7th Grade Math Quiz: Develop Non Uniform Probability Models

What this quiz covers

This quiz focuses on Develop Non Uniform Probability Models, giving you a quick way to practice the rules, question types, and explanations that matter most for 7th Grade Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $P(Red) = \frac{1}{4}$; the bag contains equal numbers of each color since there are 4 colors total
2. $P(Red) = \frac{3}{10}$; the bag likely contains 30% red marbles or 3 red marbles out of every 10 (correct answer)
3. $P(Red) = \frac{18}{42}$; this represents red marbles compared to non-red marbles in the sample
4. $P(Red) = \frac{2}{5}$; the bag likely contains 40% red marbles since blue was drawn most frequently

Explanation: When you encounter experimental probability questions, remember that probability is calculated as the number of favorable outcomes divided by the total number of trials. Here, you need to find the probability of drawing red based on the observed data. From 60 total draws, red was drawn 18 times. So $P(Red) = \frac{18}{60} = \frac{3}{10} = 0.3$, meaning there's a 30% chance of drawing red. This suggests the bag likely contains about 30% red marbles, or roughly 3 red marbles for every 10 total marbles. Choice A incorrectly assumes equal distribution just because there are 4 colors. The experimental data clearly shows unequal frequencies, so this theoretical assumption doesn't match the evidence. Choice C makes a common error by comparing red marbles to non-red marbles (18 red vs. 42 non-red). However, probability requires comparing favorable outcomes to total outcomes, not to unfavorable ones. Choice D incorrectly states that red has a $\frac{2}{5}$ probability (which would be 24 out of 60), but red was only drawn 18 times. This answer seems to confuse red's frequency with blue's frequency (24 times). The key insight is that choice B correctly calculates the experimental probability and reasonably interprets what this means about the bag's actual composition. Study tip: Always calculate experimental probability as favorable outcomes ÷ total trials, then think about what the decimal or percentage tells you about the real-world situation. Don't assume equal distribution unless the data supports it.

Question 2

1. The simulation shows systematic bias: Hearts and Clubs total 138 vs Diamonds and Spades total 102, indicating red-black imbalance
2. The simulation is significantly biased: Hearts appears 72 times vs expected 60, suggesting a 20% error rate in the generator
3. The simulation validates uniform model: all frequencies are within 10% of expected values, confirming theoretical predictions
4. The simulation shows reasonable variation from uniform: largest deviation is Hearts at $\frac{72}{240} = 0.30$ vs expected 0.25 (correct answer)

Explanation: When you encounter probability simulation questions, you need to compare observed results to theoretical expectations and evaluate whether the differences represent reasonable random variation or significant bias. In a standard deck, each suit should appear equally often. With 240 trials, you'd expect each suit 60 times (240 ÷ 4 = 60). Let's check each suit's proportion: Hearts = 72/240 = 0.30, Diamonds = 48/240 = 0.20, Clubs = 66/240 = 0.275, Spades = 54/240 = 0.225. The theoretical expectation is 0.25 for each suit. Answer D correctly identifies that Hearts shows the largest deviation (0.30 vs expected 0.25) and recognizes this as reasonable variation from the uniform model. Random simulations naturally produce some variation from theoretical values. Answer A incorrectly focuses on red vs black suits, which isn't the relevant comparison here. We're testing whether the generator produces equal frequencies for all four suits, not color balance. Answer B misinterprets the data by calling a 12-card difference (72 vs 60) a "20% error rate" and declaring it "significantly biased." This level of variation is actually quite normal in random sampling. Answer C claims all frequencies are within 10% of expected values, but this is false. Hearts (72) and Diamonds (48) both deviate by more than 10% from the expected 60. Remember: In simulation problems, perfect matches to theoretical values are actually suspicious. Real random processes show variation, so look for answers that acknowledge reasonable deviations rather than expecting exact theoretical results.

