Counting, Permutations, Combinations and Probability: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Counting Probability Worksheet Answer Key: problems 1–5

Problem 1: Fundamental counting principle

1. \(4\times 3\times 2=\)24 outfits
2. \(5\times 3\times 4=\)60 sandwiches

Problem 2: Factorials

1. \(6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=\)720
2. \(\frac{7\cdot 6\cdot 5!}{5!}=7\cdot 6=\)42
3. \(\frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}=\frac{336}{6}=\)56

Problem 3: PIN codes

Fill each box with the number of choices.

1. \(10\cdot 10\cdot 10\cdot 10=\)10,000
2. \(10\cdot 9\cdot 8\cdot 7=\)5,040

Problem 4: Permutations

1. \(8\cdot 7\cdot 6=\)336
2. \(10\cdot 9=\)90

Problem 5: Combinations

1. \(\frac{10\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 2\cdot 1}=\frac{5040}{24}=\)210
2. \(_{12}C_{10}={_{12}C_{2}=\frac{12\cdot 11}{2}=\)66

Precalculus Counting Probability Worksheet Answer Key: problems 6–10

Problem 6: Arranging letters

1. \(4!=\)24
2. 6 letters with A three times and N twice: \(\frac{6!}{3!\,2!}=\frac{720}{12}=\)60
3. 10 letters with S three times , T three times , I twice: \(\frac{10!}{3!\,3!\,2!}=\frac{3628800}{72}=\)50,400

Problem 7: Order or no order

1. Order does not matter, so a combination: \(_{12}C_{3}=\frac{12\cdot 11\cdot 10}{6}=\)220
2. Order matters, so a permutation: \(_{12}P_{3}=12\cdot 11\cdot 10=\)1,320

Problem 8: Handshakes

Each handshake is a pair, and order does not matter: \(_{6}C_{2}=\frac{6\cdot 5}{2}=15\).

15 handshakes

Problem 9: Probability with two dice

1. Pairs (2,6), (3,5), (4,4), (5,3), (6,2) give 5 of 36 outcomes: 5/36
2. Sum 10 has 3 outcomes, sum 11 has 2, sum 12 has 1, a total of 6: \(\frac{6}{36}=\)1/6

Problem 10: Probability with cards

1. \(\frac{_{13}C_{2}}{_{52}C_{2}}=\frac{78}{1326}=\)1/17
2. There are 12 face cards: \(\frac{_{12}C_{2}}{_{52}C_{2}}=\frac{66}{1326}=\)11/221

Precalculus Counting Probability Worksheet Answer Key: problems 11–15

Problem 11: Choosing a committee

Total committees: \(_{18}C_{4}=3060\). All girls: \(_{10}C_{4}=210\).

\(\frac{210}{3060}=\)7/102

Problem 12: Drawing without replacement

1. \(\frac{5}{8}\cdot\frac{4}{7}=\frac{20}{56}=\)5/14
2. \(\frac{5}{8}\cdot\frac{3}{7}=\)15/56

Problem 13: Using the multiplication rule

1. \(0.3\times 0.6=\)0.18
2. \(P(B\mid A)=\frac{0.24}{0.5}=\)0.48

Problem 14: At least one

1. \(1-(\frac{1}{2})^{3}=1-\frac{1}{8}=\)7/8
2. \(1-(\frac{5}{6})^{3}=1-\frac{125}{216}=\)91/216

Problem 15: Probability distribution

The 8 equally likely outcomes give 1, 3, 3, 1 outcomes with 0, 1, 2, 3 heads, so the probabilities are \(\frac{1}{8},\frac{3}{8},\frac{3}{8},\frac{1}{8}\).

\(E(X)=0\cdot\frac{1}{8}+1\cdot\frac{3}{8}+2\cdot\frac{3}{8}+3\cdot\frac{1}{8}=\frac{12}{8}=\)1.5

Precalculus Counting Probability Worksheet Answer Key: problems 16–20

Problem 16: Missing probability

1. Probabilities sum to 1: \(p=1-0.1-0.3-0.2=\)0.4
2. \(0(0.1)+1(0.3)+2(0.4)+3(0.2)=0.3+0.8+0.6=\)1.7
3. \(0.4+0.2=\)0.6

Problem 17: Free throws

1. \(0.8\cdot 0.8=\)0.64
2. \(2\cdot 0.8\cdot 0.2=\)0.32
3. \(P(0)=0.04\), so \(E=0(0.04)+1(0.32)+2(0.64)=\)1.6

Problem 18: Books on a shelf

1. \(5!=\)120
2. Treat the pair as one item: \(4!\) arrangements, times 2 orders inside the pair: \(24\cdot 2=\)48

Problem 19: Student council

1. \(_{6}C_{2}\cdot{_{7}C_{3}=15\cdot 35=\)525
2. \(_{13}C_{5}=1287\), so \(\frac{525}{1287}=\)175/429

Problem 20: License plates

1. \(26^{3}\cdot 10^{3}=17576\cdot 1000=\)17,576,000
2. \(26\cdot 25\cdot 24\cdot 10\cdot 9\cdot 8=15600\cdot 720=\)11,232,000

Precalculus Counting Probability Worksheet Answer Key: problems 21–25

Problem 21: Raffle

Total prize money: \(300+3(50)+6(10)=510\). Expected winnings per ticket: \(\frac{510}{200}=2.55\).

Net: \(2.55-5=\)-$2.45 per ticket.

Problem 22: Insurance

Expected claim cost: \(0.02\cdot 5000=100\).

Expected profit: \(250-100=\)$150 per policy.

Problem 23: Five hearts

\(\frac{_{13}C_{5}}{_{52}C_{5}}=\frac{1287}{2598960}=\)33/66640, which is about 0.000495.

Problem 24: Two-stage tree

First draw: \(P(G)=\frac{4}{10}\), \(P(Y)=\frac{6}{10}\). After a green, 3 of 9 are green; after a yellow, 5 of 9 are yellow.

\(P(\text{same})=\frac{4}{10}\cdot\frac{3}{9}+\frac{6}{10}\cdot\frac{5}{9}=\frac{12}{90}+\frac{30}{90}=\frac{42}{90}=\)7/15

Problem 25: A dice game

Expected winnings: \(12\cdot\frac{1}{6}+3\cdot\frac{2}{6}=\frac{18}{6}=3\).

Expected net: \(3-2.50=\)+$0.50. The game favors the player by 50 cents per play.

How to use this precalculus counting probability worksheet answer key

Read this precalculus counting probability worksheet answer key after trying each problem yourself, and redo any problem you missed.

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