A. Counting and Probability · Probability

Independent and Dependent Events

A1.6About 70 minutes

Learning goals

I can decide whether two events are independent or dependent, by reasoning and by testing P(A ∩ B) = P(A) × P(B).
I can use the multiplicative principle P(A ∩ B) = P(A) × P(B) for independent events.
I can find P(A ∩ B) for dependent events, such as drawing without replacement, by multiplying the changing probabilities.
I can draw tree diagrams with probabilities on the branches and use them to solve problems.

Words to know

independent events
Events where the occurrence of one does not change the probability of the other. A and B are independent exactly when P(A ∩ B) = P(A) × P(B).

dependent events
Events where the occurrence of one changes the probability of the other.

with replacement
Each item drawn is put back before the next draw, so the probabilities stay the same (independent draws).

without replacement
Items drawn are not put back, so the probabilities change after each draw (dependent draws).

multiplicative principle for probability
For independent events, P(A ∩ B) = P(A) × P(B), and the same for more events: P(A ∩ B ∩ C) = P(A) × P(B) × P(C).

Does one event change the other?

Two events are independent if knowing that one happened does not change the probability of the other. Coin tosses, separate die rolls and draws with replacement are independent.

Two events are dependent if one does change the probability of the other. Drawing cards or marbles without replacement is the classic example: after an ace is drawn, there are fewer aces left.

Remember from Grades 7 and 8: you used tree diagrams for two or three independent events and for dependent events without replacement. This lesson adds a precise test for independence and probability rules you will use for the rest of the course.

The multiplication rules

Independent events: P(A ∩ B) = P(A) × P(B).

This is also the definition used to test independence: if P(A ∩ B) = P(A) × P(B), the events are independent; if not, they are dependent.

Dependent events: multiply the probability of the first event by the probability of the second event given that the first has happened:

P(A ∩ B) = P(A) × P(B given A)

The probability "of B given A" is written P(B | A). It is a conditional probability, the topic of lesson 11. For draws without replacement you find it by adjusting the counts.

Worked example

Example 1: Independent events

A die is rolled and a coin is tossed. Then a spinner with 4 equal sections (one of them blue) is spun. Find (a) P(6 and heads), (b) P(6 and heads and blue), (c) P(not 6 and not blue).

The three results do not affect one another, so they are independent.
(a) P(6) × P(H) = 1/6 × 1/2 = 1/12.
(b) 1/6 × 1/2 × 1/4 = 1/48.
(c) P(not 6) = 5/6 and P(not blue) = 3/4. Multiply: 5/6 × 3/4 = 15/24 = 5/8. (The coin does not matter here.)

(a) 1/12, (b) 1/48, (c) 5/8.

Worked example

Example 2: Without replacement, with a tree diagram

A bag holds 3 red and 2 blue marbles. Two marbles are drawn without replacement. Find (a) P(both red), (b) P(one of each colour), (c) P(both the same colour).

First draw: P(R) = 3/5, P(B) = 2/5.
Second draw, given the first: after a red, 2 red and 2 blue remain, so P(R) = 2/4, P(B) = 2/4. After a blue, 3 red and 1 blue remain, so P(R) = 3/4, P(B) = 1/4.
Multiply along each branch to get the probability of each path: RR = 6/20, RB = 6/20, BR = 6/20, BB = 2/20. (Check: they add to 20/20 = 1.)
(a) P(RR) = 6/20 = 3/10.
(b) Add across the paths RB and BR (they are mutually exclusive): 6/20 + 6/20 = 12/20 = 3/5.
(c) RR or BB: 6/20 + 2/20 = 8/20 = 2/5.

(a) 3/10, (b) 3/5, (c) 2/5.

Tree diagram for drawing two marbles without replacement from 3 red and 2 blue. The second-draw probabilities depend on the first draw. Multiply along a path; add the paths you want.

Tree diagram rules

The branches leaving any point add to 1.
Multiply along a path to get the probability of that whole path ("and").
Add the probabilities of the paths that make up your event ("or"), because different paths are mutually exclusive.
All the final paths add to 1. Use this as a check.

Worked example

Example 3: With and without replacement

Two cards are drawn from a standard deck. Find P(both are aces) (a) with replacement, (b) without replacement.

(a) With replacement the draws are independent: 4/52 × 4/52 = 1/13 × 1/13 = 1/169 ≈ 0.0059.
(b) Without replacement: P(first ace) = 4/52. If it was an ace, 3 aces remain among 51 cards: P(second ace | first ace) = 3/51.
P(both aces) = 4/52 × 3/51 = 12/2652 = 1/221 ≈ 0.0045.

(a) 1/169, (b) 1/221. Without replacement is slightly less likely, because one ace is gone.