Question 3

1. Group similar frequencies: $P(1,4) = 0.40$, $P(2) = 0.325$, $P(3,5) = 0.275$ to simplify the model while preserving patterns
2. Use uniform model $P(each) = 0.20$ since theoretical probability should not depend on limited experimental observations
3. Calculate individual probabilities: $P(1) = 0.225$, $P(2) = 0.325$, $P(3) = 0.150$, $P(4) = 0.175$, $P(5) = 0.125$ based on observed frequencies (correct answer)
4. Weight by section number: $P(5) = 5 × P(1)$, $P(4) = 4 × P(1)$, etc., since larger section numbers indicate larger areas

Explanation: When you encounter probability questions involving experimental data, you need to understand the difference between theoretical and experimental probability. Since this spinner has unequal sections, you can't assume equal probabilities—you must use the actual data to create the most accurate model. The correct approach is to calculate individual probabilities based on observed frequencies (Answer C). Convert each frequency to a probability by dividing by the total number of spins: $P(1) = \frac{45}{200} = 0.225$, $P(2) = \frac{65}{200} = 0.325$, and so on. This gives you the most accurate representation of how the spinner actually behaves, since it reflects the true size differences between sections. Answer A is wrong because grouping sections with similar frequencies ignores important differences. Sections 1 and 4 may have similar frequencies by chance, but they're still separate sections with different actual probabilities. Answer B is incorrect because using a uniform model (equal probabilities) ignores the key information that sections are unequal in size. With 200 spins, you have enough data to trust the experimental results over a theoretical assumption. Answer D makes no sense because section numbers have nothing to do with section sizes. Just because a section is labeled "5" doesn't mean it's five times larger than section "1." Remember: When you have experimental data from many trials and know the outcomes aren't equally likely, use that data to calculate experimental probabilities. Don't force equal probabilities when the real-world situation clearly shows unequal outcomes.

Question 4

1. Observed $P(sum = 7) = \frac{18}{80} = 0.225$; this closely matches theoretical $P(sum = 7) = \frac{1}{6} ≈ 0.167$
2. Observed $P(sum = 7) = \frac{18}{80} = 0.225$; this is reasonably close to theoretical $P(sum = 7) = \frac{6}{36} ≈ 0.167$ (correct answer)
3. Observed $P(sum = 2) = \frac{1}{80} = 0.0125$; this exactly matches the theoretical $P(sum = 2) = \frac{1}{36} ≈ 0.028$
4. Observed $P(sum = 7) = \frac{18}{62}$; this excludes impossible outcomes and better represents the actual probability

Explanation: The observed $P(sum = 7) = \frac{18}{80} = 0.225$ compared to theoretical $P(sum = 7) = \frac{6}{36} ≈ 0.167$ shows reasonable agreement given sampling variation. Choice A incorrectly states theoretical probability as $\frac{1}{6}$. Choice C incorrectly claims the probabilities match when $0.0125 ≠ 0.028$. Choice D incorrectly uses 62 as denominator and misunderstands probability calculation.

Question 5

1. Use $P(rain) = \frac{45}{200} = 0.225$ for the entire period, recognizing this may mask seasonal variation in rainfall (correct answer)
2. Use $P(rain) = \frac{30}{100} = 0.30$ since the second half represents more current and reliable weather patterns
3. Use $P(rain) = \frac{15}{100} = 0.15$ since this represents the baseline probability without seasonal interference
4. Average the two periods: $P(rain) = \frac{0.15 + 0.30}{2} = 0.225$, which accounts for equal weighting of both time periods

Explanation: The overall probability $P(rain) = \frac{45}{200} = 0.225$ uses all available data, but the significant difference between periods (0.15 vs 0.30) suggests seasonal variation that a single probability model may not capture well. Choice B arbitrarily favors recent data. Choice C arbitrarily favors early data. Choice D incorrectly averages probabilities rather than using frequency data directly.

