Free Math Worksheets

- What height must a rectangle of width \(40\) have so that its area is \(1\)? - Connect probability with area under the density curve. - What happens to area when an interval has width \(0\)? - Distinguish continuous measurements from discrete outcomes.

1. a) A uniform density on \([0,40]\) has height \(\frac{1}{40}\), so \(f(x)=\frac{1}{40}\) for \(0\le x\le 40\) and \(f(x)=0\) otherwise. 2. b) \(P(10\le X\le 30)=\int_{10}^{30}\frac{1}{40}\,dx=\left[\frac{x}{40}\right]_{10}^{30}=\frac{30-10}{40}=0.5\). 3. c) For a continuous random variable, \(P(X=k)=\int_k^k f(x)\,dx=0\). Both probabilities are \(0\), so the statement is mathematically correct.

a) \(f(x)=\frac{1}{40}\) for \(0\le x\le 40\), and \(f(x)=0\) otherwise. b) \(\int_{10}^{30}\frac{1}{40}\,dx=0.5\) c) The statement is true because both exact-value probabilities are \(0\).

- Use the difference of cumulative distribution values at the interval endpoints. - Interpret probability as area under a density curve. - What area is associated with an interval of width \(0\)?

1. \(P(2<Z\le 6)=F(6)-F(2)=0.80-0.35=0.45\). Since \(P(Z=2)=0\), \(P(2\le Z\le 6)=0.45\) as well. 2. For a continuous random variable, \(P(Z=a)=\int_a^a f(z)\,dz=0\) because the interval has zero width. 3. Including or excluding endpoints does not change the probability. Thus, \(P(a\le Z\le b)=P(a<Z\le b)=P(a\le Z<b)=P(a<Z<b)\).

1. \(P(2<Z\le 6)=0.45\) and \(P(2\le Z\le 6)=0.45\) 2. \(P(Z=a)=\int_a^a f(z)\,dz=0\) 3. Open, closed, and half-open intervals with the same endpoints have equal probabilities for a continuous random variable.

- List the possible color combinations when order does not matter. - A mixed-color pair can occur in two orders. - Compare the probabilities of the outcomes.

1. The possible unordered color pairs are \(S=\{(B, B),(B, R),(R, R)\}\). 2. \(P((B, B))=\frac{7}{10}\cdot\frac{7}{10}=0.49\). 3. The mixed-color outcome can occur in either order, so \(P((B, R))=\frac{7}{10}\cdot\frac{3}{10}+\frac{3}{10}\cdot\frac{7}{10}=0.42\). 4. \(P((R, R))=\frac{3}{10}\cdot\frac{3}{10}=0.09\). 5. The three outcomes have different probabilities, so this sample space is not an equally likely-outcomes model.

a) \(S=\{(B, B),(B, R),(R, R)\}\) b) \(P((B, B))=0.49\), \(P((B, R))=0.42\), and \(P((R, R))=0.09\) c) The outcomes are not equally likely because their probabilities are different.

- What must the probabilities of all six outcomes add to? - Let \(p\) represent the probability of each outcome from \(1\) through \(5\). - What assumption is required before counting favorable and possible outcomes gives a probability? - Compare a fair die with a loaded die.

1. Let \(p\) be the probability of rolling any one of the numbers \(1\) through \(5\). Then \(P(6)=2p\). Since all outcome probabilities add to \(1\), \(5p+2p=1\). Thus \(7p=1\), so \(p=\frac{1}{7}\) and \(P(6)=\frac{2}{7}\approx 0.2857\). 2. The favorable-outcomes-over-possible-outcomes formula assumes that all individual outcomes are equally likely. This die is loaded, so its six outcomes are not equally likely.

1. \(P(6)=\frac{2}{7}\approx 0.2857\) 2. The formula does not apply directly because the six outcomes are not equally likely.

- When can you assume that all outcomes have the same probability? - Which numbers from \(1\) through \(6\) are prime? - Use one variable for the probability of each outcome from \(1\) through \(5\). - Which probability rule determines the value of that variable?

