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

Question 1: Sample space of coin flips (2 points)

1. \(\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}\): 8 outcomes.
2. \(A=\{HHT,HTH,THH\}\), so \(P(A)=\) \(\frac{3}{8}\).

Scoring: 1 point for all 8 outcomes (no repeats or omissions); 1 point for the set \(A\) and the probability.

Question 2: Events from a set of numbers (2 points)

1. \(A\) and \(B\): \(\{6,12\}\). \(A\) or \(B\): \(\{2,3,4,6,8,9,10,12\}\).
2. \(P(A\text{ or }B)=\frac{8}{12}=\) \(\frac{2}{3}\). \(P(\text{not }A)=\frac{6}{12}=\) \(\frac{1}{2}\).

Scoring: 1 point for both lists; 1 point for both probabilities.

Question 3: Addition rule with a Venn diagram (2 points)

1. Choir \(11+6=17\), drama \(9+6=15\): \(P(\text{choir})=\) \(\frac{17}{40}\), \(P(\text{drama})=\) \(\frac{15}{40}=\frac{3}{8}\), \(P(\text{both})=\) \(\frac{6}{40}=\frac{3}{20}\).
2. \(\frac{17}{40}+\frac{15}{40}-\frac{6}{40}=\frac{26}{40}=\) \(\frac{13}{20}\). Neither: \(\frac{14}{40}=\) \(\frac{7}{20}\).

Scoring: 1 point for the three probabilities; 1 point for the union (with the subtraction shown) and the probability of neither.

Question 4: Independent or not (2 points)

1. \(P(A)P(B)=0.3\cdot 0.6=0.18=P(A\text{ and }B)\), so the events are independent.
2. \(P(A|B)=\frac{0.2}{0.6}=\) \(\frac{1}{3}\).

Scoring: 1 point for the product comparison and conclusion; 1 point for the correct conditional probability.

Question 5: Two-way table of lunches (3 points)

1. \(\frac{64}{120}=\) \(\frac{8}{15}\).
2. The packed lunch row has \(56\) students, \(32\) in 10th: \(\frac{32}{56}=\) \(\frac{4}{7}\).
3. \(P(\text{school}\,|\,\text{9th})=\frac{36}{60}=\frac{3}{5}=0.6\), but \(P(\text{school})=\frac{8}{15}\approx 0.533\). The values differ, so the events are not independent.

Scoring: 1 point per part.

Question 6: Pens without replacement (3 points)

1. \(\frac{4}{6}\cdot\frac{3}{5}=\frac{12}{30}=\) \(\frac{2}{5}\).
2. Black-white \(\frac{4}{6}\cdot\frac{2}{5}=\frac{8}{30}\) and white-black \(\frac{2}{6}\cdot\frac{4}{5}=\frac{8}{30}\): \(\frac{16}{30}=\) \(\frac{8}{15}\).
3. Second white: \(\frac{8}{30}+\frac{2}{30}=\frac{10}{30}=\frac{1}{3}\). Then \(\frac{8/30}{10/30}=\) \(\frac{4}{5}\).

Scoring: 1 point per part; in part c, award the point for \(\frac{4}{5}\) with the correct denominator \(\frac{1}{3}\) shown.

Question 7: Free throws in a row (3 points)

1. \(0.8\cdot 0.8=\) 0.64.
2. \(0.2\cdot 0.2=\) 0.04.
3. \(1-0.04=\) 0.96 (the complement of missing both).

Scoring: 1 point per part.

Question 8: Conditional probability formula (2 points)

1. \(P(C|D)=\frac{0.25}{0.60}=\) \(\frac{5}{12}\).
2. \(P(D|C)=\frac{0.25}{0.40}=\) 0.625.

Scoring: 1 point per part; the denominator must be the probability of the given event.

Question 9: At least one six (1 point)

\(P(\text{no six on one roll})=\frac{5}{6}\), so \(P(\text{no six in two rolls})=\frac{25}{36}\). Then \(1-\frac{25}{36}=\) \(\frac{11}{36}\).

Scoring: 1 point for \(\frac{11}{36}\) with the complement idea shown.

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