10. Solutions to Section Problems

      1. Each student can choose in 4 ways and they each get to choose. So $S$ has MATH points.

        1. The first student can choose in 4 ways and the others then only have 1 section they can go in.
          $\text{Therefore }$ MATH.

        2. The first to pick has 4 ways to choose, the next has 3 sections left, and the last has 2 sections left.
          MATH (different sections) = MATH.

        3. Each has 3 ways to choose a section.
          MATH (nobody in section 1) = MATH

      2. Now $S$ has $n^{s}$ points

        1. $P$ (all in same section) = MATH

        2. $P$ (different sections) = MATH.

        3. $P$ (nobody in section 1) = MATH

      1. There are 26 ways to choose each of the 3 letters, so in all the letters can be chosen in MATH ways. If all letters are the same, there are 26 ways to choose the first letter, and only 1 way to choose the remaining 2 letters. So $P$ (all letters the same) is MATH.

      2. There are MATH ways to choose the 3 digits. The number of ways to choose all even digits is $4 \times4 \times4$. The number of ways to choose all odd digits is $5 \times5 \times5$. MATH (all even or all odd) = MATH.

      1. There are 35 symbols in all (26 letters + 9 numbers). The number of different 6-symbol passwords is $35^{6} - 26^{6}$ (we need to subtract off the $26^{6}$ arrangements in which only letters are used, since there must be at least one number). Similarly, we get the number of 7-symbol and 8-symbol passwords as $35^{7} - 26^{7}$ and $35^{8} - 26^{8}$. The total number of possible passwords is then MATH

      2. Let $N$ be the answer to part (a) (the total no. of possible passwords). Assuming you never try the same password twice, the probability you find the correct password within the first 1,000 tries is $1000/N$.

    4. There are 7! different orders

      1. We can stick the even digits together in 3! orders. This block of even digits plus the 4 odd digits can be arranged in 5! orders.
        MATH (even together) = MATH

      2. For even at ends, there are 3 ways to fill the first place, and 2 ways to fill the last place and 5! ways to arrange the middle 5 digits. For odd at ends there are 4 ways to fill the first place and 3 ways to fill the last place and 5! ways to arrange the middle 5 digits. $P$ (even or odd ends) = MATH.

    5. The number of arrangements in $S$ is $\frac{9!}{3! 2!}$

      1. $E$ at each end gives $\frac{7!}{2!}$ arrangements of the middle 7 letters.

        $L$ at each end gives $\frac{7!}{3!}$ arrangements of the middle 7 letters.

        MATH (same letter at ends) = MATH

      2. The $X, C$ and $N$ can be ``stuck'' together in $3!$ ways to form a single unit. We can then arrange the $3E$'s, $2L$'s, $T$, and $(XCN)$ in $\frac{7!}{3! 2!}$ ways.

        MATH together) =MATH.

      3. There is only 1 way to arrange the letters in the order CEEELLNTX.

        MATH (alphabetical order) = MATH

      1. The 8 cars can be chosen in MATH ways. We can choose $x$ with faulty emission controls and $(8-x)$ with good ones in MATH ways.

        MATH (at least 3 faulty) = MATH since we need $x$ = 3 or 4 or .... or 8.

      2. This assumes all MATH combinations are equally likely. This assumption probably doesn't hold since the inspector would tend to select older cars or those in bad shape.

      1. The first 6 finishes can be chosen in MATH ways. Choose 4 from numbers
        1,2, ...., 9 in MATH ways and 2 from numbers 10, ..., 15 in MATH ways.

        MATH (4 single digits in top 6) = MATH.

      2. Need 2 single digits and 2 double digit numbers in 1$^{\QTR{rm}{st}}$ 4 and then a single digit. This occurs in MATH ways.

