Ohio State Tests Geometry Probability. Practice it free.

Ohio's State Test Geometry end-of-course math subscore covering probability — understanding independence and conditional probability and using the rules of probability (Ohio's Learning Standards high-school Statistics and Probability conceptual category, Conditional Probability and the Rules of Probability). This Geometry reporting domain maps to 2 practice skills and 8 representative questions from the playable bank.

Geometry2 mapped skills
What the test measures

Probability skills

  1. Understand independence and conditional probability and use the rules of probability to compute probabilities of compound events

Standards basis

Ohio's Learning Standards (mathematics) — Common Core-derived (CCSS-M) — Ohio

How the Ohio State Tests reports it

Ohio's State Tests are Ohio-specific assessments built around Ohio's Learning Standards for Mathematics, which are CCSS-derived and retain the Common Core grade-level domains and (at high school) conceptual categories nearly verbatim. Math is tested in grades 3-8 and via end-of-course exams; the high-school graduation pathway uses Algebra I and Geometry, but students in an integrated sequence take Integrated Mathematics I and II in their place (these integrated courses are not given their own bands here — Integrated Math I draws on the same Algebra/Functions/Number & Quantity/Geometry/Statistics content as Algebra I plus early Geometry, and Integrated Math II adds the remaining Geometry/Functions/Probability content). The reporting categories below are the official math subscore categories Ohio publishes per test; because no public item-percentage blueprint was retrievable, the cluster-level Common Core domains those subscores cover are used (the standards' grade-level domains, which the reporting categories mirror).

Blueprint & weighting

Ohio publishes per-test math subscore (reporting) categories in its Subscore Definitions chart and raw-score subscale ranges, but no public per-category item-percentage blueprint was retrievable; the statistical summaries report the number of items per subscore by administration rather than a fixed blueprint weighting.

Try it now

8 free questions · 0/0 correct

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

A standard six-sided die is rolled twice. What is the probability of rolling a 66 on both rolls?

Each roll has a probability of 11 out of 6,6\textsf{,} or 16.\dfrac{1}{6}\textsf{.} The rolls are independent, so 16×16=136.\dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}\textsf{.}

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

A standard six-sided die is rolled, then a fair coin is flipped. What is the probability of rolling an even number and then getting heads?

Rolling an even number has a probability of 33 out of 6,6\textsf{,} or 12.\dfrac{1}{2}\textsf{.} Heads has a probability of 11 out of 2,2\textsf{,} or 12.\dfrac{1}{2}\textsf{.} The events are independent, so 12×12=14.\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\textsf{.}

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

A spinner has 44 equal sections and exactly 11 is red. You spin it once and flip a fair coin. What is the probability of landing on red and getting heads?

Red has a probability of 11 out of 4,4\textsf{,} or 14.\dfrac{1}{4}\textsf{.} Heads has a probability of 11 out of 2,2\textsf{,} or 12.\dfrac{1}{2}\textsf{.} The events are independent, so 14×12=18.\dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}\textsf{.}

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

A bag holds 22 red marbles and 33 blue marbles. You draw one marble, put it back, and draw again. What is the probability that both draws are red?

Each draw has a probability of 22 out of 5,5\textsf{,} or 25.\dfrac{2}{5}\textsf{.} The marble is replaced, so the draws are independent: 25×25=425.\dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25}\textsf{.}

Keep practicing

Turn probability into game time.

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