HSS.CP.A.3 math practice. Learn by doing.

HSS.CP.A.3 practice covers understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B. Work through 8 free questions with answers and explanations, then continue in the related math games.

Grade 91 mapped skills
What it covers

Skills aligned to HSS.CP.A.3

Conditional probability

Grade 916 questions
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8 free questions · 0/0 correct

Practice playeasy

From a table, 6 are defective AND from line A and 15 total are items from line A. Find the conditional probability of defective given line A (simplified fraction).

P(A∣B)=P(A and B)P(B)=615=25P(A\mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{6}{15} = \dfrac{2}{5}.

Practice playeasy

From a table, 9 are on-time AND morning flights and 24 total are morning flights. Find the conditional probability of on-time given morning (simplified fraction).

P(A∣B)=P(A and B)P(B)=924=38P(A\mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{9}{24} = \dfrac{3}{8}.

Practice playeasy

From a table, 5 are members who renew AND are premium and 12 total are premium members. Find the conditional probability of renewing given premium (simplified fraction).

P(A∣B)=P(A and B)P(B)=512=512P(A\mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{5}{12} = \dfrac{5}{12}.

Practice playeasy

From a table, 10 are customers who buy AND saw the ad and 25 total are customers who saw the ad. Find the conditional probability of buying given saw the ad (simplified fraction).

P(A∣B)=P(A and B)P(B)=1025=25P(A\mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{10}{25} = \dfrac{2}{5}.

Practice playeasy

From a table, 4 are plants that flower AND got fertilizer and 16 total are plants that got fertilizer. Find the conditional probability of flowering given fertilized (simplified fraction).

P(A∣B)=P(A and B)P(B)=416=14P(A\mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{4}{16} = \dfrac{1}{4}.

Practice playeasy

From a table, 7 are games won AND played at home and 21 total are games played at home. Find the conditional probability of winning given a home game (simplified fraction).

P(A∣B)=P(A and B)P(B)=721=13P(A\mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{7}{21} = \dfrac{1}{3}.

Practice playeasy

In a class, 1212 students play soccer, and 99 of those soccer players also play chess. A soccer player is picked at random. What is the probability that they also play chess?

Condition on the soccer players: P(chess∣soccer)=912=34P(\text{chess}\mid\text{soccer}) = \dfrac{9}{12} = \dfrac{3}{4}.

Practice playeasy

Events AA and BB are independent, with P(A)=13P(A) = \dfrac{1}{3} and P(B)=12P(B) = \dfrac{1}{2}. What is P(A∣B)P(A \mid B)?

Because AA and BB are independent, knowing BB does not change the probability of AA: P(A∣B)=P(A)=13P(A\mid B) = P(A) = \dfrac{1}{3}.

Keep playing

Practice HSS.CP.A.3 inside a game.

The placement test chooses the right difficulty while every match keeps the standard in play.

Batting Cage game coverBatting CagePlay →Blossom Leap game coverBlossom LeapPlay →Buzzer Beater game coverBuzzer BeaterPlay →Minecart Junction game coverMinecart JunctionPlay →
Tests that assess these skills:
ACTATLASNAEPNY State MathSAT
Test reporting domains
SAT All grades: Problem-Solving and Data AnalysisACT All grades: Preparing for Higher MathMAP Growth Math 6+: Statistics and ProbabilityNAEP Grade 12: Data Analysis, Statistics, and ProbabilityACT Aspire All grades: Preparing for Higher MathATLAS Algebra I: Statistics and Probability
Nearby standards
HSN.RN.A.2HSS.IC.B.4All Common Core standards