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Discrete random variables

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52380512
A spinner has three equal sections labeled \(1\), \(2\), and \(3\). The spinner is spun twice. The random variable \(S\) is the sum of the two results. a) Create a probability distribution table for \(S\). b) Find the probability that the sum is at least \(4\).

Hints

- List the ordered pairs that can result from two spins. - Determine how many equally likely outcomes there are. - Organize the outcomes in a list or grid to find each possible sum. - Translate “at least” into an inequality before selecting outcomes.

Solution

1. Each result has probability \(\frac{1}{3}\), so the two spins have \(3\cdot3=9\) equally likely ordered outcomes: \((1, 1)\), \((1, 2)\), \((1, 3)\), \((2, 1)\), \((2, 2)\), \((2, 3)\), \((3, 1)\), \((3, 2)\), and \((3, 3)\). 2. Count the outcomes for each sum: \(S=2\) occurs once, \(S=3\) twice, \(S=4\) three times, \(S=5\) twice, and \(S=6\) once. Therefore, the probabilities are \(\frac{1}{9}\), \(\frac{2}{9}\), \(\frac{3}{9}\), \(\frac{2}{9}\), and \(\frac{1}{9}\), respectively. 3. For a sum of at least \(4\), add the probabilities for \(S=4\), \(S=5\), and \(S=6\): \(P(S\ge4)=\frac{3}{9}+\frac{2}{9}+\frac{1}{9}=\frac{6}{9}=\frac{2}{3}\).

Answer

a) <table border="1"> <tr> <td>\(s\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> <td>\(5\)</td> <td>\(6\)</td> </tr> <tr> <td>\(P(S=s)\)</td> <td>\(\frac{1}{9}\)</td> <td>\(\frac{2}{9}\)</td> <td>\(\frac{3}{9}\)</td> <td>\(\frac{2}{9}\)</td> <td>\(\frac{1}{9}\)</td> </tr> </table> b) \(P(S\ge4)=\frac{2}{3}\approx66.7\%\)
52511312
Determine whether each random variable is discrete or continuous. Briefly justify each answer by describing its possible values. a) The time for a \(100\,\text{m}\) sprint, assuming unlimited measurement precision. b) The number of defective circuit boards in a batch of \(1000\). c) The exact tire pressure of a car, measured in \(\text{bar}\). d) The sum obtained when three standard number cubes are rolled at the same time.

Hints

- Decide whether the variable can take any value in an interval or only separated values. - Can the possible outcomes be counted? - Does measurement precision change the theoretical type of the variable? - Distinguish a count from a physical measurement.

Solution

1. a) Sprint time is continuous because it can take any real value within a reasonable interval. 2. b) The number of defective boards is discrete because its possible values are the integers in \(\{0,1,\ldots,1000\}\). 3. c) Tire pressure is continuous because, in theory, it can take any real value within an interval. 4. d) The sum is discrete because it can take only the integer values from \(3\) through \(18\).

Answer

a) continuous b) discrete c) continuous d) discrete
52511412
A city planning office collects data about local development. Decide whether a discrete or continuous model is appropriate for each random variable. a) The number of residents registered in each apartment building. b) The exact area of a parcel of land in square meters. c) A household's monthly electric bill, recorded to the nearest cent. d) A visitor's waiting time at a city service office when the time is recorded by rounding down to a whole number of minutes.

Hints

- Imagine marking the possible values on a number line. Are they separated points or a continuous interval? - Pay attention to rounding and minimum units such as one cent. - Is the set of possible recorded values countable, or does it contain every real value in an interval?

Solution

1. a) The number of residents is discrete because people are counted in whole numbers. 2. b) Land area is continuous because an area measurement can theoretically take any real value within an interval. 3. c) The electric bill is discrete as recorded because the possible values differ by \(\$0.01\). 4. d) The recorded waiting time is discrete because rounding down produces only whole-number values \(\{0,1,2,\ldots\}\), even though the underlying waiting time is continuous.

