Algebra 2 Conditional Probability and Independence Worksheet Solutions (PDF)

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

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This probability practice solutions 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.

Every worksheet problem on conditional probability and independence is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where treating dependent events as independent can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

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 practice solutions: exercises 1 to 10

Exercise 1 – Sample space

Outcomes: HH, HT, TH, TT. Exactly one head: HT and TH.

\(\frac{2}{4} = \frac{1}{2}\)

Exercise 2 – Two dice

All 36 outcomes for two dice; the sums of 7 are highlighted.

Six outcomes have sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

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

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

Exercise 3 – Complement

\(1 – \frac{1}{6} = \frac{5}{6}\).

\(\frac{5}{6}\)

Exercise 4 – Union of events

\(P(\text{heart}) = \frac{13}{52}\), \(P(\text{king}) = \frac{4}{52}\), \(P(\text{king of hearts}) = \frac{1}{52}\).

\(\frac{13 + 4 – 1}{52} = \frac{16}{52} = \frac{4}{13}\).

\(\frac{4}{13}\)

Exercise 5 – Mutually exclusive

Multiples of 5: 5, 10, 15, 20 (4). Multiples of 7: 7, 14 (2). No overlap.

\(\frac{4 + 2}{20} = \frac{3}{10}\).

\(\frac{3}{10}\)

Exercise 6 – Independent events

The events are independent: \(\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\).

\(\frac{1}{4}\)

Exercise 7 – Test independence

\(P(A) \cdot P(B) = 0.4 \cdot 0.5 = 0.2\), which equals \(P(A \text{ and } B)\).

Yes, they are independent

Exercise 8 – Read a two-way table

Row and column totals.

Plays a sport: \(\frac{50}{200}\). Grade 11: \(\frac{120}{200}\).

\(\frac{1}{4}\) and \(\frac{3}{5}\)

Exercise 9 – Conditional probability from a table

Restrict to Grade 11: 120 students, of whom 20 play a sport.

\(\frac{20}{120} = \frac{1}{6}\).

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

Exercise 10 – Independence in a table

\(P(\text{sport}) = \frac{50}{200} = \frac{1}{4}\) but \(P(\text{sport} | \text{Grade } 11) = \frac{1}{6}\).

The probabilities differ.

Not independent

Exercises 11 and up

Exercise 11 – Venn diagram

13 + 5 + 7 = 25 in band or choir.

\(P(\text{band or choir}) = \frac{18}{30} + \frac{12}{30} – \frac{5}{30} = \frac{25}{30}\).

\(\frac{5}{6}\)

Exercise 12 – Without replacement

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

\(P(\text{first red}) = \frac{5}{8}\). Then 4 red remain among 7: \(P(\text{second red} | \text{first red}) = \frac{4}{7}\).

\(\frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14}\).

\(\frac{5}{14}\)

Exercise 13 – With replacement

The draws are independent: \(\frac{5}{8} \cdot \frac{5}{8} = \frac{25}{64}\).

\(\frac{25}{64}\)

Exercise 14 – One of each color

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

Add: \(\frac{30}{56} = \frac{15}{28}\).

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

Exercise 15 – Permutations

Order matters: \(P(5, 3) = 5 \cdot 4 \cdot 3 = 60\).

Exercise 16 – Combinations

Order does not matter: \(C(8, 3) = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56\).

Exercise 17 – Committee probability

Favorable: \(C(4, 2) \cdot C(6, 1) = 6 \cdot 6 = 36\). Total: \(C(10, 3) = 120\).

\(\frac{36}{120} = \frac{3}{10}\).

\(\frac{3}{10}\)

Exercise 18 – Medical test

Positives: \(95 + 495 = 590\). Of them, 95 have the disease.

\(\frac{95}{590} \approx 0.161\).

About 16%, even though the test is accurate

Exercise 19 – At least one

Complement: no 6 in three rolls: \(\left(\frac{5}{6}\right)^{3} = \frac{125}{216}\).

\(1 – \frac{125}{216} = \frac{91}{216}\).

\(\frac{91}{216}\)

Exercise 20 – Find the error

Mutually exclusive means \(P(A \text{ and } B) = 0\). Independent would need \(0.3 \cdot 0.4 = 0.12\). Since \(0 \ne 0.12\), the events are not independent.

Mutually exclusive events with positive probabilities are not independent

Exercise 21 – Addition rule

\(0.7 = 0.5 + 0.4 – P(A \text{ and } B)\), so \(P(A \text{ and } B) = 0.2\).

0.2

Key facts behind this probability practice solutions

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 practice solutions

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the probability practice solutions as a PDF with the button below and keep it with your notes.

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