one 6: $P(\text{at least one 6}) = 1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$

This approach avoids calculating the probabilities of rolling exactly one 6, exactly two 6s, and exactly three 6s separately, saving valuable time during the test.