Answer

a) discrete b) continuous c) discrete d) discrete
52514112
One probability model describes the number of children per household in a city. A second model describes residents' commuting times to work, measured theoretically with exact precision in minutes. Explain why a bar graph is appropriate for the first probability distribution, while a probability density curve is appropriate for the second. In your explanation, discuss the sets of possible values.

Hints

- Can all possible numbers of children be listed individually? - What happens to the possible commuting times as measurement precision increases? - How are probabilities represented for discrete and continuous random variables?

Solution

1. The number of children is a discrete random variable. Its possible values are separated, countable numbers such as \(0,1,2,\ldots\). Each value can be assigned a probability \(P(X=k)\), which can be represented by the height of a bar. 2. Exact commuting time is a continuous random variable. It can take any real value in an appropriate interval, so its set of possible values is uncountable. 3. For a continuous random variable, \(P(X=x)=0\) at every single value. Interval probabilities are represented by areas under a density curve, making a probability density function appropriate.

Answer

The number of children is discrete and has separated, countable values, so individual probabilities can be shown as bar heights. Exact commuting time is continuous. Since individual values have probability \(0\), interval probabilities are represented by areas under a density curve.
53085312
An irregular four-sided die has faces labeled \(1\), \(2\), \(3\), and \(4\). The table gives part of the probability distribution of the random variable \(X\), the number rolled. <table> <tr> <td>\(k\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(0.18\)</td> <td>\(0.32\)</td> <td>\(p\)</td> <td>\(0.25\)</td> </tr> </table> 1. Determine the probability \(p\). 2. Find the probability that the number rolled is odd. 3. Find the probability that the number rolled is at least \(2\).

Hints

- Recall the required sum of all probabilities in a distribution. - Identify the odd face values. - Translate “at least” into an inequality. - Consider using a complement for the last part.

Solution

1. The probabilities must add to \(1\): \(0.18+0.32+p+0.25=1\). Therefore, \(0.75+p=1\), so \(p=0.25\). 2. The odd outcomes are \(1\) and \(3\). Thus, \(P(X\in\{1,3\})=0.18+0.25=0.43\). 3. The outcomes at least \(2\) are \(2\), \(3\), and \(4\). Thus, \(P(X\ge2)=0.32+0.25+0.25=0.82\). Equivalently, \(P(X\ge2)=1-P(X=1)=1-0.18=0.82\).

Answer

1. \(p=0.25\) 2. \(P(\text{odd})=0.43\) 3. \(P(X\ge2)=0.82\)
53096112
An urn contains five balls labeled \(1\) through \(5\). Two balls are drawn in order without replacement. The random variable \(X\) is the absolute difference between the two numbers drawn. Determine the set of all possible values of \(X\).

Hints

- List pairs of different numbers that could be drawn. - Find the absolute difference for each pair. - Identify the smallest and largest possible differences, then check whether every integer between them can occur.

Solution

1. Because the balls are drawn without replacement, the two numbers must be different. 2. The smallest possible absolute difference is \(|1-2|=1\), and the largest is \(|1-5|=4\). 3. Each integer difference between them is possible: \(|1-2|=1\), \(|1-3|=2\), \(|1-4|=3\), and \(|1-5|=4\). 4. Therefore, the set of possible values is \(\{1, 2, 3, 4\}\).

Answer

\(\{1, 2, 3, 4\}\)
53097712
A customer service center recorded visitor wait times. The random variable \(X\) is the wait time, in minutes, for a randomly selected visitor. The table gives cumulative probabilities. <table> <thead> <tr> <th>Wait time \(x\), in minutes</th> <th>\(P(X\le x)\)</th> </tr> </thead> <tbody> <tr><td>\(2\)</td><td>\(0.18\)</td></tr> <tr><td>\(5\)</td><td>\(0.45\)</td></tr> <tr><td>\(10\)</td><td>\(0.72\)</td></tr> <tr><td>\(15\)</td><td>\(0.88\)</td></tr> <tr><td>\(20\)</td><td>\(0.96\)</td></tr> <tr><td>\(30\)</td><td>\(1.00\)</td></tr> </tbody> </table> Use the table to determine each probability. a) A visitor waits more than \(10\) minutes. b) \(P(5<X\le20)\) c) A visitor waits more than \(2\) minutes but no more than \(10\) minutes.

