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Example 21.7. Convert the complex number to polar form.\n\n\[ \text{a)}2 + {3i}\text{, b)} - 2 - 2\sqrt{3}i\text{, c)}4 - {3i}\text{, d)} - {4i} \] | Solution. a) First, the absolute value is \( r = \left| {2 + {3i}}\right| = \sqrt{{2}^{2} + {3}^{2}} = \sqrt{13} \) . Furthermore, since \( a = 2 \) and \( b = 3 \), we have \( \tan \left( \theta \right) = \frac{3}{2} \) . To obtain \( \theta \), we\ncalculate\n\[ {\tan }^{-1}\left( \frac{3}{2}\right) \approx {56.3}^{ ... | No |
Convert the number from polar form into the standard form \( a + {bi} \) . | \[ \text{a)}3 \cdot \left( {\cos \left( {117}^{ \circ }\right) + i\sin \left( {117}^{ \circ }\right) }\right) \;\text{b)}4 \cdot \left( {\cos \left( \frac{5\pi }{4}\right) + i\sin \left( \frac{5\pi }{4}\right) }\right) \] Solution. a) Since we don’t have an exact formula for \( \cos \left( {117}^{ \circ }\right) \) or ... | Yes |
Proposition 21.9. Let \( {r}_{1}\left( {\cos \left( {\theta }_{1}\right) + i\sin \left( {\theta }_{1}\right) }\right) \) and \( {r}_{2}\left( {\cos \left( {\theta }_{2}\right) + i\sin \left( {\theta }_{2}\right) }\right) \) be two complex numbers in polar form. Then, the product and quotient of these are given by\n\n\[... | Proof. The proof uses the addition formulas for trigonometric functions \( \sin \left( {\alpha + \beta }\right) \) and \( \cos \left( {\alpha + \beta }\right) \) from proposition 18.1 on page 252\n\n\[ \n{r}_{1}\left( {\cos \left( {\theta }_{1}\right) + i\sin \left( {\theta }_{1}\right) }\right) \cdot {r}_{2}\left( {\c... | Yes |
Multiply or divide the complex numbers, and write your answer in polar and standard form. | Solution. We will multiply and divide the complex numbers using equations (21.7) and (21.8), respectively, and then convert them to standard notation \( a + {bi} \) .\n\n\[ \text{a)}5\left( {\cos \left( {11}^{ \circ }\right) + i\sin \left( {11}^{ \circ }\right) }\right) \cdot 8\left( {\cos \left( {34}^{ \circ }\right) ... | Yes |
Graph the vectors \( \overrightarrow{v},\overrightarrow{w},\overrightarrow{r},\overrightarrow{s},\overrightarrow{t} \) in the plane, where \( \overrightarrow{v} = \overrightarrow{PQ} \) with \( P\left( {6,3}\right) \) and \( Q\left( {4, - 2}\right) \), and \[ \overrightarrow{w} = \langle 3, - 1\rangle ,\;\overrightarro... |  | Yes |
Example 22.4. Find the magnitude and directional angle of the given vectors. \[ \begin{array}{lll} \text{ a) }\langle - 6,6\rangle & \text{ b) }\langle 4, - 3\rangle & \text{ c) }\langle - 2\sqrt{3}, - 2\rangle \\ \text{ d) }\langle 8,4\sqrt{5}\rangle & \text{ e) }\overline{PQ},\;\text{ where } & P\left( {9,2}\right) \... | Solution. a) We use formulas (22.2), and the calculation is in analogy with Example 21.7 The magnitude of \( \overrightarrow{v} = \langle - 6,6\rangle \) is \[ \left| \right| \overrightarrow{v}\left| \right| = \sqrt{{\left( -6\right) }^{2} + {6}^{2}} = \sqrt{{36} + {36}} = \sqrt{72} = \sqrt{{36} \cdot 2} = 6\sqrt{2}. \... | No |
Example 22.6. Multiply, and graph the vectors.\n\n\[ \text{a)}4 \cdot \langle - 2,1\rangle \;\text{b)}\left( {-3}\right) \cdot \langle - 6, - 2\rangle \] | Solution. a) The calculation is straightforward.\n\n\[ 4 \cdot \langle - 2,1\rangle = \langle 4 \cdot \left( {-2}\right) ,4 \cdot 1\rangle = \langle - 8,4\rangle \]\n\nThe vectors are displayed below. We see, that \( \langle - 2,1\rangle \) and \( \langle - 8,4\rangle \) both have the same directional angle, and the la... | Yes |
Find a unit vector in the direction of \( \overrightarrow{v} \) . | \[ \text{a)}\langle 8,6\rangle \;\text{b)}\langle - 2,3\sqrt{7}\rangle \]\n\nSolution. a) Note that the magnitude of \( \overrightarrow{v} = \langle 8,6\rangle \) is\n\n\[ \parallel \langle 8,6\rangle \parallel = \sqrt{{8}^{2} + {6}^{2}} = \sqrt{{64} + {36}} = \sqrt{100} = {10}. \]\n\nTherefore, if we multiply \( \lang... | Yes |
Perform the vector addition and simplify as much as possible.\n\n\[ \text{a)}\langle 3, - 5\rangle + \langle 6,4\rangle \;\text{b)}5 \cdot \langle - 6,2\rangle - 7 \cdot \langle 1, - 3\rangle \;\text{c)}4\overrightarrow{i} + 9\overrightarrow{j} \] | Solution. We can find the answer by direct algebraic computation.\n\na) \( \langle 3, - 5\rangle + \langle 6,4\rangle = \langle 3 + 6, - 5 + 4\rangle = \langle 9, - 1\rangle \)\n\nb) \( 5 \cdot \langle - 6,2\rangle - 7 \cdot \langle 1, - 3\rangle = \langle - {30},{10}\rangle + \langle - 7,{21}\rangle = \langle - {37},{... | Yes |
The forces \( \overrightarrow{{F}_{1}} \) and \( \overrightarrow{{F}_{2}} \) are applied to an object. Find the resulting total force \( \overrightarrow{F} = {\overrightarrow{F}}_{1} + {\overrightarrow{F}}_{2} \) . Determine the magnitude and directional angle of the total force \( \overrightarrow{F} \) . Approximate t... | a) The vectors \( {\overrightarrow{F}}_{1} \) and \( {\overrightarrow{F}}_{2} \) are given by their magnitudes and directional angles. However, the addition of vectors (in Definition 22.10) is defined in terms of their components. Therefore, our first task is to find the vectors in component form. As was stated in equa... | Yes |
Here are some examples of sequences. | For many of these sequences we can find rules that describe how to obtain the individual terms. For example, in (a), we always add the fixed number 2 to the previous number to obtain the next, starting from the first term 4. This is an example of an arithmetic sequence, and we will study those in more detail in section... | No |
Consider the sequence \( \left\{ {a}_{n}\right\} \) with \( {a}_{n} = {4n} + 3 \) . | We can calculate the individual terms of this sequence:\n\nfirst term: \( \;{a}_{1} = 4 \cdot 1 + 3 = 7 \) ,\n\nsecond term: \( {a}_{2} = 4 \cdot 2 + 3 = {11} \) ,\n\nthird term: \( \;{a}_{3} = 4 \cdot 3 + 3 = {15} \) ,\n\nfourth term: \( \;{a}_{4} = 4 \cdot 4 + 3 = {19} \) ,\n\nfifth term: \( \;{a}_{5} = 4 \cdot 5 + 3... | Yes |
Example 23.4. Find the first 6 terms of each sequence. | Solution. a) We can easily calculate the first 6 terms of \( {a}_{n} = {n}^{2} \) directly:\n\n\[ 1,4,9,{16},{25},{36},\ldots \]\n\nWe can also use the calculator to produce the terms of a sequence. To this end, we switch the calculator from function mode to sequence mode in the mode menu (press \( \left( \text{ mode }... | Yes |