        MATH (5$^{\QTR{rm}{th}}$ is $3^{\QTR{rm}{rd}}$ single digit) = MATH. (since we can choose $1^{\QTR{rm}{st}}$ 4 in MATH ways and then have 11 choices for the $5^{\QTR{rm}{th}}$)

        Alternate Solution: There are $15^{(5)}$ ways to choose the first 5 in order. We can choose in order, 2 double digit and 3 single digit finishers in $6^{(2)}9^{(3)}$ ways, and then choose which 2 of the first 4 places have double digit numbers in MATH ways.

        $\text{Therefore }P$ ($5^{\QTR{rm}{th}}$ is $3^{\QTR{rm}{rd}}$ single digit) =MATH.

      3. Choose 13 in 1 way and the other 6 numbers in MATH ways. (from 1,2, ....., 12).

        $\text{Therefore }P$ (13 is highest) = MATH.

        Alternate Solution: From the MATH ways to choose 7 numbers from 1,2, ..., 13 subtract the MATH which don't include 13 (i.e. all 7 chosen from 1,2, ..., 12).

        $\text{Therefore }P$ (13 is highest) = MATH

    8. Let MATH and MATH, and draw yourself a Venn diagram. Then

      MATH
      MATH

      (Note that the information that 40% of days with temp $>~22^{\circ} $ have no rain is not needed to solve the question).

      MATH
      This result is to be expected since 80% of days have a high temperature $\leq~22^{\circ}C$ and 30% of these days have rain.

    9. $P(ML)=.15$
      $P(M)=.45$
      $P(L)=.45$ See the Figure below:
      p181.eps
      The region outside the circles represents females to the right. To make $P(S)=1$. we need $P(FR)=.25$.

      1. MATH
        MATH
        MATH is the largest value, and this occurs when MATH.

      2. Since $ABC$ is contained within both $AC$ and $BC$ we know $P(ABC)~\leq~P(AC)$ and $P(ABC)~\leq~P(BC)$. Thus MATH iff MATH.
        While $A$ and $C$ could be mutually exclusive, it can't be determined for sure.

    11. MATH Alternatively, (look at a Venn diagram), MATH is a partition, so MATH.

      1. Points giving a total of 9 are: (3,6), (4,5), (5,4) and (6,3). The probabilities are (.1)(.3) = .03 for (3,6) and for (6,3), and (.2)(.2) = .04 for (4,5) and for (5,4).
        MATH (3, 6) or (4, 5) or (5, 4) or (6, 3)) = .03 + .04 + .04 + .03 = .14.

      2. There are MATH arrangements with 1 nine and 3 non-nines. Each arrangement has probability $(.14)(.86)^{3}$.
        MATH (nine on 1 of 4 repetitions) = MATH

    13. Let $W$ = {at least 1 woman} and $F$ = {at least 1 French speaking student}.
      MATH
      p182.eps
      Venn diagram, Problem 4.3.2

      But MATH (no woman and no French speaking student)$=P$ (all men who don't speak French)
      $P$ (woman who speaks French) =$P$ (woman)P(French\woman)= $.45\times.20=.09$.

      p183.eps
      P(woman who speaks french)

      From Venn diagram, $P$ (man without French) = .49.
      MATH MATH.

    14. From a Venn diagram: (1) MATH (2) MATH

      MATH
      MATH
      MATH
      MATH
      MATH
      MATH
      MATH

      MATH and $\overline{B}$ are independent iff $\overline{A}$ and $B$ are independent.

    15. Let $B$ = MATH and $L$ = MATH .

      MATH
      MATH

    16. Let $F$ = MATH and $H$ = MATH

      MATH

    17. Let $H$ = { defective headlights}, $M$ = {defective muffler} MATH

    18. MATH
      Differentiate with respect to $a$ on both sides:
      MATH. Multiply by $a$ to get MATH
      Let MATH. Then MATH
      Multiply by $(1-p)^{n}$:
      MATH

    19. Let $Q$ = {heads on quarter} and $D$ ={heads on dime}. Then MATH

    20. We need $f(x)\geq0$ and MATH MATHMATHMATHMATH But if $c=-1$ we have $f(1)<0$ ... impossible. MATH

    21. We are arranging $Y~F~O~O~O$ where $Y$ = {you}, $F$ = {friend}, O = {other}. There are $\frac{5!}{3!}=20$ distinct arrangements.