Hints

- Interpret each table entry as a cumulative probability. - Relate “more than \(x\)” to the complement of “at most \(x\).” - To find a probability between two cutoffs, subtract the cumulative value at the lower cutoff from the value at the upper cutoff. - Identify which visitors are included up to the upper cutoff but not up to the lower cutoff.

Solution

1. Use the complement: \(P(X>10)=1-P(X\le10)=1-0.72=0.28\). 2. Subtract cumulative probabilities: \(P(5<X\le20)=P(X\le20)-P(X\le5)=0.96-0.45=0.51\). 3. Similarly, \(P(2<X\le10)=P(X\le10)-P(X\le2)=0.72-0.18=0.54\).

Answer

a) \(P(X>10)=0.28\) b) \(P(5<X\le20)=0.51\) c) \(P(2<X\le10)=0.54\)
53115512
A spinner has five equal sections labeled \(1\), \(1\), \(2\), \(5\), and \(10\). The spinner is spun once. The random variable \(X\) assigns each section its displayed number. 1. Give the sample space \(S\) and the set of possible values of \(X\). 2. Create a probability distribution table for \(X\). 3. Briefly explain how to construct a probability histogram for this distribution and what each bar height represents when every bar has width \(1\).

Hints

- Distinguish the physical spinner sections from the numerical values assigned to them. - Count how many sections display each value. - Check that all probabilities add to \(1\). - Recall which quantities appear on the axes of a probability histogram.

Solution

1. Treat the five physical sections as distinguishable outcomes, so \(S=\{s_1,s_2,s_3,s_4,s_5\}\). The set of possible values is \(\{1,2,5,10\}\). 2. Two of the five sections show \(1\), so \(P(X=1)=\frac{2}{5}=0.4\). Each other value appears once, so \(P(X=2)=P(X=5)=P(X=10)=\frac{1}{5}=0.2\). 3. Place the possible values of \(X\) on the x-axis and probabilities on the y-axis. Draw one width-\(1\) bar at each possible value. The height, and therefore the area, of each bar equals \(P(X=k)\).

Answer

1. \(S=\{s_1,s_2,s_3,s_4,s_5\}\); the possible values are \(\{1,2,5,10\}\). 2. <table> <tr><td>\(k\)</td><td>\(1\)</td><td>\(2\)</td><td>\(5\)</td><td>\(10\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.4\)</td><td>\(0.2\)</td><td>\(0.2\)</td><td>\(0.2\)</td></tr> </table> 3. Draw a width-\(1\) bar at each possible value. Its height and area equal the corresponding probability.
52512312
A manufacturer produces metal rods. The length \(L\), in centimeters, of a rod is modeled as a continuous normal random variable with \(L\sim N(100,0.1^2)\). The number \(X\) of defective rods in a daily production run of \(500\) is modeled as a discrete binomial random variable with \(p=0.02\). 1. Explain the main difference between the sets of possible values for \(L\) and \(X\). 2. Find \(P(L=100)\). 3. Find \(P(X=10)\). 4. Explain mathematically why a continuous random variable such as \(L\) always has probability \(0\) at an exact value, while a discrete random variable such as \(X\) may have positive probability at an exact value.

Hints

- Compare a finite list of integers with an unbroken interval of real numbers. - What is the area under a curve over an interval of width \(0\)? - Recall the binomial probability formula. - Compare probability represented by area with probability assigned to individual outcomes.

Solution

1. The possible values of \(X\) are the finite set \(\{0,1,2,\ldots,500\}\). The normal random variable \(L\) can take any real value in its model. 2. For any continuous random variable, \(P(L=x)=0\) at a single value, so \(P(L=100)=0\). 3. Using the binomial probability formula, \(P(X=10)=\binom{500}{10}(0.02)^{10}(0.98)^{490}\approx 0.1264\). 4. For a continuous random variable, probability is area under a density curve: \(P(a\le L\le b)=\int_a^b f(x)\,dx\). When \(a=b\), the interval has width \(0\), so the integral is \(0\). A discrete probability distribution can assign positive probability directly to individual values.

