Expected Value and Decisions: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Expected Value Worksheet Answer Key: problems 1–5

Problem 1: Valid distributions

1. The sum is \(0.20+0.35+0.15+0.30=1.00\) and every value is between 0 and 1. Valid.
2. The sum is \(1.10\), which is greater than 1. Not valid.
3. One probability is negative (\(-0.10\)). Not valid, even though the sum is \(1\).
4. \(\frac{2}{8}+\frac{4}{8}+\frac{1}{8}+\frac{1}{8}=\frac{8}{8}=1\). Valid.

Problem 2: Find the missing probability

1. \(P(3)=1-0.10-0.30-0.20=0.40\). \(P(3)=0.40\)
2. \(P(X \ge 3)=P(3)+P(4)=0.40+0.20=0.60\). \(0.60\)

Problem 3: Distribution for three coin flips

1. HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
2. Count heads: \(X=0\) once, \(X=1\) three times , \(X=2\) three times , \(X=3\) once. So \(P(0)=\frac{1}{8}\), \(P(1)=\frac{3}{8}\), \(P(2)=\frac{3}{8}\), \(P(3)=\frac{1}{8}\) (sum \(\frac{8}{8}=1\)).
3. \(P(X \ge 2)=\frac{3}{8}+\frac{1}{8}=\frac{4}{8}=\frac{1}{2}\). \(\frac{1}{2}\)

Problem 4: Spinner prize distribution

1. Count sections: $0 appears 2 times , $1 appears 3 times , $2 appears 2 times , $5 appears 1 time. So \(P(0)=\frac{2}{8}=0.25\), \(P(1)=\frac{3}{8}=0.375\), \(P(2)=\frac{2}{8}=0.25\), \(P(5)=\frac{1}{8}=0.125\). The total is \(1\).
2. \(E=0(0.25)+1(0.375)+2(0.25)+5(0.125)=0+0.375+0.5+0.625=1.5\). The expected prize is $1.50.

Problem 5: Empirical distribution from a survey

1. Divide each count by 40: \(P(0)=\frac{8}{40}=0.20\), \(P(1)=\frac{16}{40}=0.40\), \(P(2)=\frac{10}{40}=0.25\), \(P(3)=\frac{6}{40}=0.15\). They add to \(1\).
2. \(P(X \ge 2)=0.25+0.15=0.40\). \(0.40\)

Precalculus Expected Value Worksheet Answer Key: problems 6–10

Problem 6: Expected value from a table

\(E(X)=0(0.20)+1(0.40)+2(0.25)+3(0.15)=0+0.40+0.50+0.45=1.35\).

\(E(X)=1.35\). In a large group of students chosen this way, the average number of siblings is about 1.35, although no student has exactly 1.35 siblings.

Problem 7: Fair die and unfair die

1. \(E=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=3.5\). \(3.5\)
2. The probabilities are \(P(1)=\frac{2}{6}\), \(P(2)=\frac{1}{6}\), \(P(3)=\frac{2}{6}\), \(P(6)=\frac{1}{6}\). \(E=\frac{2(1)+1(2)+2(3)+1(6)}{6}=\frac{16}{6}=\frac{8}{3}\approx 2.67\). \(\frac{8}{3}\)

Problem 8: Raffle tickets

1. Total prizes: \(50+3(10)=80\) dollars over 200 tickets. \(E=\frac{80}{200}=0.40\). $0.40
2. \(0.40-1=-0.60\). A loss of $0.60 per ticket.
3. \(25(-0.60)=-15\). A loss of $15.

Problem 9: Farmers market stall

1. The probabilities add to \(1\). \(E=0(0.10)+1(0.20)+2(0.35)+3(0.25)+4(0.10)=0+0.20+0.70+0.75+0.40=2.05\). 2.05 crates
2. \(2.05 \times 12=24.60\). $24.60 per hour

Problem 10: Bike insurance

1. \(E=0.05(400)+0.95(0)=20\). $20
2. The insurance costs $25, which is $5 more than the expected loss. By expected value, skipping the insurance is better by $5.
3. Losing $400 at once might be hard to afford, so some people pay a little extra to avoid that risk.

Precalculus Expected Value Worksheet Answer Key: problems 11–15

Problem 11: Is the dice game fair?