      $X=0$: MATH has 4 arrangements with $Y$ first and 4 with $F$ first.
      $X=1$: $YOFOO,\cdots,OOYOF$ has 3 arrangements with $Y$ first and 3 with $F$ first.
      $X=2$: $YOOFO,OYOOF$ has 2 with $Y$ first and 2 with $F$.
      $X=3$: $YOOOF$ has 1 with $Y$ first and 1 with $F$.

      $x$ 0 1 2 3
      $f(x)$ .4 .3 .2 .1
      $F(x)$ .4 .7 .9 1

    22. MATHMATHMATH

      1. Using the hypergeometric distribution, MATH MATH

      2. While we could find none tainted if $d$ is as big as 3, it is not likely to happen.
        This implies the box is not likely to have as many as 3 tainted tins.

    24. Considering order, there are $N^{(n)}$ points in $S$. We can choose which $x$ of the $n$ selections will have "success" in MATH ways. We can arrange the $x$ "successes" in their selected positions in $r^{(x)}$ ways and the $(n-x)$ "failures" in the remaining positions in $(N-r)^{(n-x)}$ ways.

      MATH with $x$ ranging from max $(0,n-(N-r))$ to min $(n,r)$

      1. Using hypergeometric, with $N=130,r=26,n=6$, MATH

      2. Using binomial as an approximation, MATH

      26. (a)

      1. $P$ (fail twice)
        = $P(A)P$ (fail twice $|A$) + $P(B)P$ (fail twice $|B$)
        = MATH.
        Where $A$ = { camera $A$ is picked }
        and $B$ = { camera $B$ is picked }

        This assumes shots are independent with a constant failure probability.

      2. $P(A|$ failed twice) = MATH

    27. We need $(x-25)$ "failures" before our 25th "success". MATH

      1. In the first $(x+17)$ selections we need to get $x$ defective (use hypergeometric distribution) and then we need a good one on the MATH draw. MATH

      2. Since 2500 is large and we're only choosing a few of them, we can approximate the hypergeometric portion of $f(x)$ using binomial MATH

    29. Using geometric, MATHMATHMATHMATHMATH At least 9.4 cars means 10 or more cars must be checked
      MATH

      1. Let $X$ be the number who don't show. Then MATHMATHMATH (To use a Poisson approximation we need $p$ near $0$. That is why we defined "success" as not showing up).

        For Poisson, MATH
        MATH

      2. Binomial requires all passengers to be independent as to showing up for the flight, and that each passenger has the same probability of showing up. Passengers are not likely independent since people from the same family or company are likely to all show up or all not show. Even strangers arriving on an earlier incoming flight would not miss their flight independently if the flight was delayed. Passengers may all have roughly the same probability of showing up, but even this is suspect. People travelling in different fare categories or in different classes (e.g. charter fares versus first class) may have different probabilities of showing up.

      1. MATHMATH

      2. MATH MATH MATH MATH Note this is a binomial probability function.

    32. Assuming the conditions for a Poisson process are met, with lines as units of ``time":

      1. $\lambda= .02$ per line; $t = 1$ line; MATH MATH

      2. MATH MATH

    33. Consider a 1 minute period with no occurrences as a ``success''. Then $X$ has a geometric distribution. The probability of ``success'' is MATHMATH (There must be $(x-1)$ failures before the first success.)

      1. MATH MATH

      2. MATH, using a geometric distribution

      3. Using a binomial distribution MATH Approximate by Poisson with MATH

        MATH ($n$ large, $p$ small)
        Thus, MATH.