Answer

1. \(X\) has finitely many separated values, while \(L\) can take any real value in its model. 2. \(P(L=100)=0\) 3. \(P(X=10)\approx 0.1264\) 4. A continuous probability is an area over an interval, and a single point has zero width; a discrete distribution can assign positive probability to an individual value.
52512412
Consider two random experiments. Experiment A: \(Y\) is the sum when two fair six-sided number cubes are rolled. Experiment B: \(Z\) is normally distributed with \(\mu=7\) and \(\sigma=2.4\). 1. Classify \(Y\) and \(Z\) as discrete or continuous, and justify each classification. 2. Find \(P(Y\le 3)\). 3. Find \(P(6.5\le Z\le 7.5)\). 4. Explain how including endpoints, such as comparing \(P(X\le k)\) with \(P(X<k)\), affects probabilities for each type of random variable.

Hints

- List the possible sums from rolling two number cubes. - Use standardization or a normal cumulative distribution function for \(Z\). - What area does one isolated point contribute under a density curve? - How many ordered rolls produce a sum of at most \(3\)?

Solution

1. \(Y\) is discrete because its possible values are the integers \(\{2,3,\ldots,12\}\). \(Z\) is continuous because a normal random variable can take any real value. 2. The outcomes with \(Y\le 3\) are \((1, 1)\), \((1, 2)\), and \((2, 1)\). There are \(6\cdot 6=36\) equally likely ordered outcomes, so \(P(Y\le 3)=\frac{3}{36}=\frac{1}{12}\approx 0.0833\). 3. Standardize the endpoints: \(z_1=\frac{6.5-7}{2.4}\approx -0.2083\) and \(z_2=\frac{7.5-7}{2.4}\approx 0.2083\). Therefore, \(P(6.5\le Z\le 7.5)=\Phi(0.2083)-\Phi(-0.2083)\approx 0.1650\). 4. For discrete \(Y\), an endpoint may have positive probability, so \(P(Y\le 3)=P(Y<3)+P(Y=3)\). For continuous \(Z\), \(P(Z=k)=0\), so including or excluding an endpoint does not change the probability.

Answer

1. \(Y\) is discrete; \(Z\) is continuous. 2. \(P(Y\le 3)=\frac{1}{12}\approx 0.0833\) 3. \(P(6.5\le Z\le 7.5)\approx 0.1650\) 4. Endpoints can change a discrete probability because \(P(X=k)\) may be positive; they do not change a continuous probability because \(P(X=k)=0\).
52512612
Classify each random variable as discrete or continuous. Also state whether its set of possible values is finite, countably infinite, or uncountably infinite. a) The angle, in degrees, at which an ideal spinner with an infinitely thin pointer stops. b) The amount of money in a piggy bank with a fixed capacity, recorded to the nearest cent. c) The exact lifetime of a radioactive particle before it decays. d) The number of daylight hours at a location, rounded to one decimal place.

Hints

- Between two possible values, can you always find more possible values? - Distinguish the physical quantity from the way it is recorded or rounded. - A countably infinite set can be listed in a sequence like the natural numbers. - Intervals of real numbers are uncountably infinite.

Solution

1. a) The angle is continuous because any real value in \([0,360)\) is possible. The set of values is uncountably infinite. 2. b) The recorded amount is discrete because values change in increments of \(\$0.01\). The fixed capacity makes the set of possible values finite. 3. c) The lifetime is continuous because it can take any positive real value. The set of values is uncountably infinite. 4. d) The rounded daylight hours are discrete because only values on a tenth-hour grid within the physical range from \(0.0\) through \(24.0\) can be recorded. The set is finite.