1. For a fair die, the six outcomes are equally likely. The prime outcomes are \(2\), \(3\), and \(5\), so \(P(A)=\frac{3}{6}=\frac{1}{2}=0.5\). 2. Let \(x\) be the probability of each outcome from \(1\) through \(5\). Then \(P(6)=3x\). 3. The probabilities must add to \(1\): \(5x+3x=1\). Thus \(8x=1\), so \(x=0.125\). 4. Therefore, \(P(1)=P(2)=P(3)=P(4)=P(5)=0.125\) and \(P(6)=0.375\).

a) Use the equally likely outcomes model. \(P(A)=\frac{1}{2}=0.5\). b) \(P(1)=P(2)=P(3)=P(4)=P(5)=0.125\), and \(P(6)=0.375\).

- Count how many cards belong to each suit and each rank. - Equal numbers of cards produce equal probabilities. - Count face cards separately from all other cards.

1. \(S_1=\{\text{clubs},\text{diamonds},\text{hearts},\text{spades}\}\). Each suit has \(13\) cards, so each outcome has probability \(\frac{13}{52}=\frac{1}{4}\). 2. \(S_2=\{A,2,3,4,5,6,7,8,9,10,J,Q,K\}\). Each rank has \(4\) cards, so each outcome has probability \(\frac{4}{52}=\frac{1}{13}\). 3. Each suit occurs the same number of times in the deck, and each rank occurs the same number of times, so the outcomes within each sample space are equally likely. 4. There are \(12\) face cards and \(40\) non-face cards. Thus, \(P(\text{face card})=\frac{12}{52}=\frac{3}{13}\), while \(P(\text{not a face card})=\frac{40}{52}=\frac{10}{13}\). The outcomes are not equally likely.

a) \(S_1=\{\text{clubs},\text{diamonds},\text{hearts},\text{spades}\}\); each probability is \(\frac{1}{4}\). b) \(S_2=\{A,2,3,4,5,6,7,8,9,10,J,Q,K\}\); each probability is \(\frac{1}{13}\). c) Each suit and each rank appears equally often in the deck. d) No. \(P(\text{face card})=\frac{3}{13}\) and \(P(\text{not a face card})=\frac{10}{13}\).

- How do you find an interval probability for a continuous uniform distribution? - Which original values round to a given value with one decimal place? - What is the distance between consecutive numbers written to \(k\) decimal places? - How does that distance change when one more decimal place is used?

1. For a uniform distribution, the probability is the interval width divided by the total width: \(P(1.15\le X<1.25)=\frac{1.25-1.15}{5}=\frac{0.10}{5}=0.02\). 2. A value rounds to \(1.2\) to one decimal place when it lies in \([1.15,1.25)\). Therefore, \(P(\text{rounds to }1.2)=0.02\). 3. The values that round to \(a\) lie in \([a-0.5\cdot 10^{-k},a+0.5\cdot 10^{-k})\), an interval of width \(10^{-k}\). Thus, \(P=\frac{10^{-k}}{5}=0.2\cdot 10^{-k}\). Each additional decimal place divides the interval width, and therefore the probability, by \(10\).

1. \(P(1.15\le X<1.25)=0.02\) 2. \(P(\text{rounds to }1.2)=0.02\) 3. \(P=\frac{10^{-k}}{5}=0.2\cdot 10^{-k}\). The probability is proportional to the width \(10^{-k}\) of the rounding interval.

- What is the area over an interval whose width is exactly \(0\)? - Recall how probability is defined using a density function. - Compare the length of a line segment with the length of one point on that segment.

1. a) For any point \(c\), \(P(X=c)=\int_c^c f(x)\,dx=0\). Therefore, \(P(a\le X\le b)=P(X=a)+P(a<X<b)+P(X=b)=P(a<X<b)\). 2. b) For a continuous random variable, probability is not found by a countable sum of point probabilities. It is found by an integral, which is a limit of area sums and may be positive over an interval even though each individual point has probability \(0\). Similarly, a line segment has positive length even though each point has length \(0\).

a) Since \(P(X=c)=\int_c^c f(x)\,dx=0\), the endpoints \(a\) and \(b\) do not change the interval probability. b) Continuous probabilities are defined by integrals, not by adding individual point probabilities. An integral over an interval can be positive even though every point has probability \(0\).