Question 6

1. $P(A)=0.70,\;P(B)=0.50,\;P(C)=0.60,\;P(D)=0.20$
2. $P(A)=\frac{70}{200}=0.35,\;P(B)=\frac{50}{200}=0.25,\;P(C)=\frac{60}{200}=0.30,\;P(D)=\frac{20}{200}=0.10$ (correct answer)
3. $P(A)=0.30,\;P(B)=0.35,\;P(C)=0.25,\;P(D)=0.10$
4. $P(A)=\frac{1}{4},\;P(B)=\frac{1}{4},\;P(C)=\frac{1}{4},\;P(D)=\frac{1}{4}$

Explanation: This question tests developing a non-uniform model from spinner frequencies over 200 spins, using relative frequencies. Non-uniform: unequal; develop by observing A70, B50, C60, D20, calculating 70/200=0.35, etc., sum=1. For example, P(A)=0.35, P(B)=0.25, P(C)=0.30, P(D)=0.10, matching data. The correct model is B, with accurate relative frequencies. Errors: dividing by 100 in A, uniform in C, swapping in D. Developing: (1) collect, (2) calculate relative, (3) assign, (4) verify sum=1. Non-uniform from sections; mistakes: wrong denominator or assuming equal.

Question 7

1. The model predicts about $0.38\times 50=19$ Sports choices, and 21 is close, so the model seems reasonable. (correct answer)
2. The model predicts about $0.38\times 50=38$ Sports choices, so 21 is too small.
3. The model predicts exactly 19 Sports choices, so 21 means the model is incorrect.
4. The model must be uniform because there are 3 activities.

Explanation: This question tests using a non-uniform model from student activity frequencies to predict in a new survey and comparing to observations. Non-uniform: unequal; model from 100 students, P(sports)=38/100=0.38, predict 0.38×50=19, observe 21 close. For example, expected ≈19 sports, 21 is reasonably close, model fits. The correct conclusion is A, assessing fit with approximation. Errors: exact expectation in B, miscalculation in C (38 instead of 19), uniform assumption in D. Comparing: expected P×n vs. observed, assess fit; non-uniform from preferences. Mistakes: demanding exactness or math errors.

Question 8

1. $\frac{27}{60}=0.45$ (correct answer)
2. $\frac{1}{3}\approx 0.33$
3. $\frac{21}{60}=0.35$
4. $\frac{60}{27}\approx 2.22$

Explanation: To build a probability model from frequencies, divide the number of days for the outcome by the total number of days: P(Pop)=27/60=0.45. Choice B assumes all three categories are equally likely, which ignores the actual data. Choice C uses the frequency for Rap (21) instead of Pop, mixing up the categories. Choice D inverts the fraction, which does not represent a valid probability.

Question 9

1. $P(R)=45,\;P(B)=30,\;P(G)=25$; non-uniform
2. $P(R)=0.45,\;P(B)=0.30,\;P(G)=0.25$; non-uniform (correct answer)
3. $P(R)=0.45,\;P(B)=0.25,\;P(G)=0.30$; uniform
4. $P(R)=\frac{1}{3},\;P(B)=\frac{1}{3},\;P(G)=\frac{1}{3}$; uniform

Explanation: This question tests developing a non-uniform probability model from observed frequencies in spinner spins, assigning probabilities based on those frequencies, and identifying if the model is uniform or non-uniform. Non-uniform models have outcomes that are not equally likely, so to develop one, observe frequencies like 100 spins with red 45, blue 30, green 25, calculate relative frequencies as 45/100=0.45, 30/100=0.30, 25/100=0.25, assign these as probabilities, and verify they sum to 1. For example, P(red)=0.45, P(blue)=0.30, P(green)=0.25, and the sum is 1.00, which checks out, making it a valid non-uniform model since probabilities differ. The correct model is B, which uses these exact probabilities and correctly identifies it as non-uniform. Common errors include assuming a uniform model like A or D, using raw frequencies as in C, or swapping probabilities as in D. Developing a model involves (1) collecting frequencies, (2) calculating relative frequencies by dividing by total trials, (3) assigning them as probabilities, and (4) verifying the sum is 1. Non-uniform models arise from weighted or biased mechanisms, and mistakes often include assuming uniformity or not normalizing frequencies.