1. Expected winnings: \(3.5\). Net: \(3.5-4=-0.5\). A loss of $0.50 per game.
2. The game is fair when the fee equals the expected winnings. $3.50

Problem 12: Fair price of a game

1. \(1-0.1-0.4=0.5\). \(0.5\)
2. \(E=10(0.1)+2(0.4)+0(0.5)=1+0.8=1.8\). $1.80

Problem 13: Betting on a roulette wheel

1. \(E=35(\frac{1}{38})+(-1)(\frac{37}{38})=\frac{35-37}{38}=-\frac{2}{38}=-\frac{1}{19}\approx -0.053\). About a $0.053 loss per bet.
2. \(1000 \times \frac{1}{19}\approx 52.63\). An expected loss of about $52.63.

Problem 14: Two free throws

1. Multiply along the branches: MM: \(0.7 \times 0.7=0.49\); Make then Miss: \(0.7 \times 0.3=0.21\); Miss then Make: \(0.21\); Miss, Miss: \(0.3 \times 0.3=0.09\).
2. \(P(0)=0.09\), \(P(1)=0.21+0.21=0.42\), \(P(2)=0.49\). The total is \(1\).
3. \(E=0(0.09)+1(0.42)+2(0.49)=0.42+0.98=1.40\). 1.4 made shots (which equals \(2 \times 0.7\)).

Problem 15: Guessing on a quiz

1. \(E=\frac{1}{4}(4)+\frac{3}{4}(-1)=1-0.75=0.25\). 0.25 points
2. Let the penalty be \(p\). \(\frac{1}{4}(4)-\frac{3}{4}p=0\), so \(1=\frac{3}{4}p\) and \(p=\frac{4}{3}\). A penalty of \(\frac{4}{3}\) points.

Precalculus Expected Value Worksheet Answer Key: problems 16–20

Problem 16: Choosing a lemonade stand spot

1. Spot A: \(0.6(80)+0.4(20)=48+8=56\). Spot B: \(0.6(70)+0.4(40)=42+16=58\).
2. Spot B, because $58 is greater than $56. Spot B also has the smaller gap between its best and worst day ($30 versus $60).

Problem 17: Tablet warranty

1. \(0.12(250)=30\). $30
2. No. The warranty costs $50, which is $20 more than the expected repair cost.
3. Solve \(p(250)=50\): \(p=\frac{50}{250}=0.20\). \(20\%\)

Problem 18: Laptop warranty break-even

1. \(0.18(600)=108\). The warranty costs $12 more than the expected repair cost.
2. \(p(600)=120\) gives \(p=\frac{120}{600}=0.20\). \(20\%\)
3. \(0.18(900)=162\), which is more than $120. Yes, the expected loss is $42 greater than the warranty price.

Problem 19: Insurance company profit

1. \(0.01(25000)=250\). $250
2. \(450-250=200\). $200
3. \(10000 \times 200=2000000\). $2,000,000

Problem 20: School fundraiser lottery

1. Total prizes: \(400+4(50)+10(10)=700\). \(E=\frac{700}{300}\approx 2.33\). $2.33
2. \(2.33-5\approx -2.67\). About a $2.67 loss per ticket.
3. Ticket sales: \(300 \times 5=1500\). \(1500-700=800\). $800

Precalculus Expected Value Worksheet Answer Key: problems 21–24

Problem 21: Empirical versus theoretical

1. \(E=\frac{1+2+3+4}{4}=2.5\). 2.5
2. \(\frac{11(1)+14(2)+12(3)+13(4)}{50}=\frac{11+28+36+52}{50}=\frac{127}{50}=2.54\). 2.54
3. The empirical result comes from only 50 spins and varies by chance. With many more spins, the mean should settle near 2.5.

Problem 22: Drawing two marbles

1. There are \(10\) equally likely pairs. Both blue: 1 pair, so \(P(0)=\frac{1}{10}\). One of each: \(3 \times 2=6\) pairs, so \(P(1)=\frac{6}{10}\). Both red: 3 pairs, so \(P(2)=\frac{3}{10}\). The total is \(1\).
2. \(E=0(\frac{1}{10})+1(\frac{6}{10})+2(\frac{3}{10})=\frac{12}{10}=1.2\). 1.2 red marbles

Problem 23: A risky investment

1. \(E=0.30(40000)+0.50(5000)+0.20(-10000)=12000+2500-2000=12500\). $12,500
2. \(12500-3000=9500\). The project has an expected gain $9,500 greater than the savings plan.
3. There is a \(20\%\) chance of losing $10,000, while the savings plan has no chance of a loss.

Problem 24: Food truck decision

1. Net results: \(900-200=700\), \(500-200=300\), \(200-200=0\). \(E=0.4(700)+0.4(300)+0.2(0)=280+120+0=400\). $400
2. The regular spot nets $350, so the festival is better by \(400-350=50\) dollars on average.
3. \(0.2\), the chance of the $200 revenue.

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