- Think of the five drawings as five positions. - First choose the positions won by Alex, then choose the position won by Jordan. - What is the probability that a winner is neither Alex nor Jordan? - Find the probability of one specific qualifying order, then account for all such orders.

1. On each drawing, \(P(A)=\frac{1}{20}\), \(P(J)=\frac{1}{20}\), and the probability that someone else wins is \(P(O)=\frac{18}{20}\). 2. Choose the \(2\) drawings won by Alex and then the \(1\) drawing won by Jordan: \(\binom{5}{2}\binom{3}{1}=10\cdot 3=30\) arrangements. 3. Each qualifying arrangement has probability \(\left(\frac{1}{20}\right)^2\left(\frac{1}{20}\right)\left(\frac{18}{20}\right)^2\). 4. Therefore, the required probability is \(30\left(\frac{1}{20}\right)^3\left(\frac{18}{20}\right)^2=\frac{243}{80{,}000}=0.0030375\).

\(\frac{243}{80{,}000}=0.0030375\), or \(0.30375\%\)

- How many yellow draws must occur? - Find the probability of one specific order, such as red, red, blue, blue, yellow, yellow. - How many ways can the positions of the three colors be assigned? - Replacement makes the draws independent and keeps the color probabilities constant.

1. The color probabilities on each draw are \(P(R)=0.2\), \(P(B)=0.3\), and \(P(Y)=0.5\). 2. If exactly \(2\) draws are red and exactly \(2\) are blue, the remaining \(2\) draws must be yellow. 3. Choose the \(2\) red positions and then the \(2\) blue positions: \(\binom{6}{2}\binom{4}{2}=15\cdot 6=90\) color sequences. 4. Each qualifying sequence has probability \((0.2)^2(0.3)^2(0.5)^2=0.0009\). 5. Therefore, the probability is \(90\cdot 0.0009=0.081\).

\(0.081\), or \(8.1\%\)

- First find the probability of each number on one roll. - For a pair of different numbers, account for both possible orders. - Check that the probabilities add to \(1\).

1. The unordered outcomes are \(S=\{(1, 1),(1, 2),(1, 3),(2, 2),(2, 3),(3, 3)\}\). 2. On one roll, \(P(1)=\frac{1}{2}\), \(P(2)=\frac{1}{3}\), and \(P(3)=\frac{1}{6}\). 3. Therefore, \(P((1, 1))=\frac{1}{4}\), \(P((1, 2))=2\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=\frac{1}{3}\), and \(P((1, 3))=2\left(\frac{1}{2}\right)\left(\frac{1}{6}\right)=\frac{1}{6}\). 4. Also, \(P((2, 2))=\frac{1}{9}\), \(P((2, 3))=2\left(\frac{1}{3}\right)\left(\frac{1}{6}\right)=\frac{1}{9}\), and \(P((3, 3))=\frac{1}{36}\). 5. These probabilities are not all equal, so assigning probability \(\frac{1}{6}\) to each unordered pair would be incorrect.

a) \(S=\{(1, 1),(1, 2),(1, 3),(2, 2),(2, 3),(3, 3)\}\) b) \(P((1, 1))=\frac{1}{4}\), \(P((1, 2))=\frac{1}{3}\), \(P((1, 3))=\frac{1}{6}\), \(P((2, 2))=\frac{1}{9}\), \(P((2, 3))=\frac{1}{9}\), and \(P((3, 3))=\frac{1}{36}\) c) The outcomes are not equally likely because their probabilities range from \(\frac{1}{36}\) to \(\frac{1}{3}\).

- Find the probability of each of the three result categories on one roll. - Choose the positions of the two \(6\)s first. - After those positions are fixed, how many positions remain for the three \(1\)s? - Multiply the number of qualifying arrangements by the probability of one arrangement.

