HSS.CP.A.3Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.CP.A.3: Conditional Probability and What Independence Means
In plain English: HSS.CP.A.3 is the Common Core statistics and probability standard that defines the conditional probability of A given B as P(A and B)/P(B), the chance of A once we know that B happened. Students also interpret independence as P(A | B) = P(A) and P(B | A) = P(B): knowing one event does not change the chance of the other. It is usually taught in Geometry or Algebra II.
Understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B.
Common Core State Standards for Mathematics · Domain: Conditional Probability and the Rules of Probability (CP) · Cluster: Understand independence and conditional probability and use them to interpret data Also written as HSS-CP.A.3 or S-CP.3 · Official standard
Students learn what "the probability of A given B" means and how to compute it: P(A | B) = P(A and B)/P(B). The idea behind the formula is that knowing B happened shrinks the sample space to B, so the question becomes "what fraction of B is also in A?" Students compute conditional probabilities from equally likely outcomes, from given probabilities and from tables of data, and they practice reading the order of the events, since P(A | B) and P(B | A) are usually different.
The second half of the lesson connects conditional probability to independence. Students interpret P(A | B) = P(A) as "knowing B does not change the chance of A", check the matching statement P(B | A) = P(B), and see why both follow from the product definition P(A and B) = P(A) · P(B) that they met in HSS.CP.A.2.
Learning Objectives
By the end of this lesson, students will be able to:
Compute P(A | B) as P(A and B)/P(B) and explain it as the fraction of B that is also in A
Tell P(A | B) apart from P(B | A) and from P(A and B), in symbols and in words
Interpret independence as P(A | B) = P(A): knowing B does not change the probability of A
Interpret independence as P(B | A) = P(B) and check both statements for a pair of events
Decide whether events defined on data are independent by comparing conditional and overall probabilities
Prior Knowledge Required
Students should already be comfortable with:
Events as subsets and the meaning of A and B HSS.CP.A.1
The product definition of independent events HSS.CP.A.2
Finding probabilities in a uniform probability model 7.SP.C.7
Dividing fractions and decimals, and reading counts from a table
Post the situation and let students work alone for three minutes.
Warm-Up Prompt
"A class has 28 students. 16 are in band, and 10 of the band students play a sport. Of the 12 students not in band, 6 play a sport. (1) If you pick any student, what is the chance they play a sport? (2) If you pick a band student, what is the chance they play a sport? (3) If you pick a student who is not in band?"
Answers: (1) 16/28 = 4/7, about 0.57. (2) 10/16 = 5/8 = 0.625. (3) 6/12 = 1/2. Ask what changed between (1) and (2): the question did not change, but the group we choose from did. Introduce the words "given that": question (2) is the probability that a student plays a sport given that the student is in band. Ask: "Does knowing a student is in band change the chance they play a sport?" Keep the answer (yes, a little) for the independence discussion later.
Direct Instruction20 minutes
Definition. For events A and B with P(B) > 0, the conditional probability of A given B is P(A | B) = P(A and B)/P(B). The bar is read "given". Knowing that B happened makes B the new sample space, and P(A | B) is the share of B that also lies in A (Diagram 1). To compute one:
Find the condition: the event after "given", "if" or "among". That event goes in the denominator.
Find P(A and B), the probability that both happen.
Find P(B), the probability of the condition.
Divide: P(A | B) = P(A and B)/P(B). With counts, this is the same as (count in both)/(count in B).
Interpret: "Among the outcomes in B, this fraction are in A."
Restricting the sample space
A number cube is rolled. Given that the number is greater than 3, what is the probability that it is even? B = {4, 5, 6} and A = {2, 4, 6}.
For two events, P(A) = 0.3, P(B) = 0.4 and P(A and B) = 0.12. Find P(A | B).
Equation: P(A | B) = 0.12/0.4 = 0.3
The order of the events matters
Same events. Find P(B | A).
Equation: P(B | A) = 0.12/0.3 = 0.4, which is not the same as P(A | B) = 0.3
Interpreting independence
Same events. Compare each conditional probability with the unconditional one.
Equation: P(A | B) = 0.3 = P(A) and P(B | A) = 0.4 = P(B), so knowing either event does not change the other: A and B are independent
Dependence in a data table
In the invented survey of 240 students below, B = "eats breakfast every day". Compare P(B | grade 9) with P(B).
