Compound Probability — And, Or, and Multiple Events

Basic probability handles one event at a time. Compound probability is what happens when you ask about two or more events together. What are the chances it rains and you forgot your umbrella? What's the probability of flipping heads or rolling a 6? As soon as you add "and" or "or" to a probability question, you're in compound probability territory.

Independent Events — The "And" Rule

Two events are independent if the outcome of one has no effect on the other. Flipping a coin and rolling a die are independent — the coin doesn't care what the die does.

For independent events, the probability of both happening is:

$$P(A \text{ and } B) = P(A) \times P(B)$$

You just multiply.

Example: What's the probability of flipping heads and rolling a 3?

$$P(\text{heads}) = \frac{1}{2} \qquad P(\text{rolling 3}) = \frac{1}{6}$$

$$P(\text{heads and 3}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$

Example: A bag has 5 red and 5 blue chips. You draw one, note the color, put it back, then draw again. What's the probability of drawing red both times?

Since you replaced the chip, the second draw is just like the first — the bag is the same. These are independent events.

$$P(\text{red}) = \frac{5}{10} = \frac{1}{2}$$

$$P(\text{red and red}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$

Extending to more than two events: The same rule keeps going. The probability of three independent events all happening is \(P(A) \times P(B) \times P(C)\).

Example: What's the probability of flipping heads three times in a row?

$$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$

Dependent Events — When It Matters What Happened First

Events are dependent when the first outcome changes the conditions for the second. The most common example: drawing from a bag without putting the first item back.

Example: A bag has 4 red and 6 blue marbles. You draw one marble and don't replace it, then draw a second. What's the probability both are red?

The first draw: 4 red out of 10 total.

$$P(\text{1st is red}) = \frac{4}{10}$$

Now there are only 9 marbles left, and only 3 of them are red (since you took one red out).

$$P(\text{2nd is red} \mid \text{1st was red}) = \frac{3}{9}$$

The vertical bar means "given that" — the probability of the second draw being red, given that the first was red.

Multiply them together:

$$P(\text{both red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$$

The key with dependent events is to update your counts after each draw. Ask yourself: what changed? How many are left in total? How many favorable outcomes are left?

Example: A deck has 52 cards. What's the probability of drawing two aces in a row (without replacement)?

First draw: 4 aces out of 52 cards.

$$P(\text{1st ace}) = \frac{4}{52}$$

After pulling an ace, there are 51 cards left and only 3 aces remain.

$$P(\text{2nd ace}) = \frac{3}{51}$$

$$P(\text{both aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$$

That's about a 0.45% chance — pretty unlikely, which makes sense.

The "Or" Rule

So far we've been finding the probability of event A and event B. What about A or B?

If two events cannot both happen at the same time (called mutually exclusive), it's simple addition:

$$P(A \text{ or } B) = P(A) + P(B)$$

Example: What's the probability of rolling a 2 or a 5 on a die?

You can't roll both at once — they're mutually exclusive.

$$P(2 \text{ or } 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$

If the events can overlap — meaning both could happen at once — you need to subtract the overlap to avoid counting it twice:

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$

Example: In a standard deck of 52 cards, what's the probability of drawing a heart or a face card?

There are 13 hearts and 12 face cards (J, Q, K in each of 4 suits). But 3 cards are both hearts and face cards (J♥, Q♥, K♥). If we just added 13 + 12, we'd count those 3 cards twice.

$$P(\text{heart or face card}) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}$$

Keeping It Straight

When you see an "and" problem, think multiply. When you see an "or" problem, think add (and check for overlap). And whenever you see "without replacement," flag it immediately — that's a dependent events situation and the denominator changes with each draw.

Work Through These

1. You flip a coin and roll a die. What is the probability of getting tails and rolling a number greater than 4? Show answer\(\frac{1}{6}\) — \(P(\text{tails}) \times P(\text{greater than 4}) = \frac{1}{2} \times \frac{2}{6} = \frac{2}{12} = \frac{1}{6}\)

2. A bag has 3 green and 7 yellow marbles. You draw one, replace it, then draw again. What is the probability of drawing yellow both times? Show answer\(\frac{49}{100}\) — \(\frac{7}{10} \times \frac{7}{10} = \frac{49}{100}\)

3. Using the same bag as #2, but now without replacement: what is the probability of drawing yellow both times? Show answer\(\frac{7}{15}\) — \(\frac{7}{10} \times \frac{6}{9} = \frac{42}{90} = \frac{7}{15}\). After removing one yellow marble, 6 yellow remain out of 9 total.

4. What is the probability of rolling a 1 or a 6 on a standard die? Show answer\(\frac{1}{3}\) — mutually exclusive, so \(\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}\)

5. In a class of 30 students, 12 play sports, 10 are in drama, and 4 do both. If a student is picked at random, what is the probability they play sports or are in drama? Show answer\(\frac{3}{5}\) — \(P(\text{sports or drama}) = \frac{12}{30} + \frac{10}{30} - \frac{4}{30} = \frac{18}{30} = \frac{3}{5}\)