Find the first 6 terms in the sequence described below. | Solution. a) The first term is explicitly given as \( {a}_{1} = 4 \) . Then, we can calculate the following terms via \( {a}_{n} = {a}_{n - 1} + 5 \) :\n\n\[ {a}_{2} = {a}_{1} + 5 = 4 + 5 = 9 \]\n\n\[ {a}_{3} = {a}_{2} + 5 = 9 + 5 = {14} \]\n\n\[ {a}_{4} = {a}_{3} + 5 = {14} + 5 = {19} \]\n\n\[ {a}_{5} = {a}_{4} + 5 = ... | No |
\[ \text{a)}\mathop{\sum }\limits_{{i = 1}}^{4}{a}_{i}\text{, for}{a}_{i} = {7i} + 3 \] | Solution. a) The first four terms \( {a}_{1},{a}_{2},{a}_{3},{a}_{4} \) of the sequence \( {\left\{ {a}_{i}\right\} }_{i \geq 1} \) are:\n\n\[ {10},{17},{24},{31} \]\n\nThe sum is therefore:\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{4}{a}_{i} = {a}_{1} + {a}_{2} + {a}_{3} + {a}_{4} = {10} + {17} + {24} + {31} = {82} \] | Yes |
Determine if the sequence is an arithmetic sequence. If so, then find the general formula for \( {a}_{n} \) in the form of equation (23.2). | Solution. a) Calculating the difference between two consecutive terms always gives the same answer \( {13} - 7 = 6,{19} - {13} = 6,{25} - {19} = 6 \), etc. Therefore the common difference \( d = 6 \) , which shows that this is an arithmetic sequence. Furthermore, the first term is \( {a}_{1} = 7 \), so that the general... | Yes |
Find the general formula of an arithmetic sequence with the given property. | Solution. a) According to equation (23.2) the general term is \( {a}_{n} = {a}_{1} + d(n - \) 1). We know that \( d = {12} \), so that we only need to find \( {a}_{1} \) . Plugging \( {a}_{6} = {68} \) into \( {a}_{n} = {a}_{1} + d\left( {n - 1}\right) \), we may solve for \( {a}_{1} \) :\n\n\[ {68} = {a}_{6} = {a}_{1}... | Yes |
Find the sum of the first 100 integers, starting from 1. In other words, we want to find the sum of \( 1 + 2 + 3 + \cdots + {99} + {100} \) . | Let \( S = 1 + 2 + 3 + \cdots + {98} + {99} + {100} \) be what we want to find. Note that\n\n\[ \n{2S} = \begin{matrix} & 1 & + & 2 & + & 3 & \cdots & + & {98} & + & {99} & + & {100} \\ + & {100} & + & {99} & + & {98} & \cdots & + & 3 & + & 2 & + & 1 \end{matrix}. \n\]\n\nNote that the second line is also \( S \) but i... | Yes |
a) Find the sum \( {a}_{1} + \cdots + {a}_{60} \) for the arithmetic sequence \( {a}_{n} = 2 + {13}\left( {n - 1}\right) \) . | Solution. a) The sum is given by the formula (23.3): \( \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i} = \frac{k}{2} \cdot \left( {{a}_{1} + {a}_{k}}\right) \) . Here, \( k = {60} \), and \( {a}_{1} = 2 \) and \( {a}_{60} = 2 + {13} \cdot \left( {{60} - 1}\right) = 2 + {13} \cdot {59} = 2 + {767} = {769} \) . We obtain a s... | Yes |
Example 24.2. Determine if the sequence is a geometric, or arithmetic sequence, or neither or both. If it is a geometric or arithmetic sequence, then find the general formula for \( {a}_{n} \) in the form (24.1) or (23.2). | a) Calculating the quotient of two consecutive terms always gives the same number \( 6 \div 3 = 2,{12} \div 6 = 2,{24} \div {12} = 2 \), etc. Therefore the common ratio is \( r = 2 \), which shows that this is a geometric sequence. Furthermore, the first term is \( {a}_{1} = 3 \), so that the general formula for the \(... | Yes |
Find the general formula of a geometric sequence with the given property. | a) Since \( \\left\\{ {a}_{n}\\right\\} \) is a geometric sequence, it is \( {a}_{n} = {a}_{1} \\cdot {r}^{n - 1} \) . We know that \( r = 4 \), so we still need to find \( {a}_{1} \) . Using \( {a}_{5} = {64000} \), we obtain:\n\n\[ \n{6400} = {a}_{5} = {a}_{1} \\cdot {4}^{5 - 1} = {a}_{1} \\cdot {4}^{4} = {256} \\cdo... | Yes |
Consider the geometric sequence \( {a}_{n} = 8 \cdot {5}^{n - 1} \), that is the sequence:\n\n\[ 8,{40},{200},{1000},{5000},{25000},{125000},\ldots \]\n\nWe want to add the first 6 terms of this sequence.\n\n\[ 8 + {40} + {200} + {1000} + {5000} + {25000} = {31248} \] | In general, it may be much more difficult to simply add the terms as we did above, and we need to use a better general method. For this, we multiply \( \left( {1 - 5}\right) \) to the sum \( \left( {8 + {40} + {200} + {1000} + {5000} + {25000}}\right) \) and simplify this using the distributive law:\n\n\[ \left( {1 - 5... | Yes |
a) Find the sum \( \mathop{\sum }\limits_{{n = 1}}^{6}{a}_{n} \) for the geometric sequence \( {a}_{n} = {10} \cdot {3}^{n - 1} \) . | Solution. a) We need to find the sum \( {a}_{1} + {a}_{2} + {a}_{3} + {a}_{4} + {a}_{5} + {a}_{6} \), and we will do so using the formula provided in equation (24.2). Since \( {a}_{n} = {10} \cdot {3}^{n - 1} \), we have \( {a}_{1} = {10} \) and \( r = 3 \), so that\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{6}{a}_{n} = {1... | Yes |
Consider the geometric sequence\n\n\[ 1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\ldots \]\n\nHere, the common ratio is \( r = \frac{1}{2} \), and the first term is \( {a}_{1} = 1 \), so that the formula for \( {a}_{n} \) is \( {a}_{n} = {\left( \frac{1}{2}\right) }^{n - 1} \) . We are interested in summing all... | We see that adding each term takes the sum closer and closer to the number 2. More precisely, adding a term \( {a}_{n} \) to the partial sum \( {a}_{1} + \cdots + {a}_{n - 1} \) cuts the distance between 2 and \( {a}_{1} + \cdots + {a}_{n - 1} \) in half. For this reason we can, in fact, get arbitrarily close to 2 , so... | Yes |
Example 24.10. Find the value of the infinite geometric series.\n\n\[ \n\\text{a)}\\mathop{\\sum }\\limits_{{j = 1}}^{\\infty }{a}_{j}\\text{, for}{a}_{j} = 5 \\cdot {\\left( \\frac{1}{3}\\right) }^{j - 1} \n\] | Solution. a) We use formula (24.4) for the geometric series \( {a}_{n} = 5 \\cdot {\\left( \\frac{1}{3}\\right) }^{n - 1} \) , that is \( {a}_{1} = 5 \\cdot {\\left( \\frac{1}{3}\\right) }^{1 - 1} = 5 \\cdot {\\left( \\frac{1}{3}\\right) }^{0} = 5 \\cdot 1 = 5 \) and \( r = \\frac{1}{3} \) . Therefore,\n\n\[ \n\\mathop... | Yes |
The fraction \( {0.55555}\ldots \) may be written as: | \[ {0.55555}\cdots = {0.5} + {0.05} + {0.005} + {0.0005} + {0.00005} + \ldots \] Noting that the sequence \[ {0.5},\underset{\times {0.1}}{\underbrace{}}{0.05},\underset{\times {0.1}}{\underbrace{}}{0.005},\underset{\times {0.1}}{\underbrace{}}{0.0005},\underset{\times {0.1}}{\underbrace{}}{0.00005},\ldots \] is a geom... | Yes |
\[ 4! = 1 \cdot 2 \cdot 3 \cdot 4 = {24} \] | \[ 4! = 1 \cdot 2 \cdot 3 \cdot 4 = {24} \] | Yes |
