Multiplication Rule of Probability

Basic

Learn how to calculate 'and' probability — chance of both events happening.

For:Secondary 3 Mathematics · Unit:Probability & statistics · Lesson 30 of 37 in the learning path

After this lesson, you can:
  • Independent (with replacement): P(A and B) = P(A) × P(B)
  • Dependent (without replacement): P(A and B) = P(A) × P(B|A)
  • 'And' uses multiplication, 'or' uses addition
數學Mathematics
Before this lesson you should know: Addition Rule of Probability

1Summary 重點摘要

Learn how to calculate 'and' probability — chance of both events happening.

2Vocabulary cards Vocabulary

M
Multiplication rule乘法法則

Method to calculate 'and' probability

I
Independent events獨立事件

One event doesn't affect the other

D
Dependent events相關事件

One event affects the other

P
P(A and B)P(A 和 B)

Probability of both events happening

P
P(A|B)

Probability of A given B happened

C
Conditional probability條件概率

Probability under a condition

W
With replacement有回置

Put back after drawing

W
Without replacement無回置

Not put back after drawing

3Step by step Steps

  1. Step 1: Independent eventsIf two events do not affect each other, so that whether one occurs does not change the probability of the other, they are independent. The multiplication rule for independent events is P(A and B) = P(A) × P(B). For example, tossing a coin and rolling a die are independent, since the coin cannot influence the die. Why multiply directly? Because the first event leaves the sample space of the second completely unchanged. A common mistake is treating two events as independent without checking whether they influence each other.
  2. Step 2: An example with independent eventsRoll a die twice and find the probability of two sixes: the probability of a six is 16 on each roll, and the two rolls do not affect each other, so P(both sixes) = 16 × 16 = 136. Why is the second roll still 16? A die has no memory; the first result does not change the second at all. A common mistake is answering 16 + 16 = 13, which belongs to 'at least one six' problems and does not suit the 'both' situation that calls for multiplication.
  3. Step 3: Dependent events and conditional probabilityIf the result of the first event changes the probability of the second, the events are dependent, and the conditional form of the rule applies: P(A and B) = P(A) × P(B|A), where P(B|A) is the probability of B given that A has occurred. The typical case is drawing without replacement: after the first draw the total is smaller, so the second probability changes. Why not multiply directly? Because the sample space of the second draw has been altered by the first. A common mistake is ignoring the words 'without replacement' and using the independent-event formula.
  4. Step 4: An example with dependent eventsDraw two cards from a deck of 52 without replacement and find the probability that both are hearts: the first heart has probability 1352; after one heart is drawn, 51 cards remain with 12 hearts, so P(second heart | first heart) = 1251. Why do both numerator and denominator change? The total falls from 52 to 51 and the hearts from 13 to 12, each dropping by one. A common mistake is reducing only the numerator, wrongly writing 1252; the denominator must reflect the remaining total.
  5. Step 5: Multiplying and simplifyingContinuing the example: P(both hearts) = 1352 × 1251. Simplifying first reduces the numbers: 1352 = 14, so 14 × 1251 = 12204, which simplifies to 117. Why simplify first? Multiplying large numbers and simplifying afterwards is error-prone, while cancelling early cuts the workload, for example the common factor 3 of 12 and 51 gives 4 and 17 at once. A common mistake is leaving 12204 unsimplified; give the answer as a fraction in lowest terms, or as a decimal if the question asks.
  6. Step 6: Telling independent from dependentThe key test is whether the first result changes the sample space of the second: drawing with replacement, or repeated dice and coin trials, keeps the trials independent; drawing without replacement changes the total and makes them dependent. For a bag with 3 red and 2 blue balls, with replacement the probability of red stays 35 every time, while without replacement it depends on the first draw. Why practise this judgement? Choosing the wrong rule ruins the whole question. A common mistake is applying the independent formula on seeing 'draw twice'; read carefully for the word 'replacement'.

4Diagrams Diagrams

概率樹形圖:擲硬幣兩次Probability tree: two coin tosses1/21/2公 H字 T1/21/21/21/2HHHTTHTT1/41/41/41/4每條路徑嘅概率 = 沿路相乘,全部加埋 = 1Each path: multiply along the way; all paths sum to 1
Fig. 1: Probability tree

5Worked example Example

Bag has 3 red balls and 2 blue balls. Draw twice (without replacement), probability both red?

  1. Step 1: Known. Total 5 balls, 3 red 2 blue. Draw twice, no replacement.
  2. Step 2: First draw red. P(first red) = 35.
  3. Step 3: After drawing, 4 balls left, 2 red. P(second red|first red) = 24 = 12.
  4. Step 4: Multiply. P(both red) = 35 × 12 = 310.
  5. Answer: Probability both red = 310. Because the first ball is not replaced, the bag has 4 balls left with 2 red, so P(second red | first red) = 24 = 12. The events are dependent, so the multiplication rule gives P = 35 × 12 = 310. State 'without replacement', derive the conditional probability, then multiply.

6Common mistakes Mistakes

✗ Treating without replacement as independent
✓ Without replacement, subtract the one drawn!
✗ Using conditional probability for independent
✓ Independent just multiply, no conditional needed.
✗ Forgetting denominator changes
✓ Without replacement, denominator decreases by 1, numerator may change too.
✗ Using addition for 'and'
✓ 'And' uses multiplication, 'or' uses addition!

7Quick check Quick check

Q1Roll dice twice, both 1?

Q2Independent multiplication formula?

Q3Draw two cards (without replacement), both Ace?

Q4'And' uses which rule?

Q5Bag 4 red 1 blue, draw twice (no replacement), both red?

8Recap Recap

  • Independent (with replacement): P(A and B) = P(A) × P(B)
  • Dependent (without replacement): P(A and B) = P(A) × P(B|A)
  • 'And' uses multiplication, 'or' uses addition
  • Without replacement, remember both numerator and denominator change
  • Rolling dice, flipping coin is always independent

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