Answer

a) continuous; uncountably infinite b) discrete; finite c) continuous; uncountably infinite d) discrete; finite
52537312
A discrete random variable \(X\) has a binomial distribution with \(n=100\) and \(p=0.5\). A continuous random variable \(Y\) is normally distributed with \(\mu=50\) and \(\sigma=5\). 1. Find \(P(45\le X\le 55)\) and \(P(45<X<55)\). 2. Find \(P(45\le Y\le 55)\) using the empirical rule or a standard normal table. 3. Explain the key difference between discrete and continuous random variables by comparing probabilities such as \(P(X=45)\) and \(P(Y=45)\).

Hints

- Use differences of cumulative binomial probabilities for each interval. - How does removing the endpoints affect a sum compared with an integral? - Recall that continuous probabilities are areas under density curves. - What area is contributed by one vertical line above a fixed x-value?

Solution

1. For the binomial random variable, \(P(45\le X\le 55)=P(X\le 55)-P(X\le 44)\approx 0.7287\). Also, \(P(45<X<55)=P(X\le 54)-P(X\le 45)\approx 0.6318\). The probabilities differ because the endpoints \(45\) and \(55\) have positive probability. 2. The interval \([45,55]\) is \([\mu-\sigma,\mu+\sigma]\). By the empirical rule, \(P(45\le Y\le 55)\approx 0.6827\). 3. A discrete random variable may assign positive probability to an individual value; here, \(P(X=45)\approx 0.0485\). For a continuous random variable, \(P(Y=45)=0\). Therefore, including or excluding endpoints changes a discrete probability but not a continuous probability.

Answer

1. \(P(45\le X\le 55)\approx 0.7287\) and \(P(45<X<55)\approx 0.6318\) 2. \(P(45\le Y\le 55)\approx 0.6827\), or about \(68.3\%\) 3. A discrete variable can have \(P(X=k)>0\), so endpoints can matter. A continuous variable has \(P(Y=k)=0\), so endpoints do not affect interval probabilities.
53096212
A contest has three problems. Each problem earns \(0\) points for an incorrect solution, \(3\) points for a partially correct solution, or \(10\) points for a fully correct solution. The random variable \(X\) is a contestant’s total score on the three problems. Give the set of all possible values of \(X\).

Hints

- Start with the possible scores on one problem. - Organize the cases by the number of \(10\)-point scores. - List all totals formed from three allowed scores. - Include each total only once in the final set.

Solution

1. The possible score on each problem is in \(\{0, 3, 10\}\). 2. With no \(10\)-point scores, the possible totals are \(0\), \(3\), \(6\), and \(9\). 3. With exactly one \(10\)-point score, the possible totals are \(10\), \(13\), and \(16\). 4. With exactly two \(10\)-point scores, the possible totals are \(20\) and \(23\). 5. With three \(10\)-point scores, the total is \(30\). 6. Combining the distinct totals gives \(\{0, 3, 6, 9, 10, 13, 16, 20, 23, 30\}\).

Answer

\(\{0, 3, 6, 9, 10, 13, 16, 20, 23, 30\}\)
53097212
An urn contains five balls labeled \(1\), \(2\), \(3\), \(4\), and \(5\). Two balls are drawn with replacement. Define the random variable \(Z\) as follows: - \(Z=10\) if the sum of the two numbers is greater than \(7\). - \(Z=5\) if the sum is exactly \(7\). - \(Z=-5\) if the sum is less than \(7\). Determine the probability distribution of \(Z\).

Hints

- Count all ordered pairs from two draws with replacement. - List the pairs whose sum is exactly \(7\). - Use a \(5 \times 5\) grid or another systematic list to avoid missing pairs. - Check that the three probabilities add to \(1\).

Solution

1. There are \(5\cdot5=25\) equally likely ordered pairs. 2. A sum of \(7\) occurs for \((2, 5)\), \((3, 4)\), \((4, 3)\), and \((5, 2)\), so \(P(Z=5)=\frac{4}{25}=0.16\). 3. A sum less than \(7\) occurs for \(15\) ordered pairs, so \(P(Z=-5)=\frac{15}{25}=0.60\). 4. A sum greater than \(7\) occurs for the remaining \(6\) ordered pairs, so \(P(Z=10)=\frac{6}{25}=0.24\).