Example 25.5. Calculate the binomial coefficients.\n\n\[ \n\\text{a)}\\left( \\begin{array}{l} 6 \\ 4 \\end{array}\\right) \\;\\text{b)}\\left( \\begin{array}{l} 8 \\ 5 \\end{array}\\right) \\;\\text{c)}\\left( \\begin{array}{l} {25} \\ {23} \\end{array}\\right) \\;\\text{d)}\\left( \\begin{array}{l} 7 \\ 1 \\end{array... | Solution. a) Many binomial coefficients may be calculated by hand, such as:\n\n\[ \n\\left( \\begin{array}{l} 6 \\ 4 \\end{array}\\right) = \\frac{6!}{4!\\left( {6 - 4}\\right) !} = \\frac{6!}{4!2!} = \\frac{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6}{1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 1 \\cdot 2} = \\frac{5 \\cdo... | No |
\[ {\left( a + b\right) }^{3} = \left( {a + b}\right) \cdot \left( {a + b}\right) \cdot \left( {a + b}\right) \] | \[ = \left( {{a}^{2} + {2ab} + {b}^{2}}\right) \cdot \left( {a + b}\right) \] \[ = {a}^{3} + 2{a}^{2}b + a{b}^{2} + {a}^{2}b + {2a}{b}^{2} + {b}^{3} \] \[ = {a}^{3} + 3{a}^{2}b + {3a}{b}^{2} + {b}^{3} \] | Yes |
Theorem 25.9 (Binomial theorem). The \( n \) th power \( {\left( a + b\right) }^{n} \) can be expanded as:\n\n\[ \n{\left( a + b\right) }^{n} = \left( \begin{array}{l} n \\ 0 \end{array}\right) {a}^{n} + \left( \begin{array}{l} n \\ 1 \end{array}\right) {a}^{n - 1}{b}^{1} + \left( \begin{array}{l} n \\ 2 \end{array}\ri... | Using the summation symbol, we may write this in short:\n\n\[ \n{\left( a + b\right) }^{n} = \mathop{\sum }\limits_{{r = 0}}^{n}\left( \begin{array}{l} n \\ r \end{array}\right) \cdot {a}^{n - r} \cdot {b}^{r} \n\] | Yes |
Expand \( {\left( a + b\right) }^{5} \). | \[ {\left( a + b\right) }^{5} = \left( \begin{array}{l} 5 \\ 0 \end{array}\right) {a}^{5} + \left( \begin{array}{l} 5 \\ 1 \end{array}\right) {a}^{4}{b}^{1} + \left( \begin{array}{l} 5 \\ 2 \end{array}\right) {a}^{3}{b}^{2} + \left( \begin{array}{l} 5 \\ 3 \end{array}\right) {a}^{2}{b}^{3} + \left( \begin{array}{l} 5 \... | Yes |
Expand the expression. a) \( {\left( {x}^{2} + 2{y}^{3}\right) }^{5} \) | Solution. a) We use the binomial theorem with \( a = {x}^{2} \) and \( b = 2{y}^{3} \) :\n\n\[ \n{\left( {x}^{2} + 2{y}^{3}\right) }^{5} = {\left( {x}^{2}\right) }^{5} + \left( \begin{array}{l} 5 \\ 1 \end{array}\right) {\left( {x}^{2}\right) }^{4}\left( {2{y}^{3}}\right) + \left( \begin{array}{l} 5 \\ 2 \end{array}\ri... | Yes |
Determine: a) the \( {x}^{4}{y}^{12} \) -term in the binomial expansion of \( {\left( 5{x}^{2} + 2{y}^{3}\right) }^{6} \) | Solution. a) In this case we have \( a = 5{x}^{2} \) and \( b = 2{y}^{3} \) . The term \( {x}^{4}{y}^{12} \) can be rewritten as \( {x}^{4}{y}^{12} = {\left( {x}^{2}\right) }^{2} \cdot {\left( {y}^{3}\right) }^{4} \), so that the full term \( \left( \begin{matrix} n \\ k - 1 \end{matrix}\right) {a}^{n - k + 1}{b}^{k - ... | Yes |
Example 1.2 (Coin Tossing) As we have noted, our intuition suggests that the probability of obtaining a head on a single toss of a coin is \( 1/2 \) . To have the computer toss a coin, we can ask it to pick a random real number in the interval \( \left\lbrack {0,1}\right\rbrack \) and test to see if this number is less... | We notice that when we tossed the coin 10,000 times, the proportion of heads was close to the \ | Yes |
Example 1.3 (Dice Rolling) We consider a dice game that played an important role in the historical development of probability. The famous letters between Pascal and Fermat, which many believe started a serious study of probability, were instigated by a request for help from a French nobleman and gambler, Chevalier de M... | One can understand this calculation as follows: The probability that no 6 turns up on the first toss is \( \left( {5/6}\right) \) . The probability that no 6 turns up on either of the first two tosses is \( {\left( 5/6\right) }^{2} \) . Reasoning in the same way, the probability that no 6 turns up on any of the first f... | Yes |
For our next example, we consider a problem where the exact answer is difficult to obtain but for which simulation easily gives the qualitative results. Peter and Paul play a game called heads or tails. In this game, a fair coin is tossed a sequence of times - we choose 40 . Each time a head comes up Peter wins 1 penny... | It is easy to settle this by simulating the game a large number of times and keeping track of the number of times that Peter’s final winnings are \( j \), and the number of times that Peter ends up being in the lead by \( k \) . The proportions over all games then give estimates for the corresponding probabilities. The... | Yes |
Example 1.5 (Horse Races) Four horses (Acorn, Balky, Chestnut, and Dolby) have raced many times. It is estimated that Acorn wins 30 percent of the time, Balky 40 percent of the time, Chestnut 20 percent of the time, and Dolby 10 percent of the time. | We can have our computer carry out one race as follows: Choose a random number \( x \) . If \( x < {.3} \) then we say that Acorn won. If \( {.3} \leq x < {.7} \) then Balky wins. If \( {.7} \leq x < {.9} \) then Chestnut wins. Finally, if \( {.9} \leq x \) then Dolby wins.\n\nThe program HorseRace uses this method to ... | Yes |
Let \( E = \{ \mathrm{{HH}},\mathrm{{HT}},\mathrm{{TH}}\} \) be the event that at least one head comes up. Then, the probability of \( E \) can be calculated as follows: | \[ P\left( E\right) = m\left( \mathrm{{HH}}\right) + m\left( \mathrm{{HT}}\right) + m\left( \mathrm{{TH}}\right) \] \[ = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}\text{.} \] | Yes |
The sample space for the experiment in which the die is rolled is the 6-element set \( \Omega = \{ 1,2,3,4,5,6\} \) . We assumed that the die was fair, and we chose the distribution function defined by\n\n\[ m\left( i\right) = \frac{1}{6},\;\text{ for }i = 1,\ldots ,6. \]\n\nIf \( E \) is the event that the result of t... | \[ = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}\text{.} \] | Yes |
Example 1.9 Three people, A, B, and C, are running for the same office, and we assume that one and only one of them wins. The sample space may be taken as the 3-element set \( \Omega = \{ \mathrm{A},\mathrm{B},\mathrm{C}\} \) where each element corresponds to the outcome of that candidate's winning. Suppose that A and ... | we see that\n\n\[ {2m}\left( \mathrm{C}\right) + {2m}\left( \mathrm{C}\right) + m\left( \mathrm{C}\right) = 1, \]\n\nwhich implies that \( {5m}\left( \mathrm{C}\right) = 1 \) . Hence,\n\n\[ m\left( \mathrm{\;A}\right) = \frac{2}{5},\;m\left( \mathrm{\;B}\right) = \frac{2}{5},\;m\left( \mathrm{C}\right) = \frac{1}{5}. \... | Yes |
