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Statistics

Conditional Probability

Lesson

Conditional probability is the probability of an event given that another event has already happened. The notation P(A∣B)P(A \mid B)reads “the probability of A given B.”

P(A∣B)=P(A and B)P(B)P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}

Practical way to think about it: knowing that BB happened restricts the sample space to just the BB outcomes. Within that restricted set, what fraction also satisfies AA?

Conditional probability shows up everywhere — most importantly in drawing without replacement, where each draw changes the pool for the next draw.

Worked example 1 — restricting the sample space

A class of 20 students: 12 girls, 8 boys. Five girls and three boys like math. Given that a randomly chosen student is a girl, what’s the probability she likes math?

P(math∣girl)=512P(\text{math} \mid \text{girl}) = \frac{5}{12}

Worked example 2 — drawing without replacement

From a standard deck, draw two cards without replacement. Given that the first was a heart, what’s the probability the second is also a heart?

After one heart is removed: 12 hearts remain out of 51 cards.

P(2nd heart∣1st heart)=1251=417\begin{aligned} P(\text{2nd heart} \mid \text{1st heart}) &= \frac{12}{51} \\ &= \frac{4}{17} \end{aligned}

How to type your answer

Fraction in lowest terms or a decimal. Examples: 5/12, 4/17, 1/2, 2/3.

Practice

Work through these. Stuck? Click Get a hint.

Warm-Up

Quick problems to get going.

Problem 1

Card is red. P(heart∣red)\text{Card is red. } P(\text{heart} \mid \text{red})

Problem 2

Die shows an even number. P(6∣even)\text{Die shows an even number. } P(6 \mid \text{even})

Problem 3

5 red, 5 blue. Two draws no replacement. P(2nd red∣1st red)\text{5 red, 5 blue. Two draws no replacement. } P(\text{2nd red} \mid \text{1st red})

Problem 4

P(5∣die>3)P(5 \mid \text{die} > 3)

Practice

Standard problems matching the lesson.

Problem 5

Class: 12 girls (5 like math), 8 boys (3 like math). P(math∣girl)\text{Class: 12 girls (5 like math), 8 boys (3 like math). } P(\text{math} \mid \text{girl})

Problem 6

Same class. P(math∣boy)\text{Same class. } P(\text{math} \mid \text{boy})

Problem 7

Same class. P(girl∣likes math)\text{Same class. } P(\text{girl} \mid \text{likes math})

Problem 8

Deck, two draws no replacement. P(2nd heart∣1st heart)\text{Deck, two draws no replacement. } P(\text{2nd heart} \mid \text{1st heart})

Problem 9

Same setup. P(2nd heart∣1st not heart)\text{Same setup. } P(\text{2nd heart} \mid \text{1st not heart})

Problem 10

P(multiple of 3∣even) on one dieP(\text{multiple of 3} \mid \text{even})\ \text{on one die}

Problem 11

4 red, 3 blue, 3 green. No replacement. P(2nd red∣1st blue)\text{4 red, 3 blue, 3 green. No replacement. } P(\text{2nd red} \mid \text{1st blue})

Problem 12

Two dice. P(sum=7∣first die=4)\text{Two dice. } P(\text{sum}=7 \mid \text{first die}=4)

Problem 13

Two dice. P(both even∣first even)\text{Two dice. } P(\text{both even} \mid \text{first even})

Problem 14

Fair coin landed heads 9 times in a row. P(heads next)\text{Fair coin landed heads 9 times in a row. } P(\text{heads next})

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 15

100 people: 60 own car, 40 own bike, 20 own both. P(bike∣car)\text{100 people: 60 own car, 40 own bike, 20 own both. } P(\text{bike} \mid \text{car})

Problem 16

Same survey. P(car∣bike)\text{Same survey. } P(\text{car} \mid \text{bike})

Problem 17

Two dice. P(sum>8∣first die=6)\text{Two dice. } P(\text{sum} > 8 \mid \text{first die}=6)

Problem 18

6 red, 4 blue. 3 draws no replacement. P(all red)\text{6 red, 4 blue. 3 draws no replacement. } P(\text{all red})

Problem 19

Same bag. P(2nd blue∣1st red)\text{Same bag. } P(\text{2nd blue} \mid \text{1st red})

Problem 20

Deck. Two draws no replacement. P(both kings)\text{Deck. Two draws no replacement. } P(\text{both kings})

Problem 21

8 marbles, 3 red. Two draws no replacement. P(both red)\text{8 marbles, 3 red. Two draws no replacement. } P(\text{both red})

Problem 22

25 students: 10 fr, 15 so. 4 fr and 9 so wear glasses. P(glasses∣sophomore)\text{25 students: 10 fr, 15 so. 4 fr and 9 so wear glasses. } P(\text{glasses} \mid \text{sophomore})