Probability of Compound Events — Print

Two or more things happening: multiply along a path, add across paths, and watch whether the first event changes the second.

What the practice questions cover

These questions use 2 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Flip a coin, then spin a pointer that lands on red or blue. How many different endings are there?

Most people say three: heads, tails, and a colour. Draw it and you can see that is wrong.

flip a coin, then spin for red or blue flip H T R B R B two choices, then two more — four endings, not three

Each of the two flips splits into two spins. Two branches, then two more on each — four endings.

H-red, H-blue, T-red, T-blue. And each is equally likely, so each has probability 14.

Notice where the 4 came from: 2 × 2. Not 2 + 2, which also happens to be four and misleads everybody.

"This and then that" multiplies. "This or that" adds. Draw the tree whenever you are unsure — it settles the question without any rule at all.

A probability tree for two draws without replacement 4/106/10 R B 3/96/9 4/95/9 R B R B RR = 4/10 × 3/9 = 2/15 RB = 4/15 BR = 4/15 BB = 1/3 multiply along a path · add across paths · all four sum to 1

Counting first

Before probability, get good at counting. If a menu has 4 mains and 3 desserts, there are 4 × 3 = 12 meals.

Multiply the number of choices at each independent stage. This is the fundamental counting principle, and it is the engine behind almost every compound probability question.

Multiply along a path

P(A and then B) = P(A) × P(B given A).

Roll two dice: P(both 6) = 16 × 16 = 136. Dice do not remember, so the second probability is unchanged.

And multiplies, or adds

"And" along a single path → multiply.
"Or" across separate paths → add.

With replacement vs without

A bag has 4 red and 6 blue marbles.

With replacement: P(two reds) = 410 × 410 = 425. You put the first marble back, so nothing changed.

Without replacement: P(two reds) = 410 × 39 = 215. One red is gone, so both the top and the bottom drop by one.

The single most common error

Forgetting to shrink the denominator on the second draw. If the problem says "without replacing", the total goes down too.

Tree diagrams

Draw a branch for each option at each stage. Write the probability on each branch. Multiply along a branch to get that path's probability; add the paths that satisfy your event.

Check yourself: all the final path probabilities must add to 1.

When order does not matter

"How many ways to choose 2 students from 5?" If order does not matter, pairs like (Ana, Ben) and (Ben, Ana) are the same team.

Count the ordered ways, 5 × 4 = 20, then divide by the 2 orderings of each pair: 202 = 10.

That is written (52) = 10 and read "5 choose 2". A round-robin tournament with 5 teams has exactly 10 games.

Worked examples

Without replacement

A drawer has 5 black socks and 3 white socks. You pull two without looking. What is P(both black)?

  1. First sock: 58 Five black out of eight socks.
  2. Second sock: 47 One black sock is gone — both numbers drop by one.
  3. 58 × 47 = 2056 = 514 Multiply along the path, then simplify.

514

Two paths, so add

Spinner A is red or blue, equally likely. Spinner B is red, blue or green, equally likely. What is P(the colours match)?

  1. Matching happens two ways: both red, or both blue. Spinner A has no green, so green can never match.
  2. Both red: 12 × 13 = 16 Multiply along that path.
  3. Both blue: 12 × 13 = 16 Same for the other path.
  4. 16 + 16 = 13 Add across separate paths.

13

Counting handshakes

Eight people at a party each shake hands with every other person exactly once. How many handshakes?

  1. Each of the 8 shakes 7 hands: 8 × 7 = 56 That counts ordered pairs.
  2. But each handshake got counted twice. Ana–Ben and Ben–Ana are the same handshake.
  3. 562 = 28 This is (82).

28 handshakes.

Words to know

Compound event
An event made of two or more simple events.
Independent
The first event does not change the second — like rolling a die twice.
Dependent
The first event changes the second — like drawing marbles without replacing them.
Tree diagram
A branching picture of all the ways a sequence can unfold.
Fundamental counting principle
If one stage has m ways and the next has n, together they have m × n ways.