    35. There are 10 tickets with all digits identical. For these there is only 1 prize. There are
      $10 \times9$ ways to pick a digit to occur twice and a different digit to occur once. These can be arranged in MATH different orders; i.e. there are 270 tickets for which 3 prizes are paid. There are $10 \times9 \times8$ ways to pick 3 different digits in order. For each of these 720 tickets there will be 3! prizes paid.

      The organization takes in $1,000. MATHMATH i.e., on average they lose $28.

    36. Let them sell $n$ tickets. Suppose $X$ show up. Then MATH. For binomial, MATH
      If $n \leq120$, revenues will be $100 X$, and MATH. This is maximized for $n = 120$. $\text{Therefore }$ Max. expected revenue is $11,640.
      For $n ~=~ 121$, revenues are $100 X$, less $500 if all 121 show up.
      i.e., MATH
      MATH = $11,724.46 is expected.
      For $n = 122$, revenues are $100 X$, less $500 if 121 show up, less $1000 if all 122 show.
      i.e., MATH
      $= ~\$11,763.77$ is expected.
      For $n = 123$, revenues are $100 X$, less $500 if 121 show, less $1,000 if 122 show, less $1500 if all 123 show.
      i.e. MATH
      MATH
      $=~ \$11,721.13$ is expected.
      $\text{Therefore }$ They should sell 122 tickets.

      1. Let $X$ be the number of words needing correction and let $T$ be the time to type the passage. Then MATH and $T = 450 + 15 X $.

        $X$ has mean $np = 18$ and variance $np (1-p) = 17.28$.
        MATH
        Var$(T)$ = VarMATH Var$(X) = 3888$.

      2. At 45 words per minute, each word takes $1\frac{1}{3}$ seconds. MATH and
        MATH
        MATH, so it takes longer on average.

      1. The marginal probability functions are:

        MATH

        Since MATH
        $\text{Therefore }X$ and $Y$ are not independent.
        e.g. MATH

      2. MATH MATH

      3. MATH

        MATH

        (e.g. MATH)

    39. MATHMATH ($y$ starts at $x~\text{since }$ no. of calls $\geq$ no. of sales). MATHMATHMATH

    40. MATH using the given identity on MATH. ($T$ has a negative binomial distribution)

      1. Use a multinomial distribution. MATH

      2. Group $C$'s and $D$'s into a single category. MATH

      3. Of the 21 non $D$'s we need $3A$'s, $11~B$'s and $7C$'s. The (conditional) probabilities for the non-$D$'s are: 1/8 for $A$, 4/8 for $B$, and $3/8$ for $C$.
        (e.g. MATH
        MATH.

    42. MATH

      $p_{1} = P$ (fewer than 5 chips) = MATH

      $p_{2} = P$ (more than 9 chips) = MATH

      1. MATH

      2. MATH

      3. Given that 7 have $>$ 9 chips, the remaining 5 are of 2 types - under 5 chips, or 5 to 9 chips MATH Using a binomial distribution,

        $P$ (3 under 5$|$7 over 9) = MATH

    43. MATH

      MATH
      MATH
      MATH
      MATH
      MATH
      MATH
      MATH

      While $\rho= 0$ indicates $X$ and $Y$ may be independent (and indeed are in this case), it does not prove that they are independent. It only indicates that there is no linear relationship between $X$ and $Y$.

