Grade 9 · Statistics & probability

Conditional probability practice

Conditional probability is a grade 9 math skill aligned to Common Core standard HSS.CP.A.3: understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 16 conditional probability problems our math games drill.

CCSS HSS.CP.A.316 questions in the bank
Sample questions

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Question 1easy

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}.

Question 2easy

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}.

Question 3easy

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}.

Question 4easy

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}.

Question 5easy

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}.

Question 6easy

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}.

Question 7easy

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}.

Question 8easy

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}.

Tests that cover conditional probability:
NAEPACTSATNY State MathATLAS
More Grade 9 skills:
Absolute value equationsCommon factor factoring (SAT)Interpreting data displaysCompound inequality situationsProblem solving in context (SAT)Discriminant analysisStatistics basicsEquivalent rational expressions (SAT)All Grade 9 practice →

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