Answer

<table border="1"> <tr> <td>\(z\)</td> <td>\(-5\)</td> <td>\(5\)</td> <td>\(10\)</td> </tr> <tr> <td>\(P(Z=z)\)</td> <td>\(0.60\)</td> <td>\(0.16\)</td> <td>\(0.24\)</td> </tr> </table>
53097812
A delivery company compares the annual mileage of two vehicle types. The random variables \(X_1\) and \(X_2\) represent annual mileage, in thousands of miles, for a delivery van and a heavy truck, respectively. The table gives cumulative relative frequencies. <table> <thead> <tr> <th>Annual mileage up to ... thousand miles</th> <th>Delivery van \(X_1\)</th> <th>Heavy truck \(X_2\)</th> </tr> </thead> <tbody> <tr><td>\(10\)</td><td>\(0.12\)</td><td>\(0.04\)</td></tr> <tr><td>\(20\)</td><td>\(0.38\)</td><td>\(0.15\)</td></tr> <tr><td>\(30\)</td><td>\(0.65\)</td><td>\(0.32\)</td></tr> <tr><td>\(40\)</td><td>\(0.84\)</td><td>\(0.58\)</td></tr> <tr><td>\(50\)</td><td>\(0.95\)</td><td>\(0.82\)</td></tr> <tr><td>\(60\)</td><td>\(1.00\)</td><td>\(1.00\)</td></tr> </tbody> </table> a) Find the probability that a randomly selected heavy truck travels more than \(40{,}000\) miles in a year. b) Compare \(P(20<X_1\le40)\) and \(P(20<X_2\le40)\). Which vehicle type is more likely to have annual mileage in this interval? c) For delivery vans, which interval has the greatest probability: \([0, 20]\), \((20, 40]\), or \((40, 60]\)?

Hints

- Each table entry already includes all observations up to that mileage. - Find an interval probability by subtracting cumulative values at its endpoints. - Read the two vehicle columns separately. - Compute the probability in each proposed interval before comparing them.

Solution

1. \(P(X_2>40)=1-P(X_2\le40)=1-0.58=0.42\). 2. For delivery vans, \(P(20<X_1\le40)=0.84-0.38=0.46\). For heavy trucks, \(P(20<X_2\le40)=0.58-0.15=0.43\). The delivery-van probability is greater. 3. For delivery vans, the three interval probabilities are \(0.38\), \(0.84-0.38=0.46\), and \(1.00-0.84=0.16\). Therefore, \((20, 40]\) has the greatest probability.

Answer

a) \(P(X_2>40)=0.42\) b) \(P(20<X_1\le40)=0.46\) and \(P(20<X_2\le40)=0.43\); delivery vans have the greater probability. c) The interval \((20, 40]\) has the greatest probability, \(0.46\).
53115612
Two fair four-sided dice labeled \(1\) through \(4\) are rolled. The random variable \(Y\) is the sum of the two results. 1. Determine the set of possible values of \(Y\). 2. Calculate the probability distribution of \(Y\) and present it in a table. 3. Use this example to explain the difference between an outcome of the experiment and a value of the random variable.

Hints

- Organize the \(16\) ordered outcomes in a grid. - Count how many outcomes produce each possible sum. - Decide whether a sum such as \(5\) is itself an outcome or a quantity calculated from an outcome. - Think of a random variable as a function from outcomes to numbers.

Solution

1. The smallest sum is \(1+1=2\), and the largest is \(4+4=8\). Thus, the possible values are \(\{2,3,4,5,6,7,8\}\). 2. There are \(4\cdot4=16\) equally likely ordered outcomes. The numbers of outcomes producing sums \(2\) through \(8\) are \(1,2,3,4,3,2,1\), respectively. Dividing by \(16\) gives the distribution. 3. An outcome is an ordered pair such as \((1, 3)\), describing the result on each die. The random variable maps that outcome to the value \(Y=4\). Several different outcomes can produce the same value of \(Y\).