1. On each roll, \(P(6)=\frac{1}{6}\), \(P(1)=\frac{1}{6}\), and \(P(\text{other})=\frac{4}{6}=\frac{2}{3}\). 2. Choose the \(2\) positions for the \(6\)s and then the \(3\) positions for the \(1\)s: \(\binom{10}{2}\binom{8}{3}=45\cdot 56\) arrangements. 3. Each qualifying arrangement has probability \(\left(\frac{1}{6}\right)^2\left(\frac{1}{6}\right)^3\left(\frac{2}{3}\right)^5\). 4. Therefore, \(P=\binom{10}{2}\binom{8}{3}\left(\frac{1}{6}\right)^5\left(\frac{2}{3}\right)^5=\frac{280}{6561}\approx 0.0427\).

\(\frac{280}{6561}\approx 0.0427\), or about \(4.27\%\)

- Match each probability base with its exponent. - A binomial coefficient can choose positions for one category. - Why are only \(4\) positions available after the first choice?

1. The exponents show that event \(E\) contains exactly \(8\) light packages, \(3\) standard packages, and \(1\) heavy package. 2. The factor \(\binom{12}{8}\) chooses the \(8\) positions occupied by light packages among the \(12\) selections. 3. After those positions are chosen, \(4\) positions remain. The factor \(\binom{4}{3}\) chooses the \(3\) positions occupied by standard packages. The final position must contain the heavy package.

Event \(E\) is “exactly \(8\) light packages, \(3\) standard packages, and \(1\) heavy package.” The factor \(\binom{12}{8}\) chooses the light-package positions, and \(\binom{4}{3}\) chooses the standard-package positions from the remaining \(4\) positions.

- What values can \(|A|\) have compared with \(n\)? - How many outcomes are in the entire sample space? - If two sets have no elements in common, how many elements are in their union?

1. Since \(|A|\) is an integer from \(0\) to \(n\) and \(n>0\), \(P(A)=\frac{|A|}{n}\ge 0\). 2. Because \(|S|=n\), \(P(S)=\frac{|S|}{n}=\frac{n}{n}=1\). 3. If \(A\cap B=\emptyset\), then \(|A\cup B|=|A|+|B|\). Therefore, \(P(A\cup B)=\frac{|A|+|B|}{n}=\frac{|A|}{n}+\frac{|B|}{n}=P(A)+P(B)\).

1. \(|A|\ge 0\), so \(P(A)=\frac{|A|}{n}\ge 0\). 2. \(P(S)=\frac{n}{n}=1\). 3. For disjoint events, \(|A\cup B|=|A|+|B|\), so \(P(A\cup B)=P(A)+P(B)\).

- Recall the relationship between an event count and its relative frequency. - Can a count of observed outcomes be negative? - Across \(N\) trials, how many outcomes belong to the sample space? - If events are disjoint, can one trial be counted in both events?

1. The count \(H_N(A)\) is nonnegative. Since \(N>0\), \(h_N(A)=\frac{H_N(A)}{N}\ge 0\). 2. Every trial produces an outcome in \(S\), so \(H_N(S)=N\). Thus \(h_N(S)=\frac{N}{N}=1\). 3. If \(A\) and \(B\) are disjoint, no trial can be counted in both events. Therefore, \(H_N(A\cup B)=H_N(A)+H_N(B)\). Dividing by \(N\) gives \(h_N(A\cup B)=\frac{H_N(A)}{N}+\frac{H_N(B)}{N}=h_N(A)+h_N(B)\).

1. \(h_N(A)\ge 0\) because \(H_N(A)\ge 0\) and \(N>0\). 2. \(H_N(S)=N\), so \(h_N(S)=1\). 3. For disjoint events, \(H_N(A\cup B)=H_N(A)+H_N(B)\), so dividing by \(N\) proves additivity.

- How can you write the entire sample space using an event and its complement? - Which axiom gives the probability of the entire sample space? - When \(A\subseteq B\), split \(B\) into two disjoint parts. - Which axiom guarantees that a probability cannot be negative?