Mistakes to watch for

Using the same denominator twice in a "without replacement" problem. After the first marble is gone there are only nine left, so the second fraction has 9 on the bottom, not 10.

Adding when you should multiply. "And" multiplies because each branch splits again; "or" adds because the branches are separate endings. Two coins give 2 × 2 = 4 outcomes, not 2 + 2.

Forgetting to divide out repeats when order does not matter. Choosing two people from five is 10 ways, not 20 — picking Ana then Ben is the same pair as Ben then Ana.

Multiplying probabilities for events that affect each other. Independence is what licenses the multiplication. Drawing two cards without replacing the first is not independent, and the second fraction must change.

The Chevalier de Méré made money betting he could roll at least one six in four throws of a die, and lost money betting on a double six in 24 throws of two dice.

He thought the two bets were equivalent, and asked Pascal why they were not. The first wins about 51.8% of the time; the second about 49.1%.

That tiny gap between the two — and one gambler's willingness to ask about it — is where the mathematics of compound events began.

Problem 1

Cam has 4 shirts and 3 pairs of pants. How many different outfits?

Worked answer

12

4 × 3 = 12

Problem 2

Two fair dice are rolled. How many outcomes are in the sample space?

Worked answer

36

6 × 6 = 36

Problem 3

Two fair coins are flipped. Find P(both heads).

Worked answer

1/4

12 × 12 = 14

Problem 4

A menu has 5 appetizers, 4 mains and 3 desserts. How many three-course meals?

Worked answer

60

5 × 4 × 3 = 60

Problem 5

A fair die is rolled twice. Find P(both sixes).

Worked answer

1/36

16 × 16 = 136

Problem 6

A bag has 4 red and 6 blue marbles. One is drawn, replaced, then another is drawn. Find P(both red).

Worked answer

4/25

P(red) = 410 each time

410 × 410 = 16100 = 425

Problem 7

Same bag (4 red, 6 blue), but the first marble is NOT replaced. Find P(both red).

Worked answer

2/15

First: 410

Second: 39 — one red gone, one marble gone

410 × 39 = 1290 = 215

Problem 8

Two fair dice are rolled. Find P(sum = 7).

Worked answer

1/6

Pairs summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — six of them

636 = 16

Problem 9

A spinner is red or blue, equally likely. A fair die is rolled. Find P(blue and an even number).

Worked answer

1/4

P(blue) = 12

P(even) = 12

12 × 12 = 14

Problem 10

Six people each shake hands with every other person exactly once. How many handshakes?

Worked answer

15

6 × 5 = 30 ordered pairs

Each handshake was counted twice

30 ÷ 2 = 15

Problem 11

How many different 2-person teams can be chosen from 7 students?

Worked answer

21

7 × 6 = 42 ordered pairs

42 ÷ 2 = 21

This is (72).

Problem 12

Two fair dice are rolled. Find P(at least one 6).

Worked answer

11/36

P(no six on one die) = 56

P(no six at all) = 56 × 56 = 2536

1 - 2536 = 1136

Problem 13

A box has 9 good and 3 defective parts. Two are drawn without replacement. Find P(both good).

Worked answer

6/11

First: 912

Second: 811

912 × 811 = 72132 = 611

Problem 14

In a league of 8 teams, each pair plays twice. How many games in total?

Worked answer

56

Pairs: 8 × 72 = 28

Each pair plays twice: 28 × 2 = 56

Problem 15

Three fair coins are flipped. Find P(exactly two heads).

Worked answer

3/8

The 8 outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Exactly two heads: HHT, HTH, THH — three of them

38

Problem 16

A drawer has 5 white and 3 black socks. Two are pulled without replacement. Find P(one of each colour).

Worked answer

15/28

White then black: 58 × 37 = 1556

Black then white: 38 × 57 = 1556

Add the two paths: 3056 = 1528

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