Theorem 1.1 The probabilities assigned to events by a distribution function on a sample space \( \Omega \) satisfy the following properties:\n\n1. \( P\left( E\right) \geq 0 \) for every \( E \subset \Omega \) .\n\n2. \( P\left( \Omega \right) = 1 \) .\n\n3. If \( E \subset F \subset \Omega \), then \( P\left( E\right)... | Proof. For any event \( E \) the probability \( P\left( E\right) \) is determined from the distribution \( m \) by\n\n\[ P\left( E\right) = \mathop{\sum }\limits_{{\omega \in E}}m\left( \omega \right) \]\n\nfor every \( E \subset \Omega \) . Since the function \( m \) is nonnegative, it follows that \( P\left( E\right)... | Yes |
Theorem 1.2 If \( {A}_{1},\ldots ,{A}_{n} \) are pairwise disjoint subsets of \( \Omega \) (i.e., no two of the \( {A}_{i} \)’s have an element in common), then\n\n\[ P\left( {{A}_{1} \cup \cdots \cup {A}_{n}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}P\left( {A}_{i}\right) . \] | Proof. Let \( \omega \) be any element in the union\n\n\[ {A}_{1} \cup \cdots \cup {A}_{n}\text{.} \]\n\nThen \( m\left( \omega \right) \) occurs exactly once on each side of the equality in the statement of the theorem. | No |
Theorem 1.3 Let \( {A}_{1},\ldots ,{A}_{n} \) be pairwise disjoint events with \( \Omega = {A}_{1} \cup \cdots \cup {A}_{n} \) , and let \( E \) be any event. Then\n\n\[ P\left( E\right) = \mathop{\sum }\limits_{{i = 1}}^{n}P\left( {E \cap {A}_{i}}\right) \] | Proof. The sets \( E \cap {A}_{1},\ldots, E \cap {A}_{n} \) are pairwise disjoint, and their union is the set \( E \) . The result now follows from Theorem 1.2. | Yes |
Theorem 1.4 If \( A \) and \( B \) are subsets of \( \Omega \), then\n\n\[ P\left( {A \cup B}\right) = P\left( A\right) + P\left( B\right) - P\left( {A \cap B}\right) . \] | Proof. The left side of Equation 1.1 is the sum of \( m\left( \omega \right) \) for \( \omega \) in either \( A \) or \( B \) . We must show that the right side of Equation 1.1 also adds \( m\left( \omega \right) \) for \( \omega \) in \( A \) or \( B \) . If \( \omega \) is in exactly one of the two sets, then it is c... | Yes |
What is the probability of getting a sum of 7 on the roll of two dice - or getting a sum of 11 ? | The first event, denoted by \( E \), is the subset\n\n\[ E = \{ \left( {1,6}\right) ,\left( {6,1}\right) ,\left( {2,5}\right) ,\left( {5,2}\right) ,\left( {3,4}\right) ,\left( {4,3}\right) \} .\n\nA sum of 11 is the subset \( F \) given by\n\n\[ F = \{ \left( {5,6}\right) ,\left( {6,5}\right) \} .\n\nConsequently,\n\n\... | Yes |
Example 1.13 A coin is tossed until the first time that a head turns up. Let the outcome of the experiment, \( \omega \), be the first time that a head turns up. Then the possible outcomes of our experiment are\n\n\[ \Omega = \{ 1,2,3,\ldots \} .\n\]\n\nNote that even though the coin could come up tails every time we h... | That this is true follows from the formula for the sum of a geometric series,\n\n\[ 1 + r + {r}^{2} + {r}^{3} + \cdots = \frac{1}{1 - r}, \]\n\nor\n\[ r + {r}^{2} + {r}^{3} + {r}^{4} + \cdots = \frac{r}{1 - r}, \]\n\n(1.2)\n\nfor \( - 1 < r < 1 \) .\n\nPutting \( r = 1/2 \), we see that we have a probability of 1 that ... | Yes |
Example 2.1 We begin by constructing a spinner, which consists of a circle of unit circumference and a pointer as shown in Figure 2.1. We pick a point on the circle and label it 0 , and then label every other point on the circle with the distance, say \( x \), from 0 to that point, measured counterclockwise. The experi... | If we proceed as we did in Chapter 1 for experiments with a finite number of possible outcomes, then we must assign the probability 0 to each outcome, since otherwise, the sum of the probabilities, over all of the possible outcomes, would not equal 1. (In fact, summing an uncountable number of real numbers is a tricky ... | No |
In this example we show how simulation can be used to estimate areas of plane figures. Suppose that we program our computer to provide a pair \( \left( {x, y}\right) \) or numbers, each chosen independently at random from the interval \( \left\lbrack {0,1}\right\rbrack \) . Then we can interpret this pair \( \left( {x,... | We can use this method to estimate the area of the region \( E \) under the curve \( y = {x}^{2} \) in the unit square (see Figure 2.2). We choose a large number of points \( \left( {x, y}\right) \) at random and record what fraction of them fall in the region \( E = \left\{ {\left( {x, y}\right) : y \leq {x}^{2}}\righ... | Yes |
Suppose that we take a card table and draw across the top surface a set of parallel lines a unit distance apart. We then drop a common needle of unit length at random on this surface and observe whether or not the needle lies across one of the lines. We can describe the possible outcomes of this experiment by coordinat... | Now the area of the rectangle is \( \pi /4 \), while the area of \( E \) is\n\n\[ \text{Area} = {\int }_{0}^{\pi /2}\frac{1}{2}\sin {\theta d\theta } = \frac{1}{2}.\ ]\n\nHence, we get\n\n\[ P\left( E\right) = \frac{1/2}{\pi /4} = \frac{2}{\pi }. \] | Yes |
Example 2.4 Suppose that we choose two random real numbers in \( \\left\\lbrack {0,1}\\right\\rbrack \) and add them together. Let \( X \) be the sum. How is \( X \) distributed? | To help understand the answer to this question, we can use the program Are-abargraph. This program produces a bar graph with the property that on each interval, the area, rather than the height, of the bar is equal to the fraction of outcomes that fell in the corresponding interval. We have carried out this experiment ... | Yes |
Example 2.5 Suppose that we choose 100 random numbers in \( \left\lbrack {0,1}\right\rbrack \), and let \( X \) represent their sum. How is \( X \) distributed? | It turns out that the type of function which does the job is called a normal density function. This type of function is sometimes referred to as a \ | No |
Example 2.7 The spinner experiment described in Example 2.1 has the interval \( \lbrack 0,1) \) as the set of possible outcomes. We would like to construct a probability model in which each outcome is equally likely to occur. We saw that in such a model, it is necessary to assign the probability 0 to each outcome. This... | If we let \( E = \left\lbrack {c, d}\right\rbrack \), then we can write the above formula in the form\n\n\[ P\left( E\right) = {\int }_{E}f\left( x\right) {dx} \]\n\nwhere \( f\left( x\right) \) is the constant function with value 1 . This should remind the reader of the corresponding formula in the discrete case for t... | Yes |