Worked example

Example 4: Testing for independence

One die is rolled. A = even, B = at most 4, C = at most 3. Are A and B independent? Are A and C independent?

A = {2, 4, 6}, P(A) = 1/2. B = {1, 2, 3, 4}, P(B) = 2/3. C = {1, 2, 3}, P(C) = 1/2.
A ∩ B = {2, 4}: P(A ∩ B) = 2/6 = 1/3. P(A) × P(B) = 1/2 × 2/3 = 1/3. Equal, so A and B are independent.
A ∩ C = {2}: P(A ∩ C) = 1/6. P(A) × P(C) = 1/2 × 1/2 = 1/4. Not equal, so A and C are dependent.

A and B are independent; A and C are dependent. Independence is not always obvious from the wording, so test it with the definition.

Each face shows which of A (even) and B (at most 4) it belongs to. Faces 2 and 4 are in both. Since P(A ∩ B) equals P(A) × P(B), knowing B happened leaves the chance of an even number unchanged: 2 of the 4 faces in B are even, which is still 1/2.

Just for fun

A dog walks into a probability class. The teacher asks, "Is your bark independent of the mail carrier's arrival?" The dog checks the data: P(bark ∩ mail) is far bigger than P(bark) × P(mail). Definitely dependent.

Hands-on activity

Socks without replacement

You need: 6 socks (4 of one colour and 2 of another; or any 6 small items in 2 types), a bag or pillowcase, paper and a pencil.

Draw a tree diagram for drawing 2 socks without replacement and calculate P(a matching pair). Then do the same with replacement.
Without looking, draw 2 socks without replacement and record whether they match. Put both back and shake. Repeat 30 times.
Repeat 30 times with replacement (draw one, record, put it back, draw again).
Compare your experimental probabilities with the theoretical ones.

What to look for: with 4 of colour X and 2 of colour Y, without replacement: P(match) = 4/6 × 3/5 + 2/6 × 1/5 = 12/30 + 2/30 = 14/30 = 7/15 ≈ 0.47. With replacement: (4/6)² + (2/6)² = 16/36 + 4/36 = 20/36 = 5/9 ≈ 0.56. Your 30-trial results will vary around these values.

Your turn

A coin is tossed 3 times. Find P(H, H, T in that order).

A die is rolled twice. Find P(an even number then a 5).

Two cards are drawn without replacement. Find P(both hearts).

Two cards are drawn with replacement. Find P(both hearts).

A box has 4 good and 2 defective fuses. Two are chosen without replacement. Find P(both good).

Using the fuse box, find P(exactly one defective).

P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Are A and B independent?

More practice: warm-up

Are two rolls of a die independent?

Are two cards drawn without replacement independent?

Independent, P(A) = 0.3, P(B) = 0.2. P(A and B)?

What do the branches from one point of a tree diagram add to?

A coin is tossed and a die is rolled. P(tails and 3)?

More practice: on your own

Three cards are drawn without replacement. Find P(all three are face cards).

A goalie stops 92% of shots, independently. Find P(she stops the next 4 shots), and P(she stops the first 3 and allows the 4th).

P(A) = 0.6, P(B) = 0.3, P(A ∪ B) = 0.72. Are A and B independent?

A bag has 5 green and 3 yellow candies. You eat one at random, then take another. Find P(the second is green).

Two dice are rolled. Let A = first die is 6 and B = sum is 7. Are A and B independent?

Challenge

Two dice are rolled. Let A = first die is 6 and C = sum is 10. Are A and C independent? Explain the difference from the previous question.

A best-of-three series: a team wins each game with probability 0.6, independently. Find P(the team wins the series).

Show that if A and B are independent, then A and B′ are also independent.

Check your understanding

Quiz

A die is rolled and a card is drawn from a deck. What is P(rolling a 4 and drawing a king)?

Two marbles are drawn without replacement from a bag of 4 red and 6 white. What is P(both red)?

Which pair of events is dependent?

P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.1. Which is true?

On a tree diagram, how do you find the probability of one complete path?

A machine part works with probability 0.95, independently of a second part that works with probability 0.9. What is P(both work)?

For parents and teachers

Big idea (A1.6, independence): independent events don't influence each other, and P(A and B) = P(A) × P(B). For dependent events, the second probability changes after the first (as with drawing without replacement). Tree diagrams organize multi-step problems: multiply along paths, add across paths.

How to help: The sock activity is quick and shows clearly why "without replacement" changes things. Ask your teen to explain to you why the second draw's probabilities change.

Common difficulties:

Multiplying probabilities for dependent events without adjusting the second probability.
Confusing independent with mutually exclusive.
Forgetting that a combination such as "one of each" can happen in two orders (RB and BR).
Deciding independence from wording alone instead of testing P(A ∩ B) = P(A) × P(B).