Probability: Sample Spaces, Independence and Conditional Probability: Algebra 2 Math Worksheet Answer Key – PDF Download

Problem 1: Sample space of two actions

1. Sample space: \(\{H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6\}\). There are 12 outcomes.
2. The event is \(\{H2,H4,H6\}\), so \(P=\frac{3}{12}=\) \(\frac{1}{4}\).

Problem 2: Events as subsets

1. \(A=\{2,4,6\}\) and \(B=\{4,5,6\}\).
2. “\(A\) and \(B\)” \(=\{4,6\}\); “\(A\) or \(B\)” \(=\{2,4,5,6\}\); “not \(A\)” \(=\{1,3,5\}\).
3. \(P(A\text{ or }B)=\frac{4}{6}=\) \(\frac{2}{3}\).

Problem 3: Use the complement

1. \(1-0.35=\) \(0.65\)
2. \(1-\frac{3}{8}=\) \(\frac{5}{8}\)
3. The only outcome with no heads is TTT, with probability \(\frac{1}{8}\). So \(1-\frac{1}{8}=\) \(\frac{7}{8}\).

Problem 4: Cards from a deck

1. \(\frac{13}{52}=\) \(\frac{1}{4}\)
2. \(\frac{4}{52}=\) \(\frac{1}{13}\)
3. There are \(3\) face cards in each of \(2\) red suits: \(\frac{6}{52}=\) \(\frac{3}{26}\)
4. \(1-\frac{13}{52}=\frac{39}{52}=\) \(\frac{3}{4}\)

Problem 5: Apply the addition rule

1. \(0.5+0.3-0.1=\) \(0.7\)
2. \(\frac{3}{8}+\frac{4}{8}-\frac{1}{8}=\frac{6}{8}=\) \(\frac{3}{4}\)

Problem 6: Find the overlap

Rearrange the addition rule: \(P(A\text{ and }B)=P(A)+P(B)-P(A\text{ or }B)=0.5+0.6-0.8=\) \(0.3\).

Problem 7: Heart or king

\(P(\text{heart})=\frac{13}{52}\), \(P(\text{king})=\frac{4}{52}\), and the king of hearts is in both: \(\frac{1}{52}\).

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

Problem 8: Read a Venn diagram

Soccer total \(12+5=17\); art total \(8+5=13\).

1. \(\frac{17}{30}\): \(\frac{17}{30}\)
2. \(\frac{13}{30}\): \(\frac{13}{30}\)
3. \(\frac{5}{30}=\) \(\frac{1}{6}\)
4. \(\frac{17}{30}+\frac{13}{30}-\frac{5}{30}=\frac{25}{30}=\) \(\frac{5}{6}\)
5. \(\frac{5}{30}=\) \(\frac{1}{6}\)

Problem 9: Two dice

1. The sum is \(8\) for \((2,6),(3,5),(4,4),(5,3),(6,2)\): \(\frac{5}{36}\). \(\frac{5}{36}\)
2. Sums \(10\), \(11\), \(12\) occur \(3+2+1=6\) ways: \(\frac{6}{36}=\) \(\frac{1}{6}\)
3. Six doubles: \(\frac{6}{36}=\) \(\frac{1}{6}\)

Problem 10: Mutually exclusive events

1. They cannot happen together: \(0\)
2. \(0.25+0.4-0=\) \(0.65\)
3. No. Independence needs \(P(A\text{ and }B)=0.25\cdot 0.4=0.1\), but the intersection is \(0\). Not independent

Problem 11: Test independence by the product

1. \(0.4\cdot 0.5=0.2=P(A\text{ and }B)\): independent
2. \(0.6\cdot 0.5=0.30\), but \(P(A\text{ and }B)=0.35\): not independent

Problem 12: Area model

The regions come from products: \(0.6\cdot0.5=0.30\), \(0.6\cdot0.5=0.30\), \(0.4\cdot0.5=0.20\), \(0.4\cdot0.5=0.20\).