Example 2.8 A game of darts involves throwing a dart at a circular target of unit radius. Suppose we throw a dart once so that it hits the target, and we observe where it lands. | To describe the possible outcomes of this experiment, it is natural to take as our sample space the set \( \Omega \) of all the points in the target. It is convenient to describe these points by their rectangular coordinates, relative to a coordinate system with origin at the center of the target, so that each pair \( ... | Yes |
What probabilities should we assign to the events \( E \) of \( \Omega \) ? If\n\n\[ E = \{ r : 0 \leq r \leq a\} \]\n\nthen \( E \) occurs if the dart lands within a distance \( a \) of the center, that is, within the circle of radius \( a \), and we saw in the previous example that under our assumptions the probabili... | More generally, if\n\n\[ E = \{ r : a \leq r \leq b\} ,\]\n\nthen by our basic assumptions,\n\n\[ P\left( E\right) = P\left( \left\lbrack {a, b}\right\rbrack \right) = P\left( \left\lbrack {0, b}\right\rbrack \right) - P\left( \left\lbrack {0, a}\right\rbrack \right)\]\n\n\[ = {b}^{2} - {a}^{2}\]\n\n\[ = \left( {b - a}... | Yes |
In the spinner experiment, we choose for our set of outcomes the interval \( 0 \leq x < 1 \), and for our density function\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }0 \leq x < 1 \\ 0, & \text{ otherwise. } \end{array}\right. \]\n\nIf \( E \) is the event that the head of the spinner falls in th... | \[ P\left( E\right) = {\int }_{0}^{1/2}{1dx} = \frac{1}{2}. \] | Yes |
In the first dart game experiment, we choose for our sample space a disc of unit radius in the plane and for our density function the function\n\n\[ f\left( {x, y}\right) = \left\{ \begin{array}{ll} 1/\pi , & \text{ if }{x}^{2} + {y}^{2} \leq 1 \\ 0, & \text{ otherwise. } \end{array}\right. \]\n\nThe probability that t... | \[ P\left( E\right) = \iint {\int }_{E}\frac{1}{\pi }{dxdy} \]\n\n\[ = \frac{1}{\pi } \cdot \left( {\text{ area of }E}\right) . \] | Yes |
In the second dart game experiment, we choose for our sample space the unit interval on the real line and for our density the function \[ f\left( r\right) = \left\{ \begin{array}{ll} {2r}, & \text{ if }0 < r < 1 \\ 0, & \text{ otherwise. } \end{array}\right. \] Then the probability that the dart lands at distance \( r,... | \[ = {b}^{2} - {a}^{2}\text{.} \] Here again, since the density is small when \( r \) is near 0 and large when \( r \) is near 1, we see that in this experiment the dart is more likely to land near the rim of the target than near the center. In terms of the bar graph of Example 2.9, the heights of the bars approximate ... | Yes |
Theorem 2.1 Let \( X \) be a continuous real-valued random variable with density function \( f\left( x\right) \) . Then the function defined by\n\n\[ F\left( x\right) = {\int }_{-\infty }^{x}f\left( t\right) {dt} \]\n\nis the cumulative distribution function of \( X \) . Furthermore, we have\n\n\[ \frac{d}{dx}F\left( x... | Proof. By definition,\n\n\[ F\left( x\right) = P\left( {X \leq x}\right) . \]\n\nLet \( E = ( - \infty, x\rbrack \) . Then\n\n\[ P\left( {X \leq x}\right) = P\left( {X \in E}\right) ,\]\n\nwhich equals\n\n\[ {\int }_{-\infty }^{x}f\left( t\right) {dt} \]\n\nApplying the Fundamental Theorem of Calculus to the first equa... | Yes |
A real number is chosen at random from \( \left\lbrack {0,1}\right\rbrack \) with uniform probability, and then this number is squared. Let \( X \) represent the result. What is the cumulative distribution function of \( X \) ? What is the density of \( X \) ? | We begin by letting \( U \) represent the chosen real number. Then \( X = {U}^{2} \) . If \( 0 \leq x \leq 1 \), then we have\n\n\[ \n{F}_{X}\left( x\right) = P\left( {X \leq x}\right) \]\n\n\[ \n= P\left( {{U}^{2} \leq x}\right) \]\n\n\[ \n= P\left( {U \leq \sqrt{x}}\right) \]\n\n\[ \n= \sqrt{x}\text{.} \]\n\nIt is cl... | Yes |
In Example 2.4, we considered a random variable, defined to be the sum of two random real numbers chosen uniformly from \( \left\lbrack {0,1}\right\rbrack \) . Let the random variables \( X \) and \( Y \) denote the two chosen real numbers. Define \( Z = X + Y \) . We will now derive expressions for the cumulative dist... | Here we take for our sample space \( \Omega \) the unit square in \( {\mathbf{R}}^{2} \) with uniform density. A point \( \omega \in \Omega \) then consists of a pair \( \left( {x, y}\right) \) of numbers chosen at random. Then \( 0 \leq Z \leq 2 \) . Let \( {E}_{z} \) denote the event that \( Z \leq z \) . In Figure 2... | Yes |
In the dart game described in Example 2.8, what is the distribution of the distance of the dart from the center of the target? What is its density? | Here, as before, our sample space \( \Omega \) is the unit disk in \( {\mathbf{R}}^{2} \), with coordinates \( \left( {X, Y}\right) \) . Let \( Z = \sqrt{{X}^{2} + {Y}^{2}} \) represent the distance from the center of the target. Let\n\n \) as the arrival times (after 5:00 P.M.) for the Lockhorns. Let \( Z = \left| {X - Y}\right| \) . Then we have \( {F}_{X}\left( x\right) = x \) and \( {F}_{Y}\left( y\right) = y \) . Moreover (see Figure 2.19),\n\n\[ \n{F... | Yes |
Consider an experiment in which a fair coin is tossed repeatedly, without stopping. We have seen in Example 1.6 that, for a coin tossed \( n \) times, the natural sample space is a binary tree with \( n \) stages. On this evidence we expect that for a coin tossed repeatedly, the natural sample space is a binary tree wi... | It is surprising to learn that, although the \( n \) -stage tree is obviously a finite sample space, the unlimited tree can be described as a continuous sample space. To see how this comes about, let us agree that a typical outcome of the unlimited coin tossing experiment can be described by a sequence of the form \( \... | Yes |
How many possible choices do you have for your complete meal? | We illustrate the possible meals by a tree diagram shown in Figure 3.1. Your menu is decided in three stages - at each stage the number of possible choices does not depend on what is chosen in the previous stages: two choices at the first stage, three at the second, and two at the third. From the tree diagram we see th... | Yes |
Example 3.2 We can show that there are at least two people in Columbus, Ohio, who have the same three initials. Assuming that each person has three initials, there are 26 possibilities for a person's first initial, 26 for the second, and 26 for the third. Therefore, there are \( {26}^{3} = {17},{576} \) possible sets o... | This number is smaller than the number of people living in Columbus, Ohio; hence, there must be at least two people with the same three initials. | Yes |
