Free Math Worksheets

- Match the number of assigned outcomes to the desired probability. - Rewrite \(\frac{3}{5}\) and \(\frac{2}{5}\) with denominator \(10\). - More than one valid assignment is possible.

1. A fair coin has two outcomes with probability \(0.5\) each, so assign \(5\) numbers to each outcome. For example, \(\{1,2,3,4,5\}\) represents heads and \(\{6,7,8,9,10\}\) represents tails. 2. In \(E_2\), \(P(\text{red})=\frac{3}{5}=\frac{6}{10}\) and \(P(\text{blue})=\frac{2}{5}=\frac{4}{10}\). Assign \(6\) numbers to red and \(4\) to blue. For example, \(1\) through \(6\) represent red, and \(7\) through \(10\) represent blue.

a) Example: \(\{1,2,3,4,5\}\) for heads and \(\{6,7,8,9,10\}\) for tails b) Example: \(1\) through \(6\) for red and \(7\) through \(10\) for blue

- Determine what happens to the available outcomes after the first draw. - Identify which simulation changes the list of outcomes accordingly.

1. Without replacement, the contents of the bag change after the first draw. 2. Simulation A keeps the probabilities unchanged, so it models drawing with replacement. 3. Simulation B removes the selected outcome, so it correctly models drawing without replacement.

Simulation B.

- Count how many equally likely outcomes must be designated as success. - Determine whether the target probability can be written as \(\frac{k}{6}\) for an integer \(k\). - Two rolls create \(36\) equally likely ordered pairs.

1. A fair die has \(6\) equally likely outcomes. Designate any \(2\) outcomes, such as \(1\) and \(2\), as success. Then \(P(\text{success})=\frac{2}{6}=\frac{1}{3}\). 2. Any event based on one roll must contain an integer number \(k\) of the \(6\) outcomes, so its probability must be \(\frac{k}{6}\). The equation \(\frac{k}{6}=0.4\) gives \(k=2.4\), not an integer, so exact simulation is impossible. 3. Two rolls produce \(36\) equally likely ordered pairs. Designate \(9\) pairs as success. For example, define success as both rolls being at most \(3\). This gives \(3\cdot3=9\) successful pairs and probability \(\frac{9}{36}=0.25\).

a) Designate two die results, such as \(1\) and \(2\), as success. b) One-roll event probabilities must be multiples of \(\frac{1}{6}\), and \(0.4\) is not such a multiple. c) Roll twice and call it a success when both results are at most \(3\); the probability is \(\frac{9}{36}=0.25\).

- Each simulated arrangement should be a uniformly random permutation. - Divide successful trials by total trials. - Distinguish long-run stabilization from a guarantee for one finite simulation.

1. In each trial, generate a random permutation of \(1\) through \(100\). Check whether any position \(i\) contains the value \(i\), record a success if so, and repeat many times. 2. The relative frequency is \(\frac{3162}{5000}=0.6324\). 3. The estimate differs from \(0.6321\) by only \(0.0003\), so this simulation happened to be very close. The law of large numbers explains why estimates tend to stabilize as the number of trials grows, but it does not guarantee such a small error in every \(5000\)-trial simulation.

a) Repeatedly generate a random permutation and check whether at least one value is in its own position. b) \(0.6324\) c) The estimate is very close to \(0.6321\), but a finite simulation has no guarantee of that exact level of accuracy.

- A fair simulation must produce all \(12\) outcomes with probability \(\frac{1}{12}\). - Combine devices whose numbers of equally likely outcomes multiply to \(12\). - Check both the range and the probability distribution.

1. The sum ranges only from \(2\) through \(10\), and the sums are not equally likely. Therefore, it is not a valid simulation. 2. Method 1: Toss a coin and roll a six-sided die. If the coin shows heads, use the die result \(1\) through \(6\). If it shows tails, add \(6\) to obtain \(7\) through \(12\). Each outcome has probability \(\frac{1}{2}\cdot\frac{1}{6}=\frac{1}{12}\). 3. Method 2: Roll a four-sided die and a six-sided die. Convert the six-sided result to one of three equally likely values by mapping \(1,2\) to \(1\), \(3,4\) to \(2\), and \(5,6\) to \(3\). If the four-sided result is \(T\) and the converted value is \(R\), output \(3(T-1)+R\). The \(4\cdot3=12\) combinations are equally likely.

1. No. The sums range from \(2\) to \(10\) and are not equally likely. 2. Use a coin with a six-sided die, or use a four-sided die with a six-sided die grouped into three equally likely outcomes.

- Use combinations because order does not matter. - A coin and die together can create \(12\) equally likely outcomes. - Reject repeated numbers to model drawing without replacement.

