Does the first event matter?
| Type | Meaning | Example |
|---|---|---|
| Independent events | the first doesn’t affect the second | flip a coin, then roll a die |
| Dependent events | the first changes the second | draw a card, keep it, draw another |
Independent events
P(A and B) = P(A) × P(B)
Example: Spin a spinner (½ red) and roll a die. P(red and 4) = ½ × ⅙ = 1/12.
Coins and dice have no memory. After 5 heads in a row, P(heads) is still ½.
With and without replacement
A bag has 5 red and 3 blue marbles. Draw two marbles.
With replacement (put the first back): independent.
P(red, then red) = 5/8 × 5/8 = 25/64
Without replacement (keep the first): dependent. After a red is taken, 4 red of 7 remain.
P(red, then red) = 5/8 × 4/7 = 20/56 = 5/14
The general multiplication rule
For dependent events:
P(A and B) = P(A) × P(B | A)
P(B | A) is a conditional probability: the probability of B given that A has already happened.
Example: cards
Draw two cards without replacement. P(both aces)?
- P(1st ace) = 4/52
- P(2nd ace | 1st ace) = 3/51
- P = 4/52 × 3/51 = 12/2,652 = 1/221
Example: committee
A class has 12 girls and 10 boys. Two students are chosen at random. P(both boys)?
10/22 × 9/21 = 90/462 = 15/77
Testing for independence
A and B are independent if P(A and B) = P(A) × P(B).
Example: In a survey, P(likes soccer) = 0.4, P(plays an instrument) = 0.5, P(both) = 0.2. Since 0.4 × 0.5 = 0.2, the events are independent.
Real-life examples
| Situation | Independent or dependent? |
|---|---|
| Rolling two dice | independent |
| Drawing names from a hat without putting them back | dependent |
| Rain today and rain tomorrow | dependent (weather systems last) |
| Your coin flip and a friend’s coin flip | independent |
Worked examples
Example 1: Roll a die twice. P(two 6s)? 1/36.
Example 2: 3 defective bulbs in a box of 20. Pick 2 without replacement. P(both defective)? 3/20 × 2/19 = 3/190.
Example 3: A bag has 4 green and 6 yellow candies. Take one, eat it, take another. P(green then yellow)? 4/10 × 6/9 = 4/15.
Common mistakes
- Treating dependent events as independent (forgetting to reduce the total).
- Thinking past results affect independent events (the gambler’s fallacy).
- Reducing only the numerator and not the denominator.
Try it yourself
- Flip a coin 4 times. P(all tails)?
- A bag has 6 red and 4 white balls. Draw 2 with replacement. P(both white)?
- Same bag, without replacement. P(both white)?
- Are "drawing a heart" and "drawing a king" from one card independent?
Answers: 1) 1/16 2) 4/25 3) 2/15 4) Yes: P(king of hearts) = 1/52 = 1/4 × 1/13.
Quick recap
- Independent events: multiply the probabilities.
- Dependent events: multiply by the conditional probability P(B | A).
- Without replacement makes events dependent.