Equation: P(B | grade 9) = 72/120 = 0.6 but P(B) = 120/240 = 0.5, so breakfast and grade are not independent
Invented survey of 240 students: daily breakfast by grade
Eats breakfast daily
Does not
Total
Grade 9
72
48
120
Grade 12
48
72
120
Total
120
120
240
Why the two independence statements work. If P(A and B) = P(A) · P(B), then P(A | B) = P(A) · P(B)/P(B) = P(A), and in the same way P(B | A) = P(B). So independence says that the condition makes no difference, in either direction. Use Diagram 1 for Example 4 and Diagram 2 for Example 5. For Example 5, also compute P(grade 9 | B) = 72/120 = 0.6 and compare it with P(grade 9) = 0.5: the other direction fails too.
Guided Practice15 minutes
Display this invented record of 180 gym members. Let W = "uses the pool" and M = "usually visits in the morning".
Invented record of 180 gym members
Uses the pool
Does not
Total
Morning
30
50
80
Evening
45
55
100
Total
75
105
180
Pairs compute and interpret four probabilities, one at a time, and must say which total goes in the denominator before dividing: P(W | M) = 30/80 = 0.375; P(W) = 75/180 ≈ 0.417; P(M | W) = 30/75 = 0.4; P(M) = 80/180 ≈ 0.444. Since P(W | M) ≠ P(W) and P(M | W) ≠ P(M), the events are not independent: morning visitors use the pool a little less often than members overall. Listen for pairs who divide by 180 for a conditional probability, and for pairs who swap 30/80 and 30/75.
Independent Practice10-15 minutes
Students work alone on two problems. (1) An invented company of 150 employees has 60 who work remotely (R), 50 who use a standing desk (D), and 20 who do both. Find P(D | R), P(D), P(R | D) and P(R), and decide whether R and D are independent. (Answers: P(D | R) = 20/60 = 1/3 = P(D) = 50/150; P(R | D) = 20/50 = 0.4 = P(R) = 60/150; independent.) (2) A spinner has 10 equal sections numbered 1-10. Let A = "at most 3" and B = "odd". Find P(A | B) and P(B | A) and compare them with P(A) and P(B). (Answers: P(A | B) = 2/5, but P(A) = 3/10; P(B | A) = 2/3, but P(B) = 1/2; not independent.)
Closure5-10 minutes
Exit ticket: P(A) = 0.6, P(B) = 0.15 and P(A and B) = 0.09. (1) Find P(A | B) and P(B | A). (2) Are A and B independent? Explain using conditional probability. (Answers: P(A | B) = 0.09/0.15 = 0.6 and P(B | A) = 0.09/0.6 = 0.15. Both equal the unconditional probabilities, so the events are independent.)
Differentiation Strategies
For Struggling Students
Before every conditional probability, have students circle the condition in the question and highlight that row or column of the table: that total is the denominator
Use counts first (count in both / count in B) and only then the decimal form P(A and B)/P(B)
Give sentence frames: "Among the ___, the fraction who ___ is ___."
For Advanced Students
Ask students to prove that if P(A | B) = P(A) and P(A), P(B) are positive, then P(B | A) = P(B)
Ask for a table of 100 people in which P(A | B) = 0.5 and P(B | A) = 0.2, and ask what that requires of P(A) and P(B)
Ask students to show that P(A | B) + P(Aᶜ | B) = 1 and to explain it with the zoomed-in square of Diagram 1
Assessment Guidance
What to Look For
Check the denominator first: a student who divides by the grand total has found P(A and B), not a conditional probability. Ask students to say every conditional probability as a sentence that starts with "Among"; a student who cannot say which group they are choosing from is likely to reverse P(A | B) and P(B | A). For independence, look for a comparison of a conditional probability with the matching unconditional one (P(A | B) with P(A)), not with some other number, and for a conclusion that says in words that the condition does or does not change the probability.
02
Classroom Activities
3 Activities
1
Shrink the Sample Space
20 minPairs
Pairs use 20 index cards numbered 1-20 to see a conditional probability physically: they remove every card outside the condition and count what is left.