Question 10

1. Because 6 happened the most, $P(6)=1$ and all other outcomes have probability 0.
2. A model from the data gives $P(6)=30$, so the die appears uniform.
3. A model from the data gives $P(6)=\frac{30}{120}=0.25$, so the die appears non-uniform. (correct answer)
4. The model is uniform because each outcome has probability $\frac{1}{6}$.

Explanation: This question tests developing a non-uniform model from biased die roll frequencies over 120 rolls and identifying non-uniformity. Non-uniform: not equal; develop by observing frequencies like 6:30, calculating P(6)=30/120=0.25, higher than 1/6≈0.167, indicating bias. For example, P(6)=0.25 vs. uniform 1/6, and increasing frequencies suggest non-uniform. The correct statement is B, recognizing the relative frequency shows non-uniformity. Errors: claiming uniform in A or C, using raw count in C, or extreme in D. Developing: (1) collect frequencies, (2) calculate relative, (3) assign, (4) verify sum=1; compare to uniform. Non-uniform from bias; mistakes: assuming fair or not calculating properly.

Question 11

1. About (36/80) x 50=22.5, so about 23 bananas
2. About (28/80) x 50=17.5, so about 18 bananas (correct answer)
3. About (28/80) x 50=28, so about 28 bananas
4. About (16/80) x 50=10, so about 10 bananas

Explanation: Out of 80 lunches, bananas were chosen 28 times, so the relative frequency for bananas is 28/80, or 0.35. Multiplying this by the next 50 lunches gives 0.35 x 50 = 17.5, which rounds to about 18 banana choices, matching Choice B. Choice A uses the apple frequency, 36/80, instead of the banana frequency, so it predicts the wrong fruit. Choice C uses the correct fraction for bananas but makes an arithmetic error, since 28/80 x 50 actually equals 17.5, not 28. Choice D uses the orange frequency, 16/80, instead of the banana frequency.

Question 12

1. $P(\text{rain})=12,\;P(\text{sun})=18$
2. $P(\text{rain})=0.4,\;P(\text{sun})=0.6$ (correct answer)
3. $P(\text{rain})=0.5,\;P(\text{sun})=0.5$
4. $P(\text{rain})=0.6,\;P(\text{sun})=0.4$

Explanation: Out of 30 total days, it rained on 12 of them and was sunny on 18, so the relative frequencies are 12 divided by 30, which equals 0.4 for rain, and 18 divided by 30, which equals 0.6 for sun, matching choice B. Choice A lists the raw day counts, 12 and 18, instead of converting them into probabilities between 0 and 1. Choice C assumes rain and sun are equally likely, which ignores the actual data showing sun occurred more often. Choice D swaps the two probabilities, assigning 0.6 to rain and 0.4 to sun instead of the other way around. Converting frequencies into probabilities always means dividing each count by the total number of observations, here 30 days.

Question 13

1. P(red)=45, P(blue)=30, P(green)=25
2. P(red)=0.45, P(blue)=0.30, P(green)=0.25 (correct answer)
3. P(red)=1/3, P(blue)=1/3, P(green)=1/3
4. P(red)=0.40, P(blue)=0.35, P(green)=0.25

Explanation: Dividing each color's spins by the total gives the relative frequencies: 45/100=0.45 for red, 30/100=0.30 for blue, and 25/100=0.25 for green, matching choice B. Choice A uses the raw counts (45, 30, 25) instead of converting them to probabilities between 0 and 1. Choice C assumes a uniform model, which ignores the unequal spin counts. Choice D shifts the values away from what the data actually shows.

