Independent and Dependent Events

“What’s the chance of rolling a six and flipping heads?” “What’s the chance both cards are aces?” Probabilities of two things happening together depend on one question: does the first event change the chances for the second? If not, the events are independent; if so, they’re dependent.

Key ideas

Independent events

Events are independent if one happening doesn’t change the probability of the other: flipping a coin and rolling a die, or rolling a die twice. For independent events:

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

Dependent events

Events are dependent if the first changes the probability of the second. Drawing two cards without replacement is the classic case: after one ace is gone, there are fewer aces and fewer cards left.

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

”$B$ given $A$” means the probability of $B$ after $A$ has happened. (More on this in conditional probability.)

With and without replacement

• With replacement: the item is put back, so the second draw is like the first. The draws are independent.
• Without replacement: the item stays out, so the second draw has one fewer item. The draws are dependent.

Tree diagrams

A tree diagram shows each stage as a set of branches, labelled with probabilities.

• Multiply along a path to get the probability of that combination.
• Add across paths when several combinations satisfy the event.
• The probabilities of all the paths add to $1$.

Drawing two marbles without replacement from a bag of $3$ red and $2$ blue.

Worked examples

Example 1: Independent events

A coin is flipped and a die is rolled. Find $P(\text{heads and } 6)$.

Solution. They’re independent:

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

Example 2: With replacement

A bag has $3$ red and $2$ blue marbles. A marble is drawn, replaced, and a second is drawn. Find $P(\text{both red})$.

Solution. With replacement, each draw has $P(\text{red}) = \tfrac{3}{5}$:

$$P(\text{both red}) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$$

Example 3: Without replacement

Using the same bag, two marbles are drawn without replacement. Use the tree diagram to find $P(\text{both red})$ and $P(\text{one of each colour})$.

Solution. After a red is drawn, $2$ of the remaining $4$ are red:

$$P(\text{RR}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$$

“One of each” happens two ways, red-then-blue or blue-then-red. Add the paths:

$$P(\text{one of each}) = \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} = \frac{6}{20} + \frac{6}{20} = \frac{3}{5}$$

Check that all paths add to $1$: $\tfrac{6}{20} + \tfrac{6}{20} + \tfrac{6}{20} + \tfrac{2}{20} = 1$. ✓

Example 4: Testing for independence

$P(A) = 0.4$, $P(B) = 0.5$, and $P(A \text{ and } B) = 0.2$. Are $A$ and $B$ independent?

Solution. $P(A) \times P(B) = 0.4 \times 0.5 = 0.2$, which equals $P(A \text{ and } B)$. So yes, they’re independent.

Common mistakes

Multiplying the original probabilities when there’s no replacement. In Example 3, the second red has probability $\tfrac{2}{4}$, not $\tfrac{3}{5}$.

Forgetting the other order. “One of each colour” includes red-blue and blue-red. Add both paths.

Multiplying when you should add. Multiply for “and” (along a path). Add for “or” between separate paths.

Mixing up independent and mutually exclusive. If two events with non-zero probabilities are mutually exclusive, they can’t be independent: one happening makes the other impossible.

Practice

1. (Warm-up) Are the events independent or dependent?

• (a) rolling a die twice
• (b) drawing two cards from a deck without replacement
• (c) whether it rains today, and whether you bring an umbrella

Solution

(a) Independent. (b) Dependent. (c) Dependent (rain makes you more likely to bring one).

2. (Warm-up) Two coins are flipped. Find $P(\text{two heads})$.

Solution

$\tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}$

3. (Warm-up) $A$ and $B$ are independent with $P(A) = 0.3$ and $P(B) = 0.6$. Find $P(A \text{ and } B)$.

Solution

$0.3 \times 0.6 = 0.18$

4. (Core) Two cards are drawn from a standard deck without replacement. Find $P(\text{both aces})$.

Solution

$$\frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$$

5. (Core) Repeat Question 4, but with the first card replaced before the second draw.

Solution

$$\frac{4}{52} \times \frac{4}{52} = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$$

6. (Core) A bus is late $20\%$ of the time, independently from day to day. Find the probability it’s late on both Monday and Tuesday, and the probability it’s on time all five school days.

Solution

$P(\text{late both}) = 0.2 \times 0.2 = 0.04$.

$P(\text{on time all five}) = 0.8^5 \approx 0.328$.

7. (Core) A bag holds $4$ green and $6$ yellow marbles. Two are drawn without replacement. Find $P(\text{same colour})$.

Solution

$$P(\text{GG}) + P(\text{YY}) = \frac{4}{10} \times \frac{3}{9} + \frac{6}{10} \times \frac{5}{9} = \frac{12}{90} + \frac{30}{90} = \frac{42}{90} = \frac{7}{15}$$

8. (Challenge) $P(A) = 0.5$, $P(B) = 0.3$, and $P(A \text{ or } B) = 0.65$. Are $A$ and $B$ independent?

Solution

From the additive principle: $P(A \text{ and } B) = 0.5 + 0.3 - 0.65 = 0.15$. And $P(A) \times P(B) = 0.15$. They match, so $A$ and $B$ are independent.

9. (Challenge) Find the probability of rolling at least one $6$ in four rolls of a fair die.

Solution

Use the complement. The rolls are independent, so:

$$P(\text{no } 6\text{s}) = \left(\frac{5}{6}\right)^4 \approx 0.482$$$$P(\text{at least one } 6) = 1 - \left(\frac{5}{6}\right)^4 \approx 0.518$$