AP Statistics Conditional Probability and Independence Worksheet Solutions (PDF)

AP Statistics math worksheet solutions: Conditional Probability and Independence – Maths PDF

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This conditional probability answers for AP Statistics covers conditional probability and independence (Common Core AP STAT 4.5, AP STAT 4.6, HSS-CP.A.2, HSS-CP.A.3, HSS-CP.A.4, HSS-CP.B.6, HSS-CP.B.8), with a free PDF to print.

Each solution identifies the condition, shows the restricted denominator or the branch multiplication, and gives an exact fraction with a decimal. Independence checks compare P(A given B) with P(A).

AP Statistics · Common Core State Standards: AP STAT 4.5, AP STAT 4.6, HSS-CP.A.2, HSS-CP.A.3, HSS-CP.A.4, HSS-CP.B.6, HSS-CP.B.8

Conditional probability answers: exercises 1 to 10

Exercise 1 – Conditional from a table

200 adults by age group and daily streaming.

First: \(\frac{48}{80}=0.60\). Second: \(\frac{48}{70}\approx 0.686\).

0.60 and about 0.686; the conditions are different.

Exercise 2 – Use the formula

\(P(A\mid B)=\frac{0.2}{0.4}=0.5\). \(P(B\mid A)=\frac{0.2}{0.5}=0.4\).

0.5 and 0.4. Note both equal the unconditional probabilities.

Exercise 3 – General multiplication rule

\(P(\text{1st red})=\frac{6}{10}\). \(P(\text{2nd red}\mid\text{1st red})=\frac{5}{9}\).

\(\frac{6}{10}\times\frac{5}{9}=\frac{30}{90}=\frac{1}{3}\).

1/3, about 0.333.

Exercise 4 – Tree diagram

Tree diagram for a medical test.

\(P(+)=0.02(0.95)+0.98(0.04)=0.0190+0.0392=0.0582\).

0.0582.

Exercise 5 – Reverse conditional

\(P(D\mid +)=\frac{0.0190}{0.0582}\approx 0.326\).

About 32.6%. Most positives are false positives because the disease is rare.

Exercise 6 – Independence check

\(P(A)P(B)=0.30\times 0.50=0.15\), which equals P(A and B).

Yes, A and B are independent.

Exercise 7 – Independent events

\(P(A\text{ and }B)=0.6\times 0.25=0.15\). \(P(A\text{ or }B)=0.6+0.25-0.15=0.70\).

0.15 and 0.70.

Exercise 8 – Exclusive vs independent

\(P(A\text{ and }B)=0\) but \(P(A)P(B)=0.12\).

They are not independent: knowing A occurred means B cannot occur.

Exercise 9 – At least one

Miss all three: \(0.2^{3}=0.008\). At least one: \(1-0.008=0.992\).

0.992.

Exercise 10 – Table independence

Tutoring and passing.

\(P(\text{passed})=\frac{50}{100}=0.5\). \(P(\text{passed}\mid\text{tutored})=\frac{30}{40}=0.75\).

Not independent: the tutored students were more likely to pass (0.75 compared with 0.5).

Exercises 11 and up

Exercise 11 – Without replacement

\(\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}=\frac{1716}{132600}=\frac{11}{850}\).

11/850, about 0.0129.

Exercise 12 – Interpret a conditional

On rainy days 35% of buses are late; overall 12% are late.

Not independent: knowing that it rains changes the probability of being late.

Exercise 13 – Pets

Pet ownership.

Cat: 30 of 100, so 0.30. Dog: 40 of 100, so 0.40. Both: 12 of 100, so 0.12.

\(0.30\times 0.40=0.12\).

0.12 equals P(cat and dog), so they are independent.

Exercise 14 – Conditional with a tree

\(P(\text{defective})=0.6(0.02)+0.4(0.05)=0.012+0.020=0.032\).

\(P(B\mid\text{defective})=\frac{0.020}{0.032}=0.625\).

0.032 and 0.625.

Exercise 15 – Find P(B | A)

\(P(B\mid A)=\frac{0.18}{0.45}=0.4\).

P(B | A) = 0.4 = P(B), so A and B are independent.

Exercise 16 – Quality check

Defective then good: \(\frac{3}{10}\times\frac{7}{9}=\frac{21}{90}\). Good then defective: \(\frac{7}{10}\times\frac{3}{9}=\frac{21}{90}\).

Sum: \(\frac{42}{90}=\frac{7}{15}\).

7/15, about 0.467.

Exercise 17 – Two tests

\(0.9(0.8)+0.1(0.3)=0.72+0.03=0.75\).

0.75.

Exercise 18 – Is it independent?

\(P(\text{junior})=\frac{15}{30}=0.5\). \(P(\text{junior}\mid\text{band})=\frac{9}{12}=0.75\).

Not independent: 0.75 is not 0.5.

Exercise 19 – Multi-step

Urgent: \(0.4(0.10)+0.6(0.02)=0.04+0.012=0.052\).

\(P(\text{work}\mid\text{urgent})=\frac{0.04}{0.052}\approx 0.769\).

About 0.769.

Exercise 20 – Draw a tree

Left-handed: \(0.4(0.30)+0.6(0.10)=0.12+0.06=0.18\).

\(P(\text{guard}\mid\text{left})=\frac{0.12}{0.18}\approx 0.667\).

0.18 and about 0.667.

Key facts behind this conditional probability answers

Key facts

P(A | B) = P(A and B) / P(B): restrict to the outcomes where B occurred.
General multiplication rule: P(A and B) = P(B) P(A | B).
Independent: P(A | B) = P(A), so P(A and B) = P(A) P(B).
Mutually exclusive events with positive probabilities are never independent.
Tree diagrams multiply along branches and add the paths that fit the event.

How to use this conditional probability answers

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.

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