Question 14

1. 32/50=0.64
2. 18/50=0.36 (correct answer)
3. 50/18≈2.78
4. 18/50=0.18

Explanation: Out of 50 days, the leaves were wet on 18 of them, so the relative frequency for wet is 18 divided by 50, which equals 0.36, matching Choice B. Choice A, 32/50 = 0.64, is actually the probability of the leaves being dry, not wet. Choice C inverts the fraction, dividing 50 by 18 instead of 18 by 50, which does not represent a probability at all. Choice D uses the correct fraction but makes an arithmetic error, since 18 divided by 50 actually equals 0.36, not 0.18.

Question 15

1. It is valid because $0.45+0.30+0.25=1.00$. (correct answer)
2. It is valid because 45, 30, and 25 add to 1.
3. It is valid because $0.45+0.30+0.25=0.90$.
4. It is valid because each probability is greater than 1.

Explanation: This question tests verifying a non-uniform probability model developed from observed frequencies, ensuring it meets probability rules. A non-uniform model means outcomes are not equally likely, developed by observing frequencies (100 trials: red 45, blue 30, green 25), calculating relative frequencies (0.45, 0.30, 0.25), assigning probabilities, and verifying they sum to 1 for validity. For example, P(red)=0.45, P(blue)=0.30, P(green)=0.25 sum to 1.00, confirming it's valid, while sums not equaling 1 or using raw counts would invalidate it. The correct verification is choice A, accurately stating the sum is 1.00. Common errors include miscalculating the sum like in B, using raw frequencies for the sum as in C, or misunderstanding probability bounds like D. To develop and verify: (1) collect frequencies, (2) calculate relative frequencies, (3) assign probabilities, (4) verify sum=1 and each is between 0 and 1. Non-uniform models reflect data patterns, and mistakes often involve sum errors or confusing frequencies with probabilities.

Question 16

1. It is valid because each probability is greater than 1.
2. It is valid because 45, 30, and 25 add to 1.
3. It is valid because $0.45+0.30+0.25=0.90$.
4. It is valid because $0.45+0.30+0.25=1.00$. (correct answer)

Explanation: A probability model is valid when all of its probabilities add up to exactly 1, and here 0.45 plus 0.30 plus 0.25 equals 1.00, matching choice D. Choice A is incorrect because a valid probability can never be greater than 1; every probability must be between 0 and 1. Choice B incorrectly adds the raw numbers 45, 30, and 25, which sum to 100, not 1, and confuses those whole numbers with the actual decimal probabilities. Choice C contains an arithmetic error, since 0.45 plus 0.30 plus 0.25 actually equals 1.00, not 0.90. Verifying a probability model always means checking that every individual probability falls between 0 and 1 and that they all sum to exactly 1.

Question 17

1. $P(A)=\frac{1}{4},\;P(B)=\frac{1}{4},\;P(C)=\frac{1}{4},\;P(D)=\frac{1}{4}$
2. $P(A)=\frac{70}{200}=0.35,\;P(B)=\frac{50}{200}=0.25,\;P(C)=\frac{60}{200}=0.30,\;P(D)=\frac{20}{200}=0.10$ (correct answer)
3. $P(A)=0.30,\;P(B)=0.35,\;P(C)=0.25,\;P(D)=0.10$
4. $P(A)=0.70,\;P(B)=0.50,\;P(C)=0.60,\;P(D)=0.20$

Explanation: This question tests developing a non-uniform model from spinner frequencies over 200 spins, using relative frequencies. Non-uniform: unequal; develop by observing A70, B50, C60, D20, calculating 70/200=0.35, etc., sum=1. For example, P(A)=0.35, P(B)=0.25, P(C)=0.30, P(D)=0.10, matching data. The correct model is B, with accurate relative frequencies. Errors: dividing by 100 in A, uniform in C, swapping in D. Developing: (1) collect, (2) calculate relative, (3) assign, (4) verify sum=1. Non-uniform from sections; mistakes: wrong denominator or assuming equal.

