11.7 Probability

Probability Concepts

Probability models let you use math to predict how likely something is to happen. They show up everywhere, from card games to medical research, and they give you a structured way to make predictions about uncertain events. This section covers the core probability rules you need, plus the counting techniques that make complex probability problems manageable.

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Probability Models for Real-World Scenarios

There are two main ways to assign probabilities:

• Theoretical probability calculates likelihood based on what should happen under ideal conditions. You divide the number of favorable outcomes by the total number of possible outcomes. This works when you can assume fairness (a balanced coin, an unbiased die).
• Experimental probability calculates likelihood based on what actually happened during repeated trials. You divide the number of times an event occurred by the total number of trials. This is what you'd use with survey results, medical studies, or any situation where you're collecting real data.

To build and use a probability model, follow these steps:

1. Identify the sample space: list every possible outcome for the scenario (for a coin flip: {heads, tails}; for a die roll: {1, 2, 3, 4, 5, 6}).
2. Assign probabilities to each outcome using either theoretical reasoning or experimental data. Every probability must be between 0 and 1, and the probabilities of all outcomes in the sample space must add up to 1.
3. Use the model to make predictions by comparing what the model says should happen to what you actually observe.

The law of large numbers connects these two approaches: as you increase the number of trials, experimental probability gets closer and closer to the theoretical probability. Flip a fair coin 10 times and you might get 70% heads. Flip it 10,000 times and you'll be very close to 50%.

Probabilities with Equally Likely Outcomes

When every outcome in the sample space has the same chance of occurring, you're working with equally likely outcomes. Fair coins, fair dice, and well-shuffled decks all fit this category.

The formula is straightforward:

$P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$

For example, the probability of rolling an even number on a fair six-sided die: the favorable outcomes are {2, 4, 6}, so $P(\text{even}) = \frac{3}{6} = \frac{1}{2}$.

Another example: the probability of drawing a heart from a standard 52-card deck is $P(\text{heart}) = \frac{13}{52} = \frac{1}{4}$, since there are 13 hearts in the deck.

Union of Events Probabilities

The union of two events A and B, written $A \cup B$, is the event that A happens, B happens, or both happen. Think of it as "A or B."

The general formula is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

You subtract $P(A \cap B)$ (the probability of both events happening) because otherwise you'd count that overlap twice.

Example: What's the probability of drawing a heart or a king from a standard deck? There are 13 hearts and 4 kings, but one card (the king of hearts) is both. So $P(\text{heart or king}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.

When two events are mutually exclusive, they can't happen at the same time, so $P(A \cap B) = 0$. The formula simplifies to:

$P(A \cup B) = P(A) + P(B)$

Rolling a 2 and rolling a 5 on a single die roll are mutually exclusive. Drawing a card that is red and drawing a card that is black are also mutually exclusive (each card is one or the other, not both).

Complement Rule in Probability

The complement of event A, written $A'$ or $A^c$, is everything in the sample space that is not A. Since an event either happens or it doesn't:

$P(A') = 1 - P(A)$

This rule is especially useful when calculating the probability of an event directly would be tedious, but calculating the probability of it not happening is easy.

Example: What's the probability of rolling at least one 6 in four rolls of a die? Calculating that directly means considering rolling exactly one 6, exactly two 6s, etc. Instead, find the complement: the probability of rolling no 6s. On each roll, $P(\text{not 6}) = \frac{5}{6}$, so $P(\text{no 6 in four rolls}) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296} \approx 0.482$. Therefore, $P(\text{at least one 6}) = 1 - 0.482 = 0.518$.

Advanced Probability Concepts

These topics go a step beyond the basics and show up frequently in honors-level problems:

• Conditional probability is the probability of event A occurring given that event B has already occurred, written $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$. For instance, the probability of drawing a second heart from a deck after already drawing one heart (without replacement) changes because the sample space has shrunk.
• Independent events are events where the occurrence of one does not affect the probability of the other. Mathematically, A and B are independent if $P(A \cap B) = P(A) \cdot P(B)$. Flipping a coin and rolling a die are independent; drawing two cards without replacement are not.
• Expected value is the long-run average outcome of a random process. For a discrete random variable, $E(X) = \sum x_i \cdot P(x_i)$. If a game pays $10 with probability 0.3 and $0 with probability 0.7, the expected value is $10(0.3) + 0(0.7) = \$3$ per play.
• Bayes' theorem lets you update a probability when you receive new information: $P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}$. This is powerful in situations like medical testing, where you want to know the probability of actually having a disease given a positive test result.

Counting Techniques

Counting Techniques for Complex Probabilities

When probability problems involve large sample spaces, counting techniques let you figure out the total number of outcomes and favorable outcomes without listing them all.

The Fundamental Counting Principle says: if one event can happen in $m$ ways and an independent second event can happen in $n$ ways, then the two events together can happen in $m \times n$ ways. This extends to any number of events.

For example, if you have 3 shirts, 4 pairs of pants, and 2 pairs of shoes, the total number of outfits is $3 \times 4 \times 2 = 24$.

Permutations count the number of ways to arrange objects when order matters. The number of permutations of $n$ distinct objects taken $r$ at a time is:

$P(n, r) = \frac{n!}{(n-r)!}$

Example: How many ways can you award 1st, 2nd, and 3rd place among 8 runners? Order matters here (1st place is different from 2nd), so $P(8, 3) = \frac{8!}{5!} = 8 \times 7 \times 6 = 336$.

Combinations count the number of ways to select objects when order does not matter. The number of combinations of $n$ distinct objects taken $r$ at a time is:

$C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

Example: How many ways can you choose a committee of 3 people from a group of 10? Since it doesn't matter who's picked first, $C(10, 3) = \frac{10!}{3! \cdot 7!} = \frac{720}{6} = 120$.

Deciding between permutations and combinations comes down to one question: does the order of selection matter?

• Arranging books on a shelf, assigning ranked positions, creating passwords → permutation (order matters)
• Choosing team members, selecting toppings, picking lottery numbers → combination (order doesn't matter)

Once you've determined the total number of outcomes and the number of favorable outcomes using these techniques, plug them into the probability formula to solve the problem.