Procedure
For each question, pairs first pull out the cards in the condition, set the rest aside, and count; then they check with P(A and B)/P(B) using all 20 cards
Questions and answers: P(even | greater than 12) = 4/8 = 1/2; P(greater than 12 | even) = 4/10 = 2/5; P(multiple of 3 | odd) = 3/10; P(prime | less than 10) = 4/9
Pairs then compare each of the first two answers with the unconditional probability: P(even) = 10/20 = 1/2 and P(greater than 12) = 8/20 = 2/5
Pairs write two sentences about what they notice
Discussion Questions
Why did the first two questions have different answers even though they use the same two events?
Knowing a card is greater than 12 did not change the chance that it is even. What does that tell you about the two events?
Is P(prime | less than 10) the same as P(prime)? What does that mean?
Modification for Distance Learning
Show the 20 numbers on a shared slide and let students gray out the numbers outside the condition before counting.
2
Which Probability Is It? Card Match
15 minGroups of 3-4
Groups match 5 description cards to 5 value cards for an invented set of 500 emails: 150 are spam, 100 contain the word "free", and 80 are spam and contain "free".
The 10 Cards
Description cards: "Of the emails that contain free, the share that are spam"; "Of the spam emails, the share that contain free"; "The share of all emails that are spam and contain free"; "The share of all emails that are spam"; "Of the emails without free, the share that are spam"
Groups first write each description in symbols, then match it to a value and show the division
Each group decides whether "spam" and "contains free" are independent and justifies the answer with a conditional probability
Discussion Questions
Why is P(spam | free) so much larger than P(free | spam)?
A filter that marks every email containing free as spam: what fraction of its marks would be right, based on these data?
Challenge Variation
Change one count so that spam and "contains free" become independent while the totals of 500 emails and 150 spam stay the same. (The number of emails with free that are spam must be 30% of the emails with free.)
3
Build an Independent Table
15 minPairs
Pairs fill in a 2-by-2 table so that two events are independent, then confirm both conditional statements of the standard.
Procedure
Scenario: an invented group of 200 phone owners, of whom 120 use a phone case (C) and 50 use a screen protector (S)
Pairs find the number who use both that makes C and S independent (120 · 50/200 = 30) and complete the table
Pairs check that P(S | C) = 30/120 = 0.25 = P(S) and P(C | S) = 30/50 = 0.6 = P(C)
Pairs change the "both" cell to 40, adjust the other cells to keep the totals, and show that both statements now fail
Discussion Questions
Why did both conditional statements fail at the same time when you changed the cell?
In the table with 40, does knowing that someone uses a case make a screen protector more or less likely?
Modification for Distance Learning
Share a spreadsheet with the totals fixed and one editable cell, with formulas that show P(S | C), P(S), P(C | S) and P(C) as the cell changes.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Conditional Probability as Zooming In on B
Drawn to scale with the numbers of Worked Examples 2-4. On the left, the unit square is the sample space; B is the column of width 0.4 and A is the dark band. Knowing that B happened throws away everything outside B, and on the right B is stretched to fill the square. The share of B covered by A is P(A and B)/P(B) = 0.3, the same as A's share of the whole square.
Diagram 2: Comparing P(B | grade) with P(B)
Bars drawn to scale from the table in Worked Example 5. If the events were independent, every conditional bar would end at the same place as the overall bar (0.5). Here knowing the grade changes the probability, so the events are not independent.
04
Homework Assignment
~30 min
HSS.CP.A.3 Homework: Conditional Probability and Independence
Directions: For each conditional probability, write it in symbols, show the division, and give one sentence of interpretation that starts with "Among". When you decide about independence, compare each conditional probability with the matching unconditional probability.
Part 1: Using the Definition (Problems 1-2)
For two events, P(A and B) = 0.24, P(A) = 0.4 and P(B) = 0.6. Find P(A | B) and P(B | A). Are A and B independent? Explain using both conditional probabilities.
A bag holds 15 marbles: 6 red (3 of them striped) and 9 blue (4 of them striped). One marble is drawn. Find P(striped | red), P(red | striped) and P(striped). Is "striped" independent of "red"?
Part 2: Conditional Probability in Data (Problems 3-4)
In an invented survey of 400 adults, 220 are under 40 and 132 of them exercise weekly, while 180 are 40 or older and 90 of them exercise weekly. (a) Make a two-way table. (b) Find P(exercise | under 40), P(exercise | 40 or older) and P(exercise). (c) Find P(under 40 | exercise). (d) Are age group and exercise independent for these adults?