1. The events \(A\) and \(A^c\) are disjoint and satisfy \(A\cup A^c=S\). By additivity, \(P(S)=P(A)+P(A^c)\). 2. Since \(P(S)=1\), \(P(A^c)=1-P(A)\). 3. If \(A\subseteq B\), then \(B=A\cup(B\setminus A)\), and the two sets on the right are disjoint. 4. By additivity, \(P(B)=P(A)+P(B\setminus A)\). 5. Nonnegativity gives \(P(B\setminus A)\ge 0\), so \(P(B)\ge P(A)\).

a) \(P(A^c)=1-P(A)\) b) \(P(B)=P(A)+P(B\setminus A)\). Since \(P(B\setminus A)\ge 0\), \(P(A)\le P(B)\).

- What do the probability axioms require about the total probability and the sign of each probability? - Express each event probability as a sum of individual outcome probabilities. - Set up equations for the three unknown probabilities. - Check whether every resulting probability lies between \(0\) and \(1\).

1. Since \(\{e_1,e_2\}\) and \(\{e_3\}\) are disjoint and their union is \(S\), \(0.3+P(e_3)=1\). Thus \(P(e_3)=0.7\). 2. Since \(P(\{e_2,e_3\})=P(e_2)+P(e_3)\), \(0.6=P(e_2)+0.7\), so \(P(e_2)=-0.1\). 3. Since \(P(\{e_1,e_2\})=P(e_1)+P(e_2)\), \(0.3=P(e_1)-0.1\), so \(P(e_1)=0.4\). 4. The values add to \(1\), but \(P(e_2)=-0.1<0\). This violates nonnegativity, so no such probability distribution exists.

No such probability distribution exists. The proposed values imply \(P(e_1)=0.4\), \(P(e_2)=-0.1\), and \(P(e_3)=0.7\). The negative value violates the nonnegativity axiom.

- Separate the source of a probability value from the rules that the value must obey. - Which statement relies on symmetry, and which relies on observations? - Could a machine's defect rate be estimated without collecting production data?

1. Statement I uses an equally likely outcomes model because a fair four-sided die has four equally likely faces. The prime outcomes are \(2\) and \(3\), so the probability is \(\frac{2}{4}=0.5\). 2. Statement II is empirical because the value comes from long-term observed production data. 3. Statement III gives the mathematical rules that every valid probability model must satisfy, regardless of whether its numerical values come from symmetry, data, or another justified model.

a) I uses an equally likely outcomes model because the die is fair and \(2\) of the \(4\) outcomes are prime. II is empirical because the value is estimated from observed production data. b) The probability axioms provide the common mathematical framework that both approaches must satisfy; they do not specify how the numerical probabilities are obtained.

- What must the probabilities of all eight cards add to? - Write one equation that includes all four numbered cards and all four other cards. - Which cards are face cards? - In an equally likely outcomes model, compare favorable outcomes with all possible outcomes.

1. The probabilities must add to \(1\): \(4p+4(1.5p)=1\). Thus \(10p=1\), so \(p=0.10\). 2. Each face card has probability \(1.5\cdot 0.10=0.15\). There are three face cards, so \(P(E)=3\cdot 0.15=0.45\). 3. If all eight cards were equally likely, \(P(E)=\frac{3}{8}=0.375\). The weighted model gives a probability that is \(0.45-0.375=0.075\) greater.

a) \(p=0.10\) b) \(P(E)=0.45\) c) The equally likely outcomes model gives \(P(E)=\frac{3}{8}=0.375\), so the weighted model is higher by \(0.075\).

- List the perfect squares from \(1\) through \(10\). - Relative frequency is the number of occurrences divided by the number of trials. - The probabilities of all possible outcomes must add to \(1\). - Add the probabilities of the odd-numbered outcomes.

1. The perfect squares from \(1\) through \(10\) are \(1\), \(4\), and \(9\). Therefore, \(P(\text{perfect square})=\frac{3}{10}=0.3\). 2. The relative frequency is \(\frac{65}{500}=0.13\). For an equal-section spinner, \(P(10)=\frac{1}{10}=0.1\), so the observed relative frequency is \(0.03\) higher. 3. The probabilities must satisfy \(9q+0.2=1\). Thus, \(q=\frac{0.8}{9}=\frac{4}{45}\). There are \(5\) odd numbers among \(1\) through \(9\), so \(P(\text{odd})=5q=\frac{4}{9}\approx 0.444\).

