Probability and Probability Models: Grade 7 Math Worksheet Answer Key – PDF Download

Problem 1: Likelihood words

Probability 0: impossible (A).

Probability 0.1: unlikely (B), close to 0.

Probability \(\frac{1}{2}\): as likely as not (C).

Probability 0.9: likely (D), close to 1.

Probability 1: certain (E).

Problem 2: Likelihood in context

1. The two sides are equally likely, so the probability is \(\frac{1}{2}\): as likely as not.
2. No face shows 7, so the probability is 0: impossible.
3. \(0.95\) is close to 1: likely.
4. \(0.03\) is close to 0: unlikely.
5. October always follows September, so the probability is 1: certain.

Problem 3: Marbles in a bag

The bag holds \(4 + 3 + 5 = 12\) marbles.

(a) \(P(\text{blue}) = \frac{3}{12} = \frac{1}{4}\).

(b) The not-green marbles are \(4 + 3 = 7\), so \(P(\text{not green}) = \frac{7}{12}\). Check: \(1 – \frac{5}{12} = \frac{7}{12}\).

Answers: (a) \(\frac{1}{4}\), (b) \(\frac{7}{12}\).

Problem 4: Cards numbered 1 to 10

There are 10 equally likely cards.

(a) Divisible by 3: 3, 6, 9. So \(P=\frac{3}{10}\).

(b) Greater than 7: 8, 9, 10. So \(P=\frac{3}{10}\).

(c) Primes: 2, 3, 5, 7. So \(P=\frac{4}{10}=\frac{2}{5}\).

Answers: (a) \(\frac{3}{10}\), (b) \(\frac{3}{10}\), (c) \(\frac{2}{5}\).

Problem 5: Number cube

(a) The primes on a cube are 2, 3, and 5, so \(P = \frac{3}{6} = \frac{1}{2}\).

(b) Numbers less than 5: 1, 2, 3, 4, so \(P = \frac{4}{6} = \frac{2}{3}\).

(c) No face is greater than 6, so \(P = \frac{0}{6} = 0\), an impossible event.

Answers: (a) \(\frac{1}{2}\), (b) \(\frac{2}{3}\), (c) 0.

Problem 6: Raffle tickets

(a) \(P(\text{Priya wins})=\frac{10}{250}=\frac{1}{25}=0.04\).

(b) \(1-\frac{1}{25}=\frac{24}{25}=0.96\).

(c) \(P(\text{Jamal wins})=\frac{15}{250}=\frac{3}{50}=0.06\).

Answers: (a) \(\frac{1}{25}\), (b) \(\frac{24}{25}\), (c) \(\frac{3}{50}\).

Problem 7: Equal sectors

There are 8 equal sectors.

(a) Numbers greater than 5: 6, 7, 8. So \(P = \frac{3}{8}\).

(b) Even numbers: 2, 4, 6, 8. So \(P = \frac{4}{8} = \frac{1}{2}\).

(c) Multiples of 3: 3 and 6. So \(P = \frac{2}{8} = \frac{1}{4}\).

Answers: (a) \(\frac{3}{8}\), (b) \(\frac{1}{2}\), (c) \(\frac{1}{4}\).

Problem 8: Complements

(a) \(1 – 0.35 = 0.65\). The probability of no rain is 0.65.

(b) \(1 – \frac{3}{8} = \frac{8}{8} – \frac{3}{8} = \frac{5}{8}\). The probability that she does not win is \(\frac{5}{8}\).

Problem 9: Letter tiles

The word has 11 letters, so there are 11 equally likely tiles. The vowels are A, E, A, I, which is 4 tiles.

(a) \(P(\text{vowel})=\frac{4}{11}\).

(b) M appears twice, so \(P=\frac{2}{11}\).

(c) \(1-\frac{4}{11}=\frac{7}{11}\).

(d) There is no Z, so \(P=0\), an impossible event.

Answers: (a) \(\frac{4}{11}\), (b) \(\frac{2}{11}\), (c) \(\frac{7}{11}\), (d) 0.

Problem 10: Rolling a cube 50 times

(a) A 4 appeared 11 times in 50 rolls, so \(\frac{11}{50} = 0.22\).

(b) Even numbers: \(9 + 11 + 9 = 29\) rolls, so \(\frac{29}{50} = 0.58\).

(c) The theoretical probability of an even number is \(\frac{3}{6} = 0.5\). The experimental value 0.58 is 0.08 higher, which is a small difference for only 50 rolls.

Answers: (a) 0.22, (b) 0.58, (c) 0.58 is close to 0.5.

Problem 11: Free throw record

(a) Total attempts: \(21+7=28\). \(P(\text{made})=\frac{21}{28}=\frac{3}{4}=0.75\).

(b) \(0.75\times 40=30\). Check: \(\frac{3}{4}\text{ of }40=30\).

Answers: (a) 0.75; (b) about 30 free throws.

Problem 12: Predict from spins

Total spins: \(18 + 10 + 8 + 4 = 40\).