How many people do we need to have in a room to make it a favorable bet (probability of success greater than \( 1/2 \) ) that two people in the room will have the same birthday? | Since there are 365 possible birthdays, it is tempting to guess that we would need about \( 1/2 \) this number, or 183. You would surely win this bet. In fact, the number required for a favorable bet is only 23 . To show this, we find the probability \( {p}_{r} \) that, in a room with \( r \) people, there is no duplic... | Yes |
Theorem 3.3 (Stirling’s Formula) The sequence \( n \) ! is asymptotically equal to\n\n\[{n}^{n}{e}^{-n}\sqrt{2\pi n}\text{.} | The proof of Stirling's formula may be found in most analysis texts. Let us verify this approximation by using the computer. The program StirlingApprox-imations prints \( n \) !, the Stirling approximation, and, finally, the ratio of these two numbers. Sample output of this program is shown in Table 3.4. Note that, whi... | No |
Example 3.5 Let \( U = \{ a, b, c\} \) . The subsets of \( U \) are\n\n\[ \phi ,\{ a\} ,\{ b\} ,\{ c\} ,\{ a, b\} ,\{ a, c\} ,\{ b, c\} ,\{ a, b, c\} . \] | In the above example, there is one subset with no elements, three subsets with exactly 1 element, three subsets with exactly 2 elements, and one subset with exactly 3 elements. Thus, \( \left( \begin{array}{l} 3 \\ 0 \end{array}\right) = 1,\left( \begin{array}{l} 3 \\ 1 \end{array}\right) = 3,\left( \begin{array}{l} 3 ... | Yes |
Theorem 3.4 For integers \( n \) and \( j \), with \( 0 < j < n \), the binomial coefficients satisfy:\n\n\[ \left( \begin{array}{l} n \\ j \end{array}\right) = \left( \begin{matrix} n - 1 \\ j \end{matrix}\right) + \left( \begin{matrix} n - 1 \\ j - 1 \end{matrix}\right) \] | Proof. We wish to choose a subset of \( j \) elements. Choose an element \( u \) of \( U \) . Assume first that we do not want \( u \) in the subset. Then we must choose the \( j \) elements from a set of \( n - 1 \) elements; this can be done in \( \left( \begin{matrix} n - 1 \\ j \end{matrix}\right) \) ways. On the o... | Yes |
Theorem 3.5 The binomial coefficients are given by the formula\n\n\[ \left( \begin{array}{l} n \\ j \end{array}\right) = \frac{{\left( n\right) }_{j}}{j!}. \] | Proof. Each subset of size \( j \) of a set of size \( n \) can be ordered in \( j \) ! ways. Each of these orderings is a \( j \) -permutation of the set of size \( n \) . The number of \( j \) -permutations is \( {\left( n\right) }_{j} \), so the number of subsets of size \( j \) is\n\n\[ \frac{{\left( n\right) }_{j}... | Yes |
How many hands have four of a kind? | There are 13 ways that we can specify the value for the four cards. For each of these, there are 48 possibilities for the fifth card. Thus, the number of four-of-a-kind hands is \( {13} \cdot {48} = {624} \) . Since the total number of possible hands is \( \left( \begin{matrix} {52} \\ 5 \end{matrix}\right) = {2598960}... | Yes |
Theorem 3.6 Given \( n \) Bernoulli trials with probability \( p \) of success on each experiment, the probability of exactly \( j \) successes is\n\n\[ b\left( {n, p, j}\right) = \left( \begin{array}{l} n \\ j \end{array}\right) {p}^{j}{q}^{n - j} \]\n\nwhere \( q = 1 - p \) . | Proof. We construct a tree measure as described above. We want to find the sum of the probabilities for all paths which have exactly \( j \) successes and \( n - j \) failures. Each such path is assigned a probability \( {p}^{j}{q}^{n - j} \) . How many such paths are there? To specify a path, we have to pick, from the... | Yes |
A fair coin is tossed six times. What is the probability that exactly three heads turn up? | \[ b\left( {6,{.5},3}\right) = \left( \begin{array}{l} 6 \\ 3 \end{array}\right) {\left( \frac{1}{2}\right) }^{3}{\left( \frac{1}{2}\right) }^{3} = {20} \cdot \frac{1}{64} = {.3125}. \] | Yes |
A die is rolled four times. What is the probability that we obtain exactly one 6 ? | We treat this as Bernoulli trials with success \( = \) \ | No |
Example 3.10 A Galton board is a board in which a large number of BB-shots are dropped from a chute at the top of the board and deflected off a number of pins on their way down to the bottom of the board. The final position of each slot is the result of a number of random deflections either to the left or the right. We... | Note that if we write 0 every time the shot is deflected to the left, and 1 every time it is deflected to the right, then the path of the shot can be described by a sequence of 0 ’s and 1’s of length \( n \), just as for the \( n \) -fold coin toss. The distribution shown in Figure 3.6 is an example of an empirical dis... | No |
Theorem 3.7 (Binomial Theorem) The quantity \( {\left( a + b\right) }^{n} \) can be expressed in the form\n\n\[ \n{\left( a + b\right) }^{n} = \mathop{\sum }\limits_{{j = 0}}^{n}\left( \begin{array}{l} n \\ j \end{array}\right) {a}^{j}{b}^{n - j}.\n\] | Proof. To see that this expansion is correct, write\n\n\[ \n{\left( a + b\right) }^{n} = \left( {a + b}\right) \left( {a + b}\right) \cdots \left( {a + b}\right) .\n\]\n\nWhen we multiply this out we will have a sum of terms each of which results from a choice of an \( a \) or \( b \) for each of \( n \) factors. When ... | Yes |
Corollary 3.1 The sum of the elements in the \( n \) th row of Pascal’s triangle is \( {2}^{n} \) . If the elements in the \( n \) th row of Pascal’s triangle are added with alternating signs, the sum is 0 . | Proof. The first statement in the corollary follows from the fact that\n\n\[ \n{2}^{n} = {\left( 1 + 1\right) }^{n} = \left( \begin{array}{l} n \\ 0 \end{array}\right) + \left( \begin{array}{l} n \\ 1 \end{array}\right) + \left( \begin{array}{l} n \\ 2 \end{array}\right) + \cdots + \left( \begin{array}{l} n \\ n \end{a... | Yes |
Theorem 3.8 Let \( P \) be a probability distribution on a sample space \( \Omega \), and let \( \\left\\{ {{A}_{1},{A}_{2},\\ldots ,{A}_{n}}\\right\\} \) be a finite set of events. Then\n\n\[ P\\left( {{A}_{1} \\cup {A}_{2} \\cup \\cdots \\cup {A}_{n}}\\right) = \\mathop{\\sum }\\limits_{{i = 1}}^{n}P\\left( {A}_{i}\\... | Proof. If the outcome \( \\omega \) occurs in at least one of the events \( {A}_{i} \), its probability is added exactly once by the left side of Equation 3.3. We must show that it is added exactly once by the right side of Equation 3.3. Assume that \( \\omega \) is in exactly \( k \) of the sets. Then its probability ... | Yes |