Answer

1. \(\{2,3,4,5,6,7,8\}\) 2. <table> <tr><td>\(k\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td></tr> <tr><td>\(P(Y=k)\)</td><td>\(\frac{1}{16}\)</td><td>\(\frac{2}{16}\)</td><td>\(\frac{3}{16}\)</td><td>\(\frac{4}{16}\)</td><td>\(\frac{3}{16}\)</td><td>\(\frac{2}{16}\)</td><td>\(\frac{1}{16}\)</td></tr> </table> 3. An outcome is a specific ordered pair, while a value of \(Y\) is the sum assigned to that outcome. Different outcomes may have the same sum.
53118312
A fair six-sided die and a fair eight-sided die, each labeled consecutively starting at \(1\), are rolled. The random variable \(X\) is the sum of the two results. a) Determine the probability distribution of \(X\) in a table. b) Find the probability that the sum is a perfect square. c) Find \(P(5<X\le9)\).

Hints

- Count the total ordered outcomes. - Use a grid to tally each possible sum. - Identify the perfect squares between \(2\) and \(14\). - Translate the compound inequality carefully.

Solution

1. There are \(6\cdot8=48\) equally likely ordered outcomes, and the possible sums are \(2\) through \(14\). 2. The numbers of outcomes producing these sums are \(1,2,3,4,5,6,6,6,5,4,3,2,1\), respectively. Divide each count by \(48\) to obtain the distribution. 3. The perfect-square sums are \(4\) and \(9\). Thus, the probability is \(\frac{3}{48}+\frac{6}{48}=\frac{3}{16}=0.1875\). 4. The sums satisfying \(5<X\le9\) are \(6\), \(7\), \(8\), and \(9\). Their probability is \(\frac{5+6+6+6}{48}=\frac{23}{48}\approx0.4792\).

Answer

a) <table> <tr><td>\(x\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(7\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td><td>\(11\)</td><td>\(12\)</td><td>\(13\)</td><td>\(14\)</td></tr> <tr><td>\(P(X=x)\)</td><td>\(\frac{1}{48}\)</td><td>\(\frac{2}{48}\)</td><td>\(\frac{3}{48}\)</td><td>\(\frac{4}{48}\)</td><td>\(\frac{5}{48}\)</td><td>\(\frac{6}{48}\)</td><td>\(\frac{6}{48}\)</td><td>\(\frac{6}{48}\)</td><td>\(\frac{5}{48}\)</td><td>\(\frac{4}{48}\)</td><td>\(\frac{3}{48}\)</td><td>\(\frac{2}{48}\)</td><td>\(\frac{1}{48}\)</td></tr> </table> b) \(P(\text{perfect square})=\frac{3}{16}=0.1875\) c) \(P(5<X\le9)=\frac{23}{48}\approx0.4792\)
53602112
The spinner shown has \(6\) equal sections. It is spun twice. The random variable \(Y\) is the product of the two numbers. Create a probability distribution table for \(Y\).
Figure for problem 536021

Hints

- Determine the probability of each number on one spin. - List the different products that two spins can produce. - A tree diagram can help organize all ordered pairs. - Several different ordered pairs produce a product of \(0\). - Check that the probabilities add to \(1\).

Solution

1. On one spin, \(P(0)=\frac{1}{6}\), \(P(1)=\frac{2}{6}=\frac{1}{3}\), and \(P(2)=\frac{3}{6}=\frac{1}{2}\). 2. The possible products are \(0\), \(1\), \(2\), and \(4\). 3. The product is \(0\) when at least one spin is \(0\), so \(P(Y=0)=1-\left(\frac{5}{6}\right)^2=\frac{11}{36}\). 4. \(P(Y=1)=\left(\frac{1}{3}\right)^2=\frac{1}{9}\). 5. \(P(Y=2)=2\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)=\frac{1}{3}\). 6. \(P(Y=4)=\left(\frac{1}{2}\right)^2=\frac{1}{4}\). 7. The probabilities add to \(1\).

