This ap statistics probability worksheet answer key solves all 24 problems of the probability and random variables worksheet step by step.
Complete worked solutions for all 24 problems follow, with the formula, the substitution and the final answer in bold. Keep unrounded values until the last step, then round to the accuracy asked. For interpretation problems, write the answer in the context of the situation.
AP Statistics Probability Worksheet Answer Key: problems 1–5
Problem 1: Complete the table
Add across each row and down each column. Rows: \(20+70+60=150\) and \(30+55+15=100\). Columns: \(20+30=50\), \(70+55=125\), \(60+15=75\). Grand total: \(150+100=250\) (also \(50+125+75=250\)).
Problem 2: Joint relative frequency
\(\frac{55}{250}=0.22\). 22% of all shoppers are 40 and over and pay by card.
Problem 3: Marginal relative frequency
\(\frac{75}{250}=0.30\). 30% of all shoppers pay by mobile payment.
Problem 4: Conditional relative frequency
The given group is the 150 shoppers under 40: \(\frac{70}{150}\approx 0.467\). About 46.7% of shoppers under 40 pay by card.
Problem 5: Reverse condition
The given group is the 50 cash shoppers: \(\frac{30}{50}=0.60\). 60% of cash shoppers are 40 and over.
The condition changes the denominator (50 instead of 150), so the two conditional proportions answer different questions.
AP Statistics Probability Worksheet Answer Key: problems 6–10
Problem 6: Compare conditional distributions
Under 40: cash \(\frac{20}{150}\approx 13.3\%\), card \(\frac{70}{150}\approx 46.7\%\), mobile \(\frac{60}{150}=40.0\%\). 40 and over: cash \(\frac{30}{100}=30\%\), card \(\frac{55}{100}=55\%\), mobile \(\frac{15}{100}=15\%\).
The distributions differ a lot (mobile is 40% versus 15%), so payment method is associated with age group.
Problem 7: Complement rule
\(P(\text{not defective})=1-0.04=0.96\). 0.96
Problem 8: Spinner
Red and blue cannot happen on the same spin, so they are mutually exclusive: \(P(\text{red or blue})=\frac{3}{8}+\frac{2}{8}=\frac{5}{8}=0.625\).
\(P(\text{not green})=1-\frac{2}{8}=\frac{6}{8}=0.75\).
0.625 and 0.75; yes, mutually exclusive.
Problem 9: Venn diagram
Spanish only: \(48-15=33\). Band only: \(36-15=21\). Neither: \(120-(33+15+21)=51\).
\(P(\text{neither})=\frac{51}{120}=0.425\). 0.425
Problem 10: General addition rule
\(P(A\text{ or }B)=0.5+0.4-0.2=0.7\). Because \(P(A\text{ and }B)=0.2\neq 0\), the events can happen together and are not mutually exclusive.
AP Statistics Probability Worksheet Answer Key: problems 11–15
Problem 11: Conditional probability
\(P(\text{honor roll}|\text{job})=\frac{0.18}{0.45}=0.40\). 0.40
Problem 12: Without replacement
\(P(\text{both red})=\frac{5}{8}\cdot\frac{4}{7}=\frac{20}{56}=\frac{5}{14}\approx 0.357\).
\(P(\text{red, then blue})=\frac{5}{8}\cdot\frac{3}{7}=\frac{15}{56}\approx 0.268\).
Problem 13: Test independence
Overall, \(P(\text{card})=\frac{36+84}{200}=0.60\). Given 9th grade, \(P(\text{card}|\text{9th})=\frac{36}{60}=0.60\).
The two probabilities are equal (and \(P(\text{10th}|\text{card})=\frac{84}{120}=0.70=P(\text{10th})\)), so the events are independent.
Problem 14: Independent events
\(P(A\text{ and }B)=0.3\times 0.6=0.18\). \(P(A\text{ or }B)=0.3+0.6-0.18=0.72\). 0.18 and 0.72
Problem 15: Tree diagram
Multiply along each branch: \(0.6\times 0.02=0.012\), \(0.6\times 0.98=0.588\), \(0.4\times 0.05=0.020\), \(0.4\times 0.95=0.380\). These sum to 1.
\(P(\text{defective})=0.012+0.020=0.032\). 0.032
AP Statistics Probability Worksheet Answer Key: problems 16–20
Problem 16: Simulation design
Let digits 0 to 6 be a make (7 of 10 digits = 70%) and 7, 8, 9 a miss. Read three digits per trial and count the makes. Repeat for many trials.