Question 18

1. $P(1)=\frac{10}{100},\;P(2)=\frac{20}{100},\;P(3)=\frac{30}{100},\;P(4)=\frac{25}{100},\;P(5)=\frac{15}{100},\;P(6)=\frac{20}{100}$
2. $P(1)=\frac{1}{6},\;P(2)=\frac{1}{6},\;P(3)=\frac{1}{6},\;P(4)=\frac{1}{6},\;P(5)=\frac{1}{6},\;P(6)=\frac{1}{6}$
3. $P(1)=10,\;P(2)=20,\;P(3)=30,\;P(4)=25,\;P(5)=15,\;P(6)=20$
4. $P(1)=\frac{10}{120},\;P(2)=\frac{20}{120},\;P(3)=\frac{30}{120},\;P(4)=\frac{25}{120},\;P(5)=\frac{15}{120},\;P(6)=\frac{20}{120}$ (correct answer)

Explanation: This question tests developing a non-uniform probability model from observed frequencies of a biased die roll, matching the data proportions. Non-uniform means numbers are not equally likely, developed from 120 rolls (1:10,2:20,3:30,4:25,5:15,6:20) by calculating fractions like 10/120, 20/120, etc., assigning these probabilities, and verifying they sum to 1. For example, P(3)=30/120=0.25, and the total sums correctly to 1.00. The correct model uses the observed relative frequencies over 120. Common errors are assuming uniformity or using counts directly. Developing: (1) collect data, (2) compute relative frequencies, (3) assign probabilities, (4) check sum. Using this model predicts future rolls, and non-uniformity captures the die's bias.

Question 19

1. $P(\text{rain})=\frac{12}{30}=0.4,\;P(\text{sun})=\frac{18}{30}=0.6$ (correct answer)
2. $P(\text{rain})=0.5,\;P(\text{sun})=0.5$
3. $P(\text{rain})=0.6,\;P(\text{sun})=0.4$
4. $P(\text{rain})=12,\;P(\text{sun})=18$

Explanation: This question tests developing a non-uniform probability model from observed weather frequencies over 30 days, using it to predict the next day's weather. Non-uniform means outcomes like rain and sun are not equally likely, so we develop the model by observing 12 rainy days and 18 sunny days, calculating relative frequencies as 12/30=0.4 for rain and 18/30=0.6 for sun, assigning these probabilities, and verifying they sum to 1. For example, P(rain)=0.4, P(sun)=0.6, and 0.4+0.6=1.00, which is valid. The correct model uses these observed proportions. Errors might include switching the probabilities or assuming equal likelihood. Developing the model requires (1) tallying frequencies, (2) computing relative frequencies, (3) setting probabilities, and (4) checking the sum. Using the model, we can predict future events, and non-uniformity reflects actual weather patterns rather than assuming fairness.

Question 20

1. $P(A)=27,\;P(B)=21,\;P(O)=12$
2. $P(A)=\frac{27}{60}=0.45,\;P(B)=\frac{21}{60}=0.35,\;P(O)=\frac{12}{60}=0.20$ (correct answer)
3. $P(A)=\frac{1}{3},\;P(B)=\frac{1}{3},\;P(O)=\frac{1}{3}$
4. $P(A)=0.27,\;P(B)=0.21,\;P(O)=0.12$

Explanation: This question tests developing a non-uniform probability model from observed fruit choice frequencies over 60 days, using relative frequencies. Non-uniform means fruits have different probabilities, developed by calculating 27/60=0.45 for apple, 21/60=0.35 for banana, 12/60=0.20 for orange, assigning these, and verifying 0.45+0.35+0.20=1.00. For example, P(apple)=0.45 directly from the data. The correct model normalizes the counts to sum to 1. Errors include using unnormalized decimals or assuming equality. Developing involves (1) tallying choices, (2) relative frequencies, (3) probabilities, (4) sum check. This model can predict future choices, reflecting student preferences as non-uniform patterns.