Algebra 2 Conditional Probability and Independence Test Solutions and Scoring (PDF)

Algebra 2 math test solutions: Conditional Probability and Independence – Maths PDF

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This probability test answers for Algebra 2 covers conditional probability and independence (Common Core S-CP.A.1, S-CP.A.2, S-CP.A.3, S-CP.A.4, S-CP.A.5, S-CP.B.6, S-CP.B.7, S-CP.B.8, S-CP.B.9), with a free PDF to print.

Use this answer key and scoring guide to grade the test on conditional probability and independence (S-CP.A.1). Each solution shows the full reasoning, and the scoring note explains how many points to award for correct setup, correct work, and a correct final answer. Partial credit rewards a sound method even when a small arithmetic slip changes the result.

Algebra 2 · Common Core State Standards: S-CP.A.1, S-CP.A.2, S-CP.A.3, S-CP.A.4, S-CP.A.5, S-CP.B.6, S-CP.B.7, S-CP.B.8, S-CP.B.9

Probability test answers: worked solutions

Question 1 – Sample space [2 pts]

Sums 10, 11, 12: 3 + 2 + 1 = 6 outcomes.

\(\frac{6}{36} = \frac{1}{6}\).

\(\frac{1}{6}\)

Scoring: 1 point for counting the outcomes, 1 point for the probability.

Question 2 – Union [2 pts]

Soccer or track: \(10 + 4 + 6 = 20\) students. \(\frac{20}{30} = \frac{2}{3}\).

\(\frac{2}{3}\)

Scoring: 1 point for counting 20 students, 1 point for the probability.

Question 3 – Two-way table [3 pts]

Totals.

\(P(\text{bike}) = \frac{60}{100} = 0.6\). \(P(\text{bike} | \text{girl}) = \frac{36}{50} = 0.72\).

The probabilities differ, so the events are not independent.

0.6; 0.72; not independent

Scoring: 1 point for each probability, 1 point for the conclusion (3 points total: 1 + 1 + 1).

Question 4 – Independence check [2 pts]

\(0.6 \cdot 0.5 = 0.30 \ne 0.35\).

No, not independent

Scoring: 1 point for the product, 1 point for the conclusion.

Question 5 – Tree diagram [3 pts]

Two draws from a bag of 5 red (R) and 3 blue (B) marbles, without replacement.

Red then blue: \(\frac{5}{8} \cdot \frac{3}{7} = \frac{15}{56}\). Blue then red: \(\frac{3}{8} \cdot \frac{5}{7} = \frac{15}{56}\).

\(\frac{15}{56} + \frac{15}{56} = \frac{15}{28}\).

\(\frac{15}{28}\)

Scoring: 1 point for each branch, 1 point for the sum.

Question 6 – Counting [2 pts]

\(P(7, 3) = 7 \cdot 6 \cdot 5 = 210\)
\(C(7, 3) = \frac{210}{6} = 35\)

(a) 210; (b) 35

Scoring: 1 point for each part.

Question 7 – At least one [2 pts]

\(P(\text{no heads}) = \left(\frac{1}{2}\right)^{4} = \frac{1}{16}\). \(1 – \frac{1}{16} = \frac{15}{16}\).

\(\frac{15}{16}\)

Scoring: 1 point for the complement, 1 point for the answer.

Question 8 – Conditional probability [2 pts]

\(\frac{30}{80} = \frac{3}{8}\).

\(\frac{3}{8}\)

Scoring: 1 point for restricting to the 80 coffee buyers, 1 point for the answer.

Question 9 – Find the error [2 pts]

After the first card is removed, the number of cards left and the chances for the second card change, so the probability of the second event depends on the first. They are dependent.

Without replacement the events are dependent

Scoring: 1 point for stating they are dependent, 1 point for the reason.

Total score

Total

20 points

Key facts behind this probability test answers

Key facts

P(not A) = 1 – P(A); P(A or B) = P(A) + P(B) – P(A and B).
P(A given B) = P(A and B) / P(B).
Independent: P(A and B) = P(A)·P(B).
Multiplication rule: P(A and B) = P(A)·P(B given A).
Permutations P(n, r) = n!/(n – r)!; combinations C(n, r) = n!/(r!(n – r)!).

How to use this probability test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

Download the probability test answers as a PDF with the button below and keep it with your notes.

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