In an invented lot of 250 cars, 50 are electric (E), 100 are white (W), and 20 are white electric cars. Find P(W | E), P(W), P(E | W) and P(E). Are E and W independent? Explain what that means about the cars.
Part 3: Interpreting Independence (Problems 5-6)
For two events, P(A) = 0.45, P(B) = 0.8 and P(B | A) = 0.8. (a) Find P(A and B). (b) Find P(A | B). (c) Explain in words why the results show that A and B are independent.
Two number cubes are rolled. Let A = "the first cube shows an even number" and B = "the sum is at least 10". Find P(B | A), P(B), P(A | B) and P(A). Are A and B independent? Explain what knowing A tells you about B.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Conditional Probability
Correct numerator and denominator every time
One denominator wrong
Denominators confused throughout
Order of Events
P(A | B) and P(B | A) kept apart and read correctly
One reversal
Order ignored
Independence
Both conditional comparisons made and concluded correctly
One comparison or the conclusion missing
No comparison
Interpretation
Clear "Among..." sentence for every answer
Some sentences missing or vague
No interpretation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work each question on paper, including the division, before you choose. Choosing an option shows the explanation, and Reset quiz clears the score so the quiz can be retaken.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which expression gives P(A | B), the probability of A given B?
Answer: C
The condition B goes in the denominator: P(A | B) = P(A and B)/P(B). Choice A divides by P(A), which gives P(B | A). Choice B is the product that equals P(A and B) only for independent events. Choice D is the reciprocal and can be greater than 1.
Question 2 of 20 · Multiple Choice
For two events, P(A and B) = 0.18 and P(B) = 0.45. What is P(A | B)?
Answer: D
P(A | B) = 0.18/0.45 = 0.4. Choice A multiplies instead of dividing. Choice B adds the two values, and choice C subtracts them.
Question 3 of 20 · Multiple Choice
For two events, P(A) = 0.5 and P(A and B) = 0.07. What is P(B | A)?
Answer: A
The condition is A, so P(B | A) = P(A and B)/P(A) = 0.07/0.5 = 0.14. Choice B multiplies 0.07 by 0.5. Choices C and D subtract or add the two values.
Question 4 of 20 · Multiple Choice
A number cube is rolled. Given that the number is odd, what is the probability that it is greater than 2?
Answer: B
The condition leaves {1, 3, 5}, and two of those (3 and 5) are greater than 2, so the answer is 2/3. By the formula, (2/6)/(3/6) = 2/3. Choice A is 2/6, the probability of "odd and greater than 2" among all six outcomes. Choice D multiplies instead of dividing.
Question 5 of 20 · Multiple Choice
For a school bus, L = "the bus is late" and W = "it rains". What does P(L | W) = 0.3 mean?
Answer: C
P(L | W) restricts attention to the days in W, the rainy days, and says 30% of them have a late bus. Choice A describes P(L and W). Choice B describes P(W | L), the reverse order. Choice D describes P(L).
Question 6 of 20 · Multiple Choice
For two events, P(A) = 0.35, P(B) = 0.7 and P(A | B) = 0.35. Which statement is true?
Answer: A
Since P(A | B) = P(A), knowing B does not change the chance of A, so the events are independent and P(B | A) = P(B) = 0.7. Choice B confuses independence with equal probabilities. Choice C forgets to multiply: P(A and B) = 0.35 · 0.7 = 0.245. Choice D copies P(A | B) instead of reversing the order.
Question 7 of 20 · Multiple Choice
For two events, P(A) = 0.8, P(B) = 0.25 and P(A and B) = 0.1. Which statement is correct?
Answer: A
P(A | B) = 0.1/0.25 = 0.4, and 0.4 ≠ P(A) = 0.8, so knowing B lowers the chance of A: not independent. Choice B computes 0.1/0.8, which is P(B | A). Choice C computes the product P(A) · P(B) instead of the conditional probability. Choice D assumes the answer instead of computing it.
Question 8 of 20 · Multiple Choice
One card is drawn from a standard 52-card deck. What is P(king | face card)? (The face cards are the jacks, queens and kings.)
Answer: C
There are 12 face cards and 4 of them are kings, so P(king | face card) = 4/12 = 1/3. By the formula, (4/52)/(12/52) = 1/3. Choice A gives P(king) for the whole deck, ignoring the condition. Choice B is P(face card | king), the reverse order: every king is a face card. Choice D counts only one king among the 12 face cards.