\(\frac{8+8}{20}=0.80\). The estimate is 0.80.
Problem 17: Valid distribution
Probabilities sum to 1: \(0.1+0.3+k+0.2=1\), so \(k=0.4\).
\(P(X\ge 2)=0.4+0.2=0.6\). \(k=0.4\) and \(0.6\)
Problem 18: Mean of a random variable
\(\mu_X=0(0.2)+1(0.5)+2(0.2)+3(0.1)=0+0.5+0.4+0.3=1.2\). 1.2 pets
Problem 19: Standard deviation of a random variable
\(\sigma_X^{2}=(0-1.2)^{2}(0.2)+(1-1.2)^{2}(0.5)+(2-1.2)^{2}(0.2)+(3-1.2)^{2}(0.1)\)
\(=0.288+0.020+0.128+0.324=0.76\), so \(\sigma_X=\sqrt{0.76}\approx 0.87\). About 0.87 pets
Problem 20: Raffle
\(\mu_X=300\cdot\frac{1}{200}+50\cdot\frac{2}{200}+10\cdot\frac{4}{200}+0\cdot\frac{193}{200}=1.50+0.50+0.20=2.20\) dollars.
Net gain: \(2.20-5=-\$2.80\).
The mean prize is $2.20, and the mean net gain is −$2.80. Over many tickets, a buyer loses an average of $2.80 per ticket, which is how the school raises money.
AP Statistics Probability Worksheet Answer Key: problems 21–24
Problem 21: Claw machine
\(\mu_X=0+0.30+0.60+0.45+0.40=1.75\).
\(\sigma_X^{2}=(1.75)^{2}(0.15)+(0.75)^{2}(0.30)+(0.25)^{2}(0.30)+(1.25)^{2}(0.15)+(2.25)^{2}(0.10)\)
\(=0.459375+0.16875+0.01875+0.234375+0.50625=1.3875\), so \(\sigma_X=\sqrt{1.3875}\approx 1.18\).
Within one SD: \(1.75\pm 1.18\) gives the interval from 0.57 to 2.93, which contains 1 and 2: \(0.30+0.30=0.60\). \(\mu=1.75\), \(\sigma\approx 1.18\), probability 0.60
Problem 22: Die game
\(P(9)=\frac{1}{6},\ P(3)=\frac{2}{6},\ P(0)=\frac{3}{6}\).
\(\mu_X=9\cdot\frac{1}{6}+3\cdot\frac{2}{6}=1.5+1=2.5\).
\(\sigma_X^{2}=(6.5)^{2}\frac{1}{6}+(0.5)^{2}\frac{2}{6}+(2.5)^{2}\frac{3}{6}=7.0417+0.0833+3.125=10.25\), so \(\sigma_X=\sqrt{10.25}\approx 3.20\).
Net gain: \(2.5-3=-\$0.50\). The game is not fair: the player loses $0.50 per game on average.
Problem 23: Screening test
With the condition: 50 people, of whom \(0.90(50)=45\) test positive and 5 test negative. Without it: 950 people, of whom \(0.08(950)=76\) test positive and 874 negative.
| Positive | Negative | Total | |
|---|---|---|---|
| Condition | 45 | 5 | 50 |
| No condition | 76 | 874 | 950 |
| Total | 121 | 879 | 1,000 |
\(P(\text{positive})=\frac{121}{1000}=0.121\).
\(P(\text{condition}|\text{positive})=\frac{45}{121}\approx 0.372\).
0.121 and about 0.372: most positive tests are false alarms because the condition is rare.
Problem 24: Battery inspection
\(P(0)=\frac{8}{10}\cdot\frac{7}{9}=\frac{28}{45}\), \(P(2)=\frac{2}{10}\cdot\frac{1}{9}=\frac{1}{45}\), \(P(1)=2\cdot\frac{8}{10}\cdot\frac{2}{9}=\frac{16}{45}\). They sum to \(\frac{45}{45}=1\).
\(\mu_X=0\cdot\frac{28}{45}+1\cdot\frac{16}{45}+2\cdot\frac{1}{45}=\frac{18}{45}=0.4\).
\(\sigma_X^{2}=(0.4)^{2}\frac{28}{45}+(0.6)^{2}\frac{16}{45}+(1.6)^{2}\frac{1}{45}\approx 0.0996+0.1280+0.0569=0.2844\), so \(\sigma_X\approx 0.53\).
\(\mu_X=0.4\) and \(\sigma_X\approx 0.53\)
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AP Statistics Probability and Random Variables worksheet answer key
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