1. The number of combinations is \(\binom{12}{3}=220\). 2. Generate one number from \(1\) through \(12\) by tossing the coin and rolling the die: use the die result after heads and add \(6\) after tails. Repeat until three distinct numbers have been generated. If a number repeats within the same simulated drawing, discard that result and generate another number. Because each remaining number is equally likely at every accepted stage, every three-number set is equally likely. The final three-number set is the simulated lottery outcome.

1. \(220\) 2. Use the coin to choose \(1\)–\(6\) or \(7\)–\(12\), use the die to select within that range, and reject repeats until three distinct numbers are obtained.

- Count the total number of letters and the number of occurrences of each letter. - A sequence of \(n\) letters contains \(n-1\) overlapping adjacent pairs. - Use independence to find the probability of each ordered pair. - Compare each observed pair count with its expected count.

1. The sequence contains \(9\) occurrences of \(A\) and \(11\) occurrences of \(B\). Thus, the relative frequencies are \(\frac{9}{20}=0.45\) for \(A\) and \(\frac{11}{20}=0.55\) for \(B\). 2. Counting the \(19\) adjacent pairs gives counts of \(2\) for \(AA\), \(7\) for \(AB\), \(6\) for \(BA\), and \(4\) for \(BB\). The corresponding relative frequencies are \(\frac{2}{19}\approx 0.105\), \(\frac{7}{19}\approx 0.368\), \(\frac{6}{19}\approx 0.316\), and \(\frac{4}{19}\approx 0.211\). 3. Because consecutive letters are independent and each letter has probability \(0.5\), every ordered pair has probability \(0.5\cdot 0.5=0.25\). The expected count for each pair is \(19\cdot 0.25=4.75\). Compared with this value, \(AA\) and \(BB\) occur less often, while \(AB\) and \(BA\) occur more often.

1. The relative frequencies are \(0.45\) for \(A\) and \(0.55\) for \(B\). 2. The counts are \(2\) for \(AA\), \(7\) for \(AB\), \(6\) for \(BA\), and \(4\) for \(BB\); the relative frequencies are approximately \(0.105\), \(0.368\), \(0.316\), and \(0.211\), respectively. 3. The expected count is \(4.75\) for each pair. The observed counts of \(AA\) and \(BB\) are lower, and those of \(AB\) and \(BA\) are higher.

- A sequence of \(24\) results contains \(23\) overlapping adjacent pairs. - For independent fair tosses, find the probability of each two-toss pair. - Compare the expected pair counts with the observed counts. - Consider both the frequencies of individual outcomes and the pattern of consecutive outcomes.

1. The sequence contains \(12\) heads and \(12\) tails, so the relative frequency of each outcome is \(\frac{12}{24}=0.5\). 2. There are \(23\) overlapping adjacent pairs. The counts are \(0\) for \(HH\), \(12\) for \(HT\), \(11\) for \(TH\), and \(0\) for \(TT\). 3. For a fair coin, each ordered pair has probability \(0.25\), so the expected count of \(HH\) is \(23\cdot 0.25=5.75\), and the expected count of \(TT\) is also \(5.75\). The sequence contains neither pair and alternates perfectly throughout. Although its single-outcome frequencies are balanced, its pair pattern is extremely atypical of independent fair-coin tosses and suggests deliberate alternation rather than random generation.

1. The relative frequencies of heads and tails are both \(0.5\). 2. The counts are \(0\) for \(HH\), \(12\) for \(HT\), \(11\) for \(TH\), and \(0\) for \(TT\). 3. The sequence is extremely atypical of fair, independent coin tosses. The expected counts of \(HH\) and \(TT\) are each \(5.75\), but both observed counts are \(0\).

- Assign the ten equally likely digits in the same proportions as the colors in the bag. - Red can occur exactly once in either of two positions.

1. One valid assignment is red \(\{0, 1, 2\}\) and blue \(\{3, 4, 5, 6, 7, 8, 9\}\). 2. A favorable ordered pair has its first digit in the red set and its second digit in the blue set, or vice versa. 3. There are \(3\cdot7+7\cdot3=42\) favorable ordered pairs out of \(100\). 4. Therefore, the simulated probability is \(\frac{42}{100}=0.42\).

For example, assign \(\{0, 1, 2\}\) to red and \(\{3, 4, 5, 6, 7, 8, 9\}\) to blue. The favorable ordered pairs are \((r, b)\) and \((b, r)\), where \(r\in\{0, 1, 2\}\) and \(b\in\{3, 4, 5, 6, 7, 8, 9\}\). There are \(42\) such pairs, so the probability is \(0.42\).