Question 9 of 20 · Multiple Choice
One card is drawn from a standard 52-card deck. What is P(face card | spade)?
Answer: B
Of the 13 spades, 3 are face cards (jack, queen, king), so P(face card | spade) = 3/13. Choice A is P(face card and spade), which divides by all 52 cards. Choice C is P(spade | face card) = 3/12, the reverse order. Choice D counts only one face card among the spades.
Question 10 of 20 · Multiple Choice
In an invented group of 120 students, 50 take chemistry and 20 of them also take physics. Of the 70 who do not take chemistry, 28 take physics. Which statement is true?
Answer: A
P(physics | chemistry) = 20/50 = 0.4, and P(physics) = 48/120 = 0.4, so knowing a student takes chemistry does not change the chance of physics. The other direction agrees: P(chemistry | physics) = 20/48 = 5/12 = 50/120. Choice B divides by the grand total. Choice C uses the chemistry total for a probability given physics. Choice D confuses independence with not overlapping.
Question 11 of 20 · Multiple Choice
At a restaurant, 200 customers were recorded: 90 ordered dessert, 100 ordered coffee, and 60 ordered both. What is P(dessert | coffee)?
Answer: C
Among the 100 coffee customers, 60 ordered dessert: P(dessert | coffee) = 60/100 = 0.6. Choice A is 60/90, which is P(coffee | dessert). Choice B is 60/200 = P(dessert and coffee), and choice D is 90/200 = P(dessert).
Question 12 of 20 · Multiple Choice
A and B are independent, P(A) = 0.7 and P(B) = 0.2. What is P(B | A)?
Answer: B
For independent events, knowing A does not change the probability of B, so P(B | A) = P(B) = 0.2. By the formula, (0.7 · 0.2)/0.7 = 0.2. Choice A is P(A and B). Choice C is P(A), and choice D divides 0.2 by 0.7.
Question 13 of 20 · Multiple Choice
For two events, P(A | B) = 0.5 and P(B) = 0.3. What is P(A and B)?
Answer: D
Multiply both sides of P(A | B) = P(A and B)/P(B) by P(B): P(A and B) = 0.5 · 0.3 = 0.15. Choice A divides 0.3 by 0.5, and choice C divides 0.5 by 0.3, which is not even a probability. Choice B adds the two numbers.
Question 14 of 20 · Multiple Choice
Which information shows that events A and B are NOT independent?
Answer: D
In choice D, knowing B raises the probability of A from 0.42 to 0.6, so the events are not independent. Choices A and B each show that the condition does not change a probability, which is what independence means. Choice C is the product definition of independence.
Question 15 of 20 · Short Answer
In an invented survey of 160 teens, 40 own a smartwatch (W), 60 run regularly (R), and 24 do both. Find P(R | W), P(R), P(W | R) and P(W). Are W and R independent? Interpret.
P(R | W) = 24/40 = 0.6 and P(R) = 60/160 = 0.375. P(W | R) = 24/60 = 0.4 and P(W) = 40/160 = 0.25. The conditional probabilities differ from the unconditional ones, so the events are not independent: among these teens, smartwatch owners are more likely to run than teens overall.
Question 16 of 20 · Short Answer
Two number cubes are rolled. Find P(the sum is 9 | the first cube shows 5) and P(the sum is 9). Then find P(the first cube shows 5 | the sum is 9). Are the two events independent?
Given a first cube of 5, the sum is 9 only when the second cube is 4: P = 1/6. The sum is 9 in 4 of 36 outcomes: (3, 6), (4, 5), (5, 4), (6, 3), so P(sum is 9) = 1/9. Given a sum of 9, one of those 4 outcomes has a first cube of 5: P = 1/4, but P(first cube shows 5) = 1/6. The events are not independent.
Question 17 of 20 · Short Answer
A report gives P(A | B) = 0.36 and P(B | A) = 0.72 for two events. Which event is more likely, A or B, and by what factor? Explain using the definition of conditional probability.
Both conditional probabilities have the same numerator: P(A and B) = 0.36 · P(B) and P(A and B) = 0.72 · P(A). So 0.36 · P(B) = 0.72 · P(A), which gives P(B) = 2 · P(A). B is twice as likely as A. The larger event gives the smaller conditional probability, because it is the bigger denominator.