Example 3.12 We return to the hat check problem discussed in Section 3.1, that is, the problem of finding the probability that a random permutation contains at least one fixed point. Recall that a permutation is a one-to-one map of a set \( A = \left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right\} \) onto itself. Let \( {A... | \[ P\left( {{A}_{i} \cap {A}_{j}}\right) = \frac{\left( {n - 2}\right) !}{n!} = \frac{1}{n\left( {n - 1}\right) }. \] The number of terms of this form in the right side of Equation 3.3 is \[ \left( \begin{array}{l} n \\ 2 \end{array}\right) = \frac{n\left( {n - 1}\right) }{2!}. \] Hence, the second term of Equation 3.3... | Yes |
In the quantum mechanical model of the helium atom, various parameters can be used to classify the energy states of the atom. In the triplet spin state \( \left( {S = 1}\right) \) with orbital angular momentum \( 1\left( {L = 1}\right) \), there are three possibilities, \( 0,1 \), or 2, for the total angular momentum \... | The theory gives these assignments because these frequencies were observed in experiments and further parameters were developed in the theory to allow these frequencies to be predicted. | No |
Theorem 3.9 Let \( a \) and \( b \) be two positive integers. Let \( {S}_{a, b} \) be the set of all ordered pairs in which the first entry is an \( a \) -shuffle and the second entry is a \( b \) -shuffle. Let \( {S}_{ab} \) be the set of all \( {ab} \) -shuffles. Then there is a 1-1 correspondence between \( {S}_{a, ... | Proof. The easiest way to describe the required correspondence is through the idea of an unshuffle. An \( a \) -unshuffle begins with a deck of \( n \) cards. One by one, cards are taken from the top of the deck and placed, with equal probability, on the bottom of any one of \( a \) stacks, where the stacks are labelle... | Yes |
Theorem 3.10 If \( D \) is any ordering that is the result of applying an \( a \) -shuffle and then a \( b \) -shuffle to the identity ordering, then the probability assigned to \( D \) by this pair of operations is the same as the probability assigned to \( D \) by the process of applying an \( {ab} \) -shuffle to the... | Proof. Call the sample space of \( a \) -shuffles \( {S}_{a} \) . If we label the stacks by the integers from 0 to \( a - 1 \), then each cut-interleaving pair, i.e., shuffle, corresponds to exactly one \( n \) -digit base \( a \) integer, where the \( i \) th digit in the integer is the stack of which the \( i \) th c... | Yes |
If an ordering of length \( n \) has \( r \) rising sequences, then the number of cut-interleaving pairs under an \( a \) -shuffle of the identity ordering which lead to the ordering is \n\n\[ \left( \begin{matrix} n + a - r \\ n \end{matrix}\right) . \] | Proof. To see why this is true, we need to count the number of ways in which the cut in an \( a \) -shuffle can be performed which will lead to a given ordering with \( r \) rising sequences. We can disregard the interleavings, since once a cut has been made, at most one interleaving will lead to a given ordering. Sinc... | Yes |
Theorem 3.12 Let \( a \) and \( n \) be positive integers. Then\n\n\[ \n{a}^{n} = \mathop{\sum }\limits_{{r = 1}}^{a}\left( \begin{matrix} n + a - r \\ n \end{matrix}\right) A\left( {n, r}\right) .\n\]\n\n(3.5)\n\nThus,\n\n\[ \nA\left( {n, a}\right) = {a}^{n} - \mathop{\sum }\limits_{{r = 1}}^{{a - 1}}\left( \begin{mat... | Proof. The second equation can be used to calculate the values of the Eulerian numbers, and follows immediately from the Equation 3.5. The last equation is a consequence of the fact that the only ordering of \( \{ 1,2,\ldots, n\} \) with one rising sequence is the identity ordering. Thus, it remains to prove Equation 3... | Yes |
An experiment consists of rolling a die once. Let \( X \) be the outcome. Let \( F \) be the event \( \{ X = 6\} \), and let \( E \) be the event \( \{ X > 4\} \) . We assign the distribution function \( m\left( \omega \right) = 1/6 \) for \( \omega = 1,2,\ldots ,6 \) . Thus, \( P\left( F\right) = 1/6 \) . | Now suppose that the die is rolled and we are told that the event \( E \) has occurred. This leaves only two possible outcomes: 5 and 6 . In the absence of any information, we would still regard these outcomes to be equally likely, so the probability of \( F \) becomes \( 1/2 \), making \( P\left( {F \mid E}\right) = 1... | Yes |
Example 4.2 In the Life Table (see Appendix C), one finds that in a population of 100,000 females, \( {89.835}\% \) can expect to live to age 60, while \( {57.062}\% \) can expect to live to age 80 . Given that a woman is 60 , what is the probability that she lives to age 80 ? | This is an example of a conditional probability. In this case, the original sample space can be thought of as a set of 100,000 females. The events \( E \) and \( F \) are the subsets of the sample space consisting of all women who live at least 60 years, and at least 80 years, respectively. We consider \( E \) to be th... | Yes |
Example 4.4 (Example 4.1 continued) Let us return to the example of rolling a die. Recall that \( F \) is the event \( X = 6 \), and \( E \) is the event \( X > 4 \) . Note that \( E \cap F \) is the event \( F \) . So, the above formula gives | \[ P\left( {F \mid E}\right) = \frac{P\left( {F \cap E}\right) }{P\left( E\right) } \] \[ = \frac{1/6}{1/3} \] \[ = \frac{1}{2} \] in agreement with the calculations performed earlier. | Yes |
Suppose we wish to calculate \( P\left( {I \mid B}\right) \) . | Using the formula, we obtain\n\n\[ P\left( {I \mid B}\right) = \frac{P\left( {I \cap B}\right) }{P\left( B\right) } \]\n\n\[ = \frac{P\left( {I \cap B}\right) }{P\left( {B \cap I}\right) + P\left( {B \cap {II}}\right) } \]\n\n\[ = \frac{1/5}{1/5 + 1/4} = \frac{4}{9}\text{.} \] | Yes |
Theorem 4.1 Two events \( E \) and \( F \) are independent if and only if\n\n\[ P\left( {E \cap F}\right) = P\left( E\right) P\left( F\right) . \] | Proof. If either event has probability 0 , then the two events are independent and the above equation is true, so the theorem is true in this case. Thus, we may assume that both events have positive probability in what follows. Assume that \( E \) and \( F \) are independent. Then \( P\left( {E \mid F}\right) = P\left(... | Yes |