Answer

<table> <tr><td>\(y\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(4\)</td></tr> <tr><td>\(P(Y=y)\)</td><td>\(\frac{11}{36}\)</td><td>\(\frac{1}{9}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{4}\)</td></tr> </table>
52575012
An urn contains three balls labeled \(1\), \(2\), and \(3\). A ball is drawn three times with replacement. Let the results be \(z_1\), \(z_2\), and \(z_3\), and define the partial sums \(s_1=z_1\), \(s_2=z_1+z_2\), and \(s_3=z_1+z_2+z_3\). The random variable \(X\) is the number of these three partial sums that are even. Determine the probability distribution of \(X\) and present it in a table.

Hints

- Build a tree that records only whether each draw is odd or even. - At each endpoint, determine the parity of \(s_1\), \(s_2\), and \(s_3\), then count the even partial sums. - The probabilities of an even draw and an odd draw are not equal. - Check that all probabilities in the distribution add to \(1\).

Solution

1. A draw is even with probability \(\frac{1}{3}\) and odd with probability \(\frac{2}{3}\). Let \(E\) and \(O\) represent the parity of each draw. 2. Track the parity of the partial sums for each draw-parity pattern. The pattern \(OEE\) gives no even partial sums, so \(P(X=0)=\frac{2}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}=\frac{2}{27}\). 3. The patterns \(OEO\), \(OOO\), and \(EOE\) each give exactly one even partial sum. Their total probability is \(\frac{4}{27}+\frac{8}{27}+\frac{2}{27}=\frac{14}{27}\). 4. The patterns \(OOE\), \(EOO\), and \(EEO\) each give exactly two even partial sums. Their total probability is \(\frac{4}{27}+\frac{4}{27}+\frac{2}{27}=\frac{10}{27}\). 5. The pattern \(EEE\) gives three even partial sums, so \(P(X=3)=\frac{1}{27}\). 6. The probabilities add to \(\frac{2+14+10+1}{27}=1\).

Answer

<table border="1"> <tr> <td>\(k\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(\frac{2}{27}\)</td> <td>\(\frac{14}{27}\)</td> <td>\(\frac{10}{27}\)</td> <td>\(\frac{1}{27}\)</td> </tr> </table>
53097612
In a table tennis match between Maya and Noah, Maya wins each game with probability \(0.6\). The match is best-of-five, so the first player to win \(3\) games wins the match. a) Determine the probability distribution of the random variable \(X\), the total number of games played. b) Find the probability that Maya wins the match.

Hints

- The two players do not have equal game-winning probabilities. - To win in exactly \(4\) games, the winner must win game \(4\) and exactly \(2\) of the first \(3\) games. - Add Maya’s probabilities of winning in \(3\), \(4\), and \(5\) games. - Check that the probabilities of all possible match lengths add to \(1\).

Solution

1. The possible match lengths are \(3\), \(4\), and \(5\) games. Let \(p=0.6\) and \(q=0.4\). 2. A three-game match is a sweep by either player, so \(P(X=3)=p^3+q^3=0.216+0.064=0.28\). 3. A four-game match requires the winner to have exactly \(2\) wins in the first \(3\) games and then win game \(4\). Thus, \(P(X=4)=\binom{3}{2}p^3q+\binom{3}{2}q^3p=0.2592+0.1152=0.3744\). 4. A five-game match requires a \(2\)-to-\(2\) split after four games, so \(P(X=5)=\binom{4}{2}p^2q^2=0.3456\). 5. Maya can win in \(3\), \(4\), or \(5\) games. Her total win probability is \(p^3+\binom{3}{2}p^3q+\binom{4}{2}p^3q^2=0.68256\).

Answer

a) \(P(X=3)=0.28\), \(P(X=4)=0.3744\), and \(P(X=5)=0.3456\). b) Maya wins the match with probability \(0.68256\approx68.3\%\).

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