Question 18 of 20 · Short Answer
For two events, P(A) = 0.9, P(B) = 0.5 and P(A and B) = 0.45. Use conditional probability to show, in two ways, that A and B are independent.
P(A | B) = 0.45/0.5 = 0.9 = P(A), so knowing B does not change the probability of A. P(B | A) = 0.45/0.9 = 0.5 = P(B), so knowing A does not change the probability of B. Both statements hold, so A and B are independent.
Question 19 of 20 · Short Answer
A card is drawn from 12 cards numbered 1-12. Let A = "even" and B = "a multiple of 3". Find P(A | B) and P(B | A), and decide whether A and B are independent.
B = {3, 6, 9, 12}, and two of these (6 and 12) are even: P(A | B) = 2/4 = 1/2, which equals P(A) = 6/12. A = {2, 4, 6, 8, 10, 12}, and two of these are multiples of 3: P(B | A) = 2/6 = 1/3, which equals P(B) = 4/12. A and B are independent.
Question 20 of 20 · Short Answer
For two events, P(A) = 0.2, P(B) = 0.75 and P(A | B) = 0.2. Find P(A and B) and P(B | A), and explain what the results say about A and B.
P(A and B) = P(A | B) · P(B) = 0.2 · 0.75 = 0.15.P(B | A) = 0.15/0.2 = 0.75, which equals P(B). Since P(A | B) = P(A) and P(B | A) = P(B), A and B are independent: knowing either one does not change the chance of the other.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.CP.A.3 mean?
HSS.CP.A.3 means students understand conditional probability and use it to describe independence. The conditional probability of A given B is P(A and B)/P(B). Two events are independent when P(A | B) = P(A) and P(B | A) = P(B), so knowing one event tells you nothing about the chance of the other.
What is the formula for conditional probability?
The formula is P(A | B) = P(A and B)/P(B), for P(B) > 0. With counts from a table or a list of equally likely outcomes, it becomes (number of outcomes in both A and B)/(number of outcomes in B).
How do you read P(A | B)?
Read it as "the probability of A given B". The event after the bar is the condition, the thing we already know happened. In a word problem, the condition usually follows the words "given that", "if" or "among".
Is P(A | B) the same as P(B | A)?
Usually not. They share the numerator P(A and B) but have different denominators. For one card from a deck, P(red | heart) = 1 because every heart is red, but P(heart | red) = 1/2. Mixing up the two directions is a frequent error in word problems.
How does conditional probability show that two events are independent?
A and B are independent when P(A | B) = P(A): learning that B happened does not change the probability of A. The same is then true in the other direction, P(B | A) = P(B). These statements follow from the product definition in HSS.CP.A.2, since P(A and B)/P(B) = P(A) · P(B)/P(B) = P(A).
Why does HSS.CP.A.3 mention both P(A | B) = P(A) and P(B | A) = P(B)?
Because independence is a two-way relationship. When both events have positive probability, one statement implies the other, so checking one is enough to decide. The standard names both so that students see that knowing A tells nothing about B, and knowing B tells nothing about A.
What is the difference between P(A and B) and P(A | B)?
P(A and B) is a fraction of the whole sample space: the share of all outcomes in both events. P(A | B) is a fraction of B only. In the email example of Activity 2, P(spam and free) = 0.16 of all emails, while P(spam | free) = 0.8 of the emails that contain "free".
What happens if P(B) = 0?
Then P(A | B) is not defined, because the formula would divide by zero. It also makes sense in words: you cannot ask what happens among the outcomes of B if B never happens.
Is conditional probability on the SAT?
Yes. The digital SAT's Problem-Solving and Data Analysis domain includes probability and conditional probability, usually asked with a two-way table, such as "given that a person is in group X, what is the probability that...". The key skill is choosing the right row or column total as the denominator.
What mistakes do students make with conditional probability?
Dividing by the grand total, which gives P(A and B) instead of P(A | B)
Reversing the order and computing P(B | A)
Deciding independence by comparing P(A | B) with P(B), or with P(A and B)
Assuming P(A | B) + P(B | A) = 1
07
Related Standards
6 standards
These standards connect to HSS.CP.A.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSS.CP.A.1Prerequisite
Describe events as subsets of a sample space, and with or, and, not
