Conditional probability and independence
In an OpenStax sample of 100 people, 13 are left-handed, but among the 48 females only 4 are, about 0.083.
Most rainy days are cloudy. Are most cloudy days rainy?
Here is a made-up month of 30 days. It rained on 5 of them, and 4 of those 5 were cloudy. Twelve days were cloudy in all. So most rainy days were cloudy, 4 out of 5. It is tempting to turn that around and say that most cloudy days were rainy. They were not: only 4 of the 12 cloudy days had rain, a third. Both questions put the same 4 days on top of the fraction. They divide by different groups, and choosing the group is what conditional probability is about.
New information changes probabilities. If you know a card is a face card, the chance that it is a king rises from 4/52 to 4/12. The probability of A given that B has happened is written P(A | B) and read "A given B". The condition shrinks the sample space: instead of counting among all outcomes, you count only among the outcomes in B. As a formula, P(A | B) = P(A AND B) ÷ P(B), as long as P(B) is not zero.
OpenStax gives a die example. Let A = the face is 2 or 3 and B = the face is even. Given B, the possible faces are only 2, 4 and 6, and of those only 2 is in A. So P(A | B) = 1/3. The formula agrees: P(A AND B) = 1/6, P(B) = 3/6, and (1/6) ÷ (3/6) = 1/3. Without the condition, P(A) would be 2/6 = 1/3 as well, which is a coincidence worth noticing: here knowing B did not change the chance of A.
One table, three answers for left-handedness
Conditional probabilities are easiest to see in a two-way table. OpenStax Table 3.1 sorts a random sample of 100 people by sex assigned at birth and handedness. Overall, 13 of the 100 are left-handed, so P(L) = 0.13. Among the 52 males, 9 are left-handed, so P(L | M) = 9/52, about 0.173. Among the 48 females, 4 are, so P(L | F) = 4/48, about 0.083. Turning the condition around gives a different number: of the 13 left-handed people, 4 are female, so P(F | L) = 4/13, about 0.308. P(A | B) and P(B | A) are rarely equal, and swapping them is the weather mistake this lesson opened with.
| Right-handed | Left-handed | Total | |
|---|---|---|---|
| Males | 43 | 9 | 52 |
| Females | 44 | 4 | 48 |
| Total | 87 | 13 | 100 |
In this sample, handedness and sex are not independent
Events A and B are independent when P(A | B) = P(A): learning that B happened tells you nothing new about A. In the handedness sample, P(L | M) = 0.173 is not equal to P(L) = 0.13, so in these 100 people handedness and sex are not independent. Coin flips are the classic independent events. Whatever the first flip shows, the second is heads with probability 1/2. Dependence appears when the first result changes what is left. Draw two cards without replacement: P(first is an ace) = 4/52, and given that, only 3 aces remain among 51 cards, so P(second is an ace | first is an ace) = 3/51. The chance of two aces is 4/52 × 3/51 = 1/221, about 0.0045.
In 100 fair flips, a run of six or more is the usual result
Independence also explains why long streaks are not a sign that something is wrong. In 100 flips of a fair coin, the longest run of the same face is 6 or more about 81% of the time and 8 or more about 31% of the time; the average longest run is about 7. People who are asked to write down a "random" sequence usually avoid runs longer than three or four, which is one way teachers spot invented coin-flip homework.
Drag to turn. Scroll or pinch to zoom.
| P(A | B) | P(A AND B) ÷ P(B): count A only among outcomes in B |
| Independent | P(A | B) = P(A) |
| Without replacement | Later draws depend on earlier ones |
| P(A | B) vs P(B | A) | Usually different; do not swap them |
- Table 3.1. What is P(L), the probability a person is left-handed?
- Table 3.1. What is P(L | M)?
- Table 3.1. What is P(L | F)?
- Table 3.1. What is P(F | L)?
- Table 3.1. P(R | M) to 4 decimal places?
- In Table 3.1, are "left-handed" and "male" independent?
- Yes, because both are about people
- No, because P(L | M) ≈ 0.173 differs from P(L) = 0.13
- Yes, because the table adds up to 100
- It cannot be checked
- One die. A = 2 or 3, B = even. What is P(A | B)?
- Two cards without replacement. P(second is an ace | first is an ace)?
- What is P(two aces) when drawing two cards without replacement?
- Two dice. Given that the first die shows 3, what is P(sum = 8)?
- "Every player on the football team wears trainers to school, so a student in trainers is almost certainly on the team." That swaps P(trainers | team) with P(team | trainers). In a made-up class, all 6 team players wear trainers, so P(trainers | team) = 1. But if 24 students wear trainers, P(team | trainers) = 6/24 = 1/4.