Experimental \(P(\text{red}) = \frac{18}{40} = \frac{9}{20} = 0.45\).

Red in 300 spins: \(0.45 \times 300 = 135\).

Yellow: \(\frac{4}{40} = 0.1\), and \(0.1 \times 300 = 30\).

Answers: 0.45; about 135 reds and about 30 yellows.

Problem 13: Valid probability model?

A valid model has every probability between 0 and 1, and the probabilities add to 1.

Table A: \(0.40+0.35+0.25=1.00\). Valid.

Table B: \(0.5+0.3+0.3=1.1\), which is greater than 1. Not valid.

Table C: \(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{6}{6}=1\). Valid.

Table D: the sum is \(0.6-0.1+0.5=1\), but a probability cannot be negative. Not valid.

Problem 14: Non-uniform model

The sectors are 3, 2, and 1 parts of 6, so \(P(A) = \frac{3}{6} = \frac{1}{2}\), \(P(B) = \frac{2}{6} = \frac{1}{3}\), and \(P(C) = \frac{1}{6}\).

Sum: \(\frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1\).

\(P(A \text{ or } B) = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}\).

Answer: \(\frac{5}{6}\).

Problem 15: Compare with a prediction

(a) A fair cube is expected to show each face \(\frac{1}{6} \times 60 = 10\) times .

(b) Observed minus 10: face 1: \(8 – 10 = -2\); face 2: \(12 – 10 = 2\); face 3: \(9 – 10 = -1\); face 4: \(11 – 10 = 1\); faces 5 and 6: 0.

(c) The counts are within 2 of 10, so the results are consistent with a fair cube. The counts add to 60, as they should.

Answers: (a) 10; (b) -2, 2, -1, 1, 0, 0; (c) yes, reasonable.

Problem 16: Eighty spins

Total spins: \(24+20+16+20=80\).

(a) \(P(\text{red})=\frac{24}{80}=\frac{3}{10}=0.3\).

(b) \(\frac{20+20}{80}=\frac{40}{80}=\frac{1}{2}\).

(c) \(P(\text{green})=\frac{16}{80}=\frac{1}{5}\), and \(\frac{1}{5}\times 500=100\).

(d) Observed minus 20: red \(+4\), blue \(0\), green \(-4\), yellow \(0\). No count is more than 4 away from 20, so equal sectors are reasonable.

Answers: (a) \(\frac{3}{10}\), (b) \(\frac{1}{2}\), (c) about 100, (d) 4, 0, 4 and 0 away; reasonable.

Problem 17: Predict 120 rolls

1. \(P=\frac{1}{6}\), and \(\frac{1}{6}\times 120=20\).
2. \(P=\frac{3}{6}=\frac{1}{2}\), and \(\frac{1}{2}\times 120=60\).
3. The faces 5 and 6 give \(P=\frac{2}{6}=\frac{1}{3}\), and \(\frac{1}{3}\times 120=40\).
4. The factors of 6 on the cube are 1, 2, 3, 6, so \(P=\frac{4}{6}=\frac{2}{3}\), and \(\frac{2}{3}\times 120=80\).

Answers: (a) about 20 (b) about 60 (c) about 40 (d) about 80. Real results will be close to, but not exactly, these predictions.

Problem 18: Is the coin fair?

(a) A fair coin has \(P(\text{heads})=\frac{1}{2}\), so \(\frac{1}{2}\times 40=20\) heads are expected.

(b) \(\frac{26}{40}=\frac{13}{20}=0.65\).

(c) The observed count is \(26-20=6\) above the expected count. A difference of 6 in only 40 flips can happen by chance with a fair coin, so the evidence is not strong. Ethan should flip the coin many more times before deciding.

Answers: (a) 20 heads; (b) 0.65; (c) not enough evidence.

Problem 19: Paper cup model

(a) Divide each count by 100: down \(\frac{52}{100} = 0.52\), side 0.41, up 0.07. Check: \(0.52 + 0.41 + 0.07 = 1.00\).

(b) \(1 – 0.41 = 0.59\), or \(0.52 + 0.07 = 0.59\).

(c) In 200 tosses: down \(0.52 \times 200 = 104\), side \(0.41 \times 200 = 82\), up \(0.07 \times 200 = 14\). These add to 200.

Answers: (a) 0.52, 0.41, 0.07; (b) 0.59; (c) about 104, 82, and 14.

Problem 20: Design a spinner

1. Red: \(\frac{1}{2}\times 12=6\). Blue: \(\frac{1}{3}\times 12=4\). Green: \(\frac{1}{6}\times 12=2\). Check: \(6+4+2=12\).
2. Red: \(\frac{1}{2}\times 600=300\). Blue: \(\frac{1}{3}\times 600=200\). Green: \(\frac{1}{6}\times 600=100\). Check: \(300+200+100=600\).

Answers: (a) 6 red, 4 blue, 2 green; (b) about 300, 200 and 100 spins.

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