1. \(P(\text{perfect square})=0.3\) 2. The relative frequency is \(0.13\), which is \(0.03\) higher than the theoretical probability \(0.1\). 3. \(q=\frac{4}{45}\), and \(P(\text{odd})=\frac{4}{9}\approx 0.444\).

- The probabilities of all outcomes in a sample space must add to \(1\). - Compare an equally likely-outcomes model with the given unequal probability model. - Compare the observed relative frequency of \(6\) with its theoretical probability in each model.

1. The probabilities must add to \(1\), so \(3\cdot0.1+3x=1\). Therefore, \(3x=0.7\) and \(x=\frac{7}{30}\approx 0.233\). 2. Since \(E=\{4,5,6\}\), the first model gives \(P(E)=\frac{7}{30}+0.1+\frac{7}{30}=\frac{17}{30}\approx 0.567\). Under the equally likely-outcomes model, \(P(E)=\frac{3}{6}=0.5\). 3. The relative frequency is \(\frac{140}{600}=\frac{7}{30}\approx 0.233\). This equals \(P(6)\) in the first model, while the equally likely-outcomes model gives \(P(6)=\frac{1}{6}\approx 0.167\). Therefore, the first model better matches this observation. One outcome frequency alone does not validate the entire model.

1. \(x=\frac{7}{30}\approx 0.233\) 2. First model: \(P(E)=\frac{17}{30}\approx 0.567\); equally likely-outcomes model: \(P(E)=0.5\) 3. The first model better matches the observed frequency of \(6\), but this observation alone does not confirm the complete model.

- Use ordered pairs for the two rolls. - Count how many ordered pairs produce each sum. - Check nonnegativity and that the probabilities total \(1\).

1. \(S_A=\{(i, j): i, j\in\{1, 2, 3, 4\}\}\), so \(|S_A|=4^2=16\). The die is fair and the rolls are independent, so each ordered pair has probability \(\frac{1}{16}\). 2. \(S_B=\{2,3,4,5,6,7,8\}\). Counting ordered pairs gives probabilities \(\frac{1}{16},\frac{2}{16},\frac{3}{16},\frac{4}{16},\frac{3}{16},\frac{2}{16},\frac{1}{16}\), respectively. 3. Every probability is nonnegative, and their sum is \(\frac{1+2+3+4+3+2+1}{16}=1\). Event probabilities formed by adding the relevant outcome probabilities are additive for disjoint events. 4. In \(S_A\), every outcome is equally likely, so probabilities can be found by counting favorable ordered pairs. The unequal probabilities of sums then follow from how many ordered pairs produce each sum.

a) \(S_A=\{(i, j): i, j\in\{1, 2, 3, 4\}\}\), \(|S_A|=16\), and every pair has probability \(\frac{1}{16}\). b) For sums \(2,3,4,5,6,7,8\), the probabilities are \(\frac{1}{16},\frac{2}{16},\frac{3}{16},\frac{4}{16},\frac{3}{16},\frac{2}{16},\frac{1}{16}\). c) Yes. The probabilities are nonnegative and sum to \(1\). d) The equally likely ordered pairs make probability calculations a counting problem.

- Which individual standard places the strongest limit on the maximum? - Consider the probabilities of failing each standard. - What happens to the minimum number of perfect parts if the three types of failures occur on different parts?

1. The probability of meeting all three standards cannot exceed the smallest individual probability, so \(P(A\cap B\cap C)\le 0.80\). 2. Let \(A^c\), \(B^c\), and \(C^c\) represent failure to meet each standard. Their probabilities are \(0.10\), \(0.15\), and \(0.20\). 3. The probability of failing at least one standard is at most \(0.10+0.15+0.20=0.45\). Therefore, the probability of meeting all three is at least \(1-0.45=0.55\). 4. Both bounds are attainable, so \(P(A\cap B\cap C)\in[0.55,0.80]\).

The probability is at least \(55\%\) and at most \(80\%\).