1. The violet region: \(0.30\)
2. \(0.30+0.30+0.20=\) \(0.80\) (or \(0.6+0.5-0.3\))
3. The gray region: \(0.20\)

Problem 13: Coin and die

1. \(\frac{1}{2}\cdot\frac{1}{6}=\) \(\frac{1}{12}\)
2. \(\frac{1}{2}\cdot\frac{3}{6}=\) \(\frac{1}{4}\)
3. Numbers \(5\) and \(6\): \(\frac{1}{2}\cdot\frac{2}{6}=\) \(\frac{1}{6}\)

Problem 14: Read a two-way table

1. \(\frac{50}{100}=\) \(\frac{1}{2}\)
2. \(\frac{54}{100}=\) \(\frac{27}{50}\)
3. \(\frac{30}{100}=\) \(\frac{3}{10}\)
4. \(\frac{50+54-30}{100}=\frac{74}{100}=\) \(\frac{37}{50}\)

Problem 15: Conditional from a table

1. Use the grade 9 column (\(50\) students): \(\frac{30}{50}=\) \(\frac{3}{5}\)
2. Use the pet row (\(54\) students): \(\frac{30}{54}=\) \(\frac{5}{9}\)
3. Grade 10 column: \(\frac{24}{50}=\) \(\frac{12}{25}\)

Problem 16: Conditional probability formula

1. \(\frac{0.12}{0.4}=\) \(0.3\)
2. \(\frac{0.12}{0.5}=\) \(0.24\)
3. \(P(A\text{ and }B)=P(A|B)P(B)=0.75\cdot 0.2=\) \(0.15\)

Problem 17: Read a tree diagram

1. \(\frac{3}{5}\cdot\frac{2}{4}=\frac{6}{20}=\) \(\frac{3}{10}\)
2. Green then yellow \(\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{10}\) plus yellow then green \(\frac{2}{5}\cdot\frac{3}{4}=\frac{3}{10}\): \(\frac{6}{10}=\) \(\frac{3}{5}\)
3. Read the second branch after green: \(\frac{2}{4}=\frac{1}{2}\)

Problem 18: Independence in a table

1. \(\frac{60}{100}=\) \(0.6\)
2. \(\frac{24}{40}=\) \(0.6\)
3. Yes. \(P(\text{soccer}\,|\,\text{grade }7)=P(\text{soccer})=0.6\). Also \(0.6\cdot 0.4=0.24=P(\text{soccer and grade }7)\). Independent

Problem 19: Free throws

1. \(0.7\cdot 0.7=\) \(0.49\)
2. \(0.3\cdot 0.3=\) \(0.09\)
3. Make then miss, or miss then make: \(2(0.7)(0.3)=\) \(0.42\) (check: \(0.49+0.09+0.42=1\))

Problem 20: Language and art classes

1. Both: \(20\); Spanish only \(50\); art only \(25\). \(\frac{70+45-20}{120}=\frac{95}{120}=\) \(\frac{19}{24}\)
2. \(\frac{120-95}{120}=\frac{25}{120}=\) \(\frac{5}{24}\)
3. \(\frac{20}{70}=\) \(\frac{2}{7}\) and \(\frac{20}{45}=\) \(\frac{4}{9}\)
4. \(P(\text{Spanish})\cdot P(\text{art})=\frac{7}{12}\cdot\frac{3}{8}=\frac{7}{32}\approx 0.219\), but \(P(\text{both})=\frac{1}{6}\approx 0.167\). Not independent

Problem 21: Flu shot study

1. \(\frac{12}{200}=\) \(0.06\)
2. \(\frac{45}{300}=\) \(0.15\)
3. Total who caught the flu: \(12+45=57\). \(\frac{12}{57}=\) \(\frac{4}{19}\)
4. No. The flu rate is \(0.06\) with a shot and \(0.15\) without, so the shot changes the probability. Not independent

Problem 22: Marbles without replacement

1. \(\frac{6}{10}\cdot\frac{5}{9}=\frac{30}{90}=\) \(\frac{1}{3}\)
2. \(\frac{6}{10}\cdot\frac{4}{9}=\frac{24}{90}=\) \(\frac{4}{15}\)
3. Both blue: \(\frac{4}{10}\cdot\frac{3}{9}=\frac{12}{90}\). Same color: \(\frac{30}{90}+\frac{12}{90}=\frac{42}{90}=\) \(\frac{7}{15}\)

Problem 23: Weekend weather

1. \(0.4\cdot 0.3=\) \(0.12\)
2. \(0.4+0.3-0.12=\) \(0.58\)
3. \(0.6\cdot 0.7=\) \(0.42\). Check: \(1-0.58=0.42\).

Problem 24: Defective parts

1. \(P(\text{defective})=0.6\cdot 0.02+0.4\cdot 0.05=0.012+0.020=\) \(0.032\)
2. \(P(A|\text{defective})=\frac{0.6\cdot 0.02}{0.032}=\frac{0.012}{0.032}=\) \(\frac{3}{8}=0.375\)

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