- "In the made-up month, P(rain | cloudy) = 4/30." 4/30 is P(rain AND cloudy). The condition says to count only the 12 cloudy days, so P(rain | cloudy) = 4/12 = 1/3. Divide by the size of the condition group, not by everything.
- "Five paper slips, two of them winners. Two people each draw one without putting it back, so P(both win) = 2/5 × 2/5 = 4/25." After the first winner is taken, 1 winner is left among 4 slips. P(both win) = 2/5 × 1/4 = 1/10. The second draw depends on the first, so the second factor must be a conditional probability.
- "P(A | B) is high, so B causes A." A conditional probability only describes counts. In Table 3.1, P(L | M) is higher than P(L | F), and the table says nothing about why. The Korean curriculum also tells teachers not to read a conditional probability as an order in time or as a cause.
- P(A | B) = P(A AND B) ÷ P(B): the condition shrinks the sample space.
- P(A | B) and P(B | A) are different questions.
- Independent: P(A | B) = P(A). Coin flips yes; cards without replacement no.
- In 100 fair flips, a longest run around 7 is normal.
Sources for this course (9)
- Curriculum: 교육부 고시 제2022-33호 [별책 8] 수학과 교육과정 (ncic.re.kr/inv/org/download.do?year=2022&seq=10003559&orgType=ogi4, PDF read 2026-10-19: 초등학교 5~6학년군 [6수04-04]~[6수04-06] 가능성을 말과 0, 1/2, 1 같은 수로 나타내기; 중학교 1~3학년군 [9수04-03] 상대도수, [9수04-05] 경우의 수(두 경우의 수를 더하거나 곱하는 정도), [9수04-06] 확률(상대도수로서의 의미와 경우의 수의 비율로서의 의미, 각 경우의 가능성이 같다는 가정), 용어 상대도수·사건·확률; 공통수학1 [10공수1-03-01]~[10공수1-03-03] 합의 법칙·곱의 법칙·순열·조합, 기호 nCr; 확률과 통계 [12확통01-03] 이항정리, 용어 이항계수·파스칼의 삼각형; [12확통02-01]~[12확통02-06] 확률의 기본 성질·덧셈정리·여사건의 확률·조건부확률·독립과 종속·곱셈정리, 용어 시행·통계적 확률·수학적 확률·여사건·배반사건·조건부확률·종속·독립·독립시행과 기호 P(A), P(B|A); [12확통03-03] 이항분포, [12확통03-04] 정규분포와 이항분포의 관계, [12확통03-05]·[12확통03-06] 임의추출·표본평균과 모평균, 용어 큰 수의 법칙·이항분포·정규분포·모평균·표본평균과 기호 B(n, p), N(m, σ²), X̄)
- Textbook: OpenStax, Introductory Statistics 2e by Barbara Illowsky and Susan Dean, 3.1 terminology, equally likely outcomes, dime and nickel, law of large numbers, Belgian one euro coin 56% heads in 250 trials, conditional probability, Table 3.1 handedness of 100 people (openstax.org/books/introductory-statistics-2e/pages/3-1-terminology)
- 4.3 binomial distribution
- 7.1 central limit theorem for sample means, example mean 90, SD 15, n = 25, standard error 3 (.../pages/7-1-the-central-limit-theorem-for-sample-means-averages)
- Yut rules for the Korean example only: 국립민속박물관 한국민속대백과사전 「윷놀이」, 도·개·걸·윷·모 by how many of the four sticks land 뒤집어진 (folkency.nfm.go.kr/kr/topic/detail/1595, read 2026-10-19)
- Kerrich story: John Kerrich, 10,000 tosses, 5,067 heads (JMP case study JMP014 by DeWayne Derryberry, jmp.com/en/academic/case-study-library/is-a-coin-fair, citing Baldi and Moore; Wikipedia "John Edmund Kerrich")
- Galileo story: Sopra le scoperte dei dadi, sum 10 in 27 ways and 9 in 25 ways with three dice (English translation, University of York history of statistics, york.ac.uk/depts/maths/histstat/galileo.pdf)
- Pascal story: Pascal to Fermat, 29 July 1654, one six in four throws "as 671 to 625" (American Physical Society, aps.org/apsnews/2009/07/pascal-letters-fermat-points; Gaston Sanchez, De Mere's games; Statistical Science 2024 "Antoine Gombaud, Chevalier de Méré")
- Galton story: Francis Galton, Natural Inheritance of 1889, the quincunx (Wikipedia "Galton board")
Checked: 2026-10-17