Example 4.7 Suppose that we have a coin which comes up heads with probability \( p \), and tails with probability \( q \) . Now suppose that this coin is tossed twice. Using a frequency interpretation of probability, it is reasonable to assign to the outcome \( \left( {H, H}\right) \) the probability \( {p}^{2} \), to ... | We have \( P\left( E\right) = \) \( {p}^{2} + {pq} = p, P\left( F\right) = {pq} + {q}^{2} = q \) . Finally \( P\left( {E \cap F}\right) = {pq} \), so \( P\left( {E \cap F}\right) = \) \( P\left( E\right) P\left( F\right) \) . | Yes |
In the coin-tossing example above, let \( {X}_{i} \) denote the outcome of the \( i \) th toss. Then the joint random variable \( \bar{X} = \) \( \left( {{X}_{1},{X}_{2},{X}_{3}}\right) \) has eight possible outcomes. Suppose that we now define \( {Y}_{i} \), for \( i = 1,2,3 \), as the number of heads which occur in t... | We now illustrate the assignment of probabilities to the various outcomes for the joint random variables \( \bar{X} \) and \( \bar{Y} \) . In the first case, each of the eight outcomes should be assigned the probability \( 1/8 \), since we are assuming that we have a fair coin. In the second case, since \( {Y}_{i} \) h... | Yes |
Example 4.12 (Example 4.10 continued) We now consider the assignment of probabilities in the above example. In the case of the random variable \( \bar{X} \), the probability of any outcome \( \left( {{a}_{1},{a}_{2},{a}_{3}}\right) \) is just the product of the probabilities \( P\left( {{X}_{i} = {a}_{i}}\right) \), fo... | Definition 4.4 The random variables \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) are mutually independent if \[ P\left( {{X}_{1} = {r}_{1},{X}_{2} = {r}_{2},\ldots ,{X}_{n} = {r}_{n}}\right) \] \[ = P\left( {{X}_{1} = {r}_{1}}\right) P\left( {{X}_{2} = {r}_{2}}\right) \cdots P\left( {{X}_{n} = {r}_{n}}\right) \] for any choic... | Yes |
In a group of 60 people, the numbers who do or do not smoke and do or do not have cancer are reported as shown in Table 4.1. Let \( \Omega \) be the sample space consisting of these 60 people. A person is chosen at random from the group. Let \( C\left( \omega \right) = 1 \) if this person has cancer and 0 if not, and \... | Note that we would also see this from the fact that\n\n\[ \nP\left( {C = 1 \mid S = 1}\right) = \frac{3}{13} = {.23}, \n\]\n\n\[ \nP\left( {C = 1}\right) = \frac{1}{6} = {.167}. \n\] | Yes |
Example 4.16 A doctor is trying to decide if a patient has one of three diseases \( {d}_{1},{d}_{2} \), or \( {d}_{3} \) . Two tests are to be carried out, each of which results in a positive (+) or a negative (-) outcome. There are four possible test patterns \( + + , + - \) , -+, and --. National records have indicat... | From this data, we can estimate the prior probabilities for each of the diseases and, given a particular disease, the probability of a particular test outcome. For example, the prior probability of disease \( {d}_{1} \) may be estimated to be \( {3215}/{10},{000} = \) .3215. The probability of the test result \( + - \)... | Yes |
A doctor gives a patient a test for a particular cancer. Before the results of the test, the only evidence the doctor has to go on is that 1 woman in 1000 has this cancer. Experience has shown that, in 99 percent of the cases in which cancer is present, the test is positive; and in 95 percent of the cases in which it i... | We are given that \( \operatorname{prior}\left( \text{cancer}\right) = {.001} \) and \( \operatorname{prior}\left( \text{not cancer}\right) = {.999} \) . We know also that \( P\left( {+ \mid \text{cancer}}\right) = {.99}, P\left( {- \mid \text{cancer}}\right) = {.01}, P\left( {+ \mid \text{not cancer}}\right) = {.05} \... | Yes |
In the spinner experiment (cf. Example 2.1), suppose we know that the spinner has stopped with head in the upper half of the circle, \( 0 \leq x \leq 1/2 \) . What is the probability that \( 1/6 \leq x \leq 1/3 \) ? | Here \( E = \left\lbrack {0,1/2}\right\rbrack, F = \left\lbrack {1/6,1/3}\right\rbrack \), and \( F \cap E = F \) . Hence\n\n\[ P\left( {F \mid E}\right) = \frac{P\left( {F \cap E}\right) }{P\left( E\right) } \]\n\n\[ = \frac{1/6}{1/2} \]\n\n\[ = \frac{1}{3} \]\n\nwhich is reasonable, since \( F \) is \( 1/3 \) the siz... | Yes |
In the dart game (cf. Example 2.8), suppose we know that the dart lands in the upper half of the target. What is the probability that its distance from the center is less than \( 1/2 \) ? | Here \( E = \{ \left( {x, y}\right) : y \geq 0\} \), and \( F = \left\{ {\left( {x, y}\right) : {x}^{2} + {y}^{2} < {\left( 1/2\right) }^{2}}\right\} \) . Hence,\n\n\[ P\left( {F \mid E}\right) = \frac{P\left( {F \cap E}\right) }{P\left( E\right) } = \frac{\left( {1/\pi }\right) \left\lbrack {\left( {1/2}\right) \left(... | Yes |
What is the probability that there is no emission in a further \( s \) seconds, given that the clock reads \( r \) seconds and is still running? | Let \( G\left( t\right) \) be the probability that the next particle is emitted after time \( t \) . Then\n\n\[ G\left( t\right) = {\int }_{t}^{\infty }\lambda {e}^{-{\lambda x}}{dx} \]\n\n\[ = - {\left. {e}^{-{\lambda x}}\right| }_{t}^{\infty } = {e}^{-{\lambda t}}. \]\n\nLet \( E \) be the event \ | Yes |
In the dart game (see Example 4.18), let \( E \) be the event that the dart lands in the upper half of the target \( \\left( {y \\geq 0}\\right) \) and \( F \) the event that the dart lands in the right half of the target \( \\left( {x \\geq 0}\\right) \). Then \( P\\left( {E \\cap F}\\right) \) is the probability that... | \[ P\\left( {E \\cap F}\\right) = \\frac{1}{\\pi }{\\int }_{E \\cap F}{1dxdy} \] \[ = \\operatorname{Area}\\left( {E \\cap F}\\right) \] \[ = \\operatorname{Area}\\left( E\\right) \\operatorname{Area}\\left( F\\right) \] \[ = \\left( {\\frac{1}{\\pi }{\\int }_{E}{1dxdy}}\\right) \\left( {\\frac{1}{\\pi }{\\int }_{F}{1d... | Yes |
In this example, we define three random variables, \( {X}_{1},{X}_{2} \), and \( {X}_{3} \). We will show that \( {X}_{1} \) and \( {X}_{2} \) are independent, and that \( {X}_{1} \) and \( {X}_{3} \) are not independent. Choose a point \( \omega = \left( {{\omega }_{1},{\omega }_{2}}\right) \) at random from the unit ... | We have already seen (see Example 2.13) that\n\n\[ \n{F}_{1}\left( {r}_{1}\right) = P\left( {-\infty < {X}_{1} \leq {r}_{1}}\right) \]\n\n\[ = \sqrt{{r}_{1}},\;\text{if}0 \leq {r}_{1} \leq 1\text{,} \]\n\nand similarly,\n\n\[ {F}_{2}\left( {r}_{2}\right) = \sqrt{{r}_{2}} \]\n\nif \( 0 \leq {r}_{2} \leq 1 \) . Now we ha... | Yes |
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