        1. $x$ 2 4 6
          $f_{1}(x)$ 3/8 3/8 1/4

          $y$ -1 1
          $f_{2}(y)$ $\frac{3}{8} + p$ $\frac{5}{8} - p$

          MATH

          MATH

          CovMATH

          Therefore $p =5/72$

        1. If $X$ and $Y$ are independent then Cov$(X,Y)=0$, and so $p$ must be 5/72. But if $p=5/72$ then MATH $\text{Therefore }X$ and $Y$ cannot be independent for any value of $p$

      2. MATHMATHMATHMATHMATHMATH

      3. Let MATH

        MATHMATHMATHMATH Pairs are independent if they have no common points, but may not be independent if the pairs are adjacent. MATHMATHMATHMATHMATHMATHMATHMATH

      4. MATH
        MATH
        MATH

      5. MATH
        MATH
        MATH
        Also, MATH for $j \neq i \pm1$ and MATH
        MATH

      6. Using $X_{i}$ as defined,
        MATH since $X_{i} = X_{i}^{2}$
        MATH since only 1 cut is needed
        MATH since 2 cuts are needed.
        VarMATH = VarMATH
        VarMATH = VarMATH = VarMATH
        Cov MATH if $j \neq i \pm1$ since there are no common pieces and cuts are independent.
        MATH
        (product is 0 if either $x_{i}$ or $x_{i + 1}$ is a 0)
        MATH MATH MATH MATH
        Var MATH Var MATH CovMATH
        MATH
        $~~~~= 6.9516$
        $\text{Therefore }$ s.d.MATH

        1. MATH
          MATH

        2. MATH

        3. MATH

        4. MATH; MATH MATH

        5. MATH
          MATH
          MATH

        1. MATH
          MATH

        2. MATH

        3. Let $m$ be the median. Then MATH
          MATH and so the median is 1

      9. MATH.
        If MATH is a random number between $0$ and $1$, then MATH For $y = .27125$ we get MATH

      10. Let the time to disruption be $X$. MATH MATH. Take natural logs.
        MATH hours.

        1. $F(x) = P$ (distance $\leq x$) = $1 - P$ (distance $> x$)
          = $1 - P$ (0 flaws or 1 flaw within radius $x$)
          The number of flaws has a Poisson distribution with mean MATH
          MATH
          MATH

        2. MATH
          Let $y=\lambda\pi x^{2}$. Then $dy=2\lambda\pi xdx$, so MATH
          MATH
          MATH

        1. MATH.

          \begin{tabbing} = \=~~ $P ( - .8 < Z < 1.1 )$ \\ = \> $ F (1.1) - F (-.8)$ \\ = \> $F (1.1) - [ 1 - F (.8) ]$ \\ = \> $.8643 - (1 - .7881 ) = .6524$ \end{tabbing} (see the figure)

          prob9.4.1.eps

        2. MATH
          $P(2Y>X)=P(2Y-X>0)$
          MATH

        3. MATH
          MATH
          MATH
          $=~1-.9332=.0668$ (see the figure)

          prob9.4.1c.eps

      13. MATH

        MATH (about 2/3)

        MATH

        MATH (about 95%)
        Similarly, MATH (over 99%)

        1. MATH
          MATH
          MATH
          MATH

        2. MATH
          MATH
          MATH
          MATH
          MATH. Take $n=2165$ observations.

      15. Let $X$ be the number germinating.
        Then $X~\sim ~Bi(100,.8)$.
        MATH. (see the figure)

        prob9.5.1.eps
        Approximate using a normal distribution with $\mu=np=80$ and MATH.
        MATH
        MATH
        MATH
        Possible variations on this solution include calculating
        $F(1.375)$ as MATH and realizing that $X\leq100$ means
        MATH. However
        MATH so we get the same answer as before.

      16. Let $X_{i}$ be the cost associated with inspecting part $i$
        MATH
        MATH
        MATH
        By the central limit theorem
        MATH approx.
        Since $\sum X_{i}$ increases in $10 increments,
        MATH

      \setcounter{figure}{0}

      11. Answers to End of Chapter Problems

      Chapter 2:

      1. Label the profs $A,B,C$ and $D$. MATH

      2. 1/4

      3. MATH $~~$ (b) $\frac{1}{4}$;

      4. (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5); 0.4;

      5. MATH

      6. MATH (b) MATH (c) MATH

      7. .018 (b) .020 (c) 18/78 = .231

      8. .978

      Chapter 3:

      1. (a) 4/7 $~~$ (b) 5/42 $~~$ (c) 5/21;

      2. (a) (i) MATH (ii) MATH
        (b) All $n^{r}$ outcomes are equally likely. That is, all $n$ floors are equally likely to be selected, and each passenger's selection is unrelated to each other person's selection. Both assumptions are doubtful since people may be travelling together (e.g. same family) and the floors may not have equal traffic (e.g. more likely to use the stairs for going up 1 floor than for 10 floors);

      3. (a) 5/18 $~~$ (b) 5/72;

      4. MATH

      5. (a) 1/50,400 $~$ (b) 7/45;

      6. (a) 1/6 (b) 0.12;

      7. Values for $r=20,40$ and 60 are .589, .109 and .006.

      8. (a) $\frac{1}{n}$ (b) $\frac{2}{n}$

      9. MATH

      10. (a) (i) .0006 (ii) .0024 (b) MATH

      11. (a) MATH (b) 15

      12. (a) MATH (b) MATH (c) MATH (d) MATH

      13. MATH

      Chapter 4:
      4.1 .75, .6, .65, 0, 1, .35, 1

      4.2 $A$- .01, $B$ - .72, $C$ - $.9^{3}$, $D$ - $.5^{3}$, $E$ - $.5^{2}$

      4.3 $\frac{1}{6}$

      4.4 (a) 0.0576 (b) 0.4305 (c) 0.0168 (d) 0.5287

      4.5 .44

      4.6 0.7354

      4.7 (a) .3087 $~~$ (b) .1852; $~~$

      4.9 .342

      4.10 (a) .1225, .175 (b) .395

      4.11 MATH

      4.14 (a) MATH (b) MATH (c) $\frac{24p}{1+24p}$

      4.15 .9, .061, .078

      4.16 (a) .024 (b) 8 on any one wheel and 1 on the others

      4.17 (a) .995 and .005 (b) .001

      4.18 (a) .99995 (b) .99889 (c\) MATH

      4.19 (a) $\frac{r}{r+1999}$; $0.005,$ $0.0148,0.0476$ (b) 2.1%

      Chapter 5:
      5.4 .545

      Chapter 6:

      6.1 (a) .623, .251; for males, .408, .103 (b) .166

      6.2 (a) MATH (b) MATH

      6.4 MATH

      6.5 (a) .0800 (b) .171 (c) .00725

      6.6 (a) .010 (b) .864

      6.7 (a)$\frac{4}{15}$ (b) MATH (c) .0176

      6.8 0.9989

      6.9 (a) .0758 (b) .0488 (c) MATH (d) $\lambda=.12$

      6.10 (a) .0769 (b) 0.2019; 0.4751

      6.11 (a) 0.2753 (b) 0.1966 (c) 0.0215

      6.12 (b) enables us to approximate hypergeometric distribution by binomial distribution when $n$ is large and $p$ near 0.

      6.13 (a) MATH (b) (Could probably argue for other answers also). Budworms probably aren't distributed at a uniform rate over the forest and may not occur singly

      6.14 (a) .2048 (b) .0734 (c) .428 (d) .1404

      6.15 MATH

      6.16 (a) .004264; .006669 (b) .0032 (c) (i) MATH $~~$ (ii) $9.336\times10^{-5}$
      On the first 1399 attempts we essentially have a binomial distribution with $n=1399$ (large) and $p=.004264$ (near $0$)

      6.17 (a) MATH (b) $\lambda\leq0.866$ bubbles per $m^{2}$

      6.18 MATH

      6.19 (a) $(1-p)^{y}$ (b) $Y=0$ (c) $p/[1-(1-p)^{3}]$ (d) MATH for $r=0,1,2$

      6.20 (a) .555 (b) .809; .965 (c) .789; .946 (d) $n=1067$

      6.21 (a) MATH (b) .002, .051, .350, .797 (c) MATH

      Chapter 7:

      7.1 2.775; 2.574375

      7.2 -$3

      7.3 $16.90

      7.4 (a) 3 cases (b) 32 cases

      7.5 (a) - 10/37 dollars in both cases (b) .3442; 0.4865

      7.6 $.94

      7.7 (b) MATH, which gives MATH for $k=1,5,10$

      7.8 50

      7.9 (a) MATH; (b) MATH

      7.11 (a) Expand $M(t)$ in a power series in powers of $e^{t},$ i.e. MATHThen $P(X=j)=$coefficient of MATH (b) Similarly MATH Then MATH

      Chapter 8:

      8.1 (a) no MATH (b) 0.3 and 1/3

      8.2 (a) mean = 0.15, variance = 0.15

      8.3 (a) No (b) 0.978 (c) .05

      8.4 (a) MATH (b) MATH

      8.5 (b) - .10 dollars (c) $d=.95/n$

      8.7 MATH; (b) note e.g. that MATH, but $f(0,3)=0$

      8.8 (a) MATH and $y=0,1$ (b) MATH;
      MATH (c) 49/120 and 1/2

      8.9 (a) MATH (b) $e^{-4}$ (c) MATH
      (d) MATH

      8.10 (b) .468

      8.11 (a) MATH (b) MATH (c) MATH

      8.12 (a) 1.76 (b) MATH

      8.13 (a) Multinomial (b) .4602 (c) $5700

      8.15 207.867

      8.16 (a) $Bi(n,p+q)$ (b) $n(p+q)$ and $n(p+q)(1-p-q)$ (c) $-npq$

      8.17 (a) MATH (b) 0
      (c) no. e.g. MATH and $V=1)=0$

      8.19 -1

      8.20 (a) 1.22 (b) 17.67%

      8.21 MATH

        1. The transition matrix is MATH from which, solving MATH and rescaling so that the sum of the probabilities is one, we obtain MATH the long run fraction of time spent in cities A,B,C respectively.

        2. By arguments similar to those in section 8.3, the limiting matrix has rows all identically $\pi^{\prime}$ where the vector $\pi ^{\prime}$ are the stationary probabilities satisfying MATH and MATH The solution is MATH and the limit is MATH

        3. With $Z=X+Y,$ MATH and since this is the MGF of a Poisson(MATH distribution, this must be the distribution of $Z.$

        4. If today is raining, the probability of Rain, Nice, Snow three days from now is obtainable from the first row of the matrix $P^{3},$ i.e. $(0.406$ $0.203$ $\ 0.391).$The probabilities of the three states in five days, given (1) today is raining (ii) today is nice (iii) today is snowing are the three rows of the matrix $P^{5}.$ In this case call rows are identical to three decimals; they are all equal the equilibrium distribution MATH $0.200$ $0.400).$

          1. If a>b, and both parties raise then the probability B wins is MATH and the probability A wins is 1 minus this or MATH. If $a\leq b,$ then the probability A wins is MATH

          5. (a)

          1. In the special case b=1, the expected winnings of A are MATH For a=1,2,...,13 this gives expected winnings of 0, 0.38462, 0.69231, 0.92308, 1.0769, 1.1538, 1.1538, 1.0769, 0.92308, 0.69231, 0.38462, 0, -0.46154 respectively, maximum for a=6 or 7.

          2. By a similar calculation as in part (b), A's possible winnings per game for a=1,2,...,13 are, respectively, -0.38462, -0.34615, -0.30769, -0.26923, -0.23077, -0.19231, -0.15385, -0.11538, -0.076923, -0.038462, 0, 0, -0.076923. There is no strategy that provides a positive expected return. The optimal is the break-even strategy a=11 or 12. (Note: in this two-person zero-sum game, a= 11 or 12 and b=11 or 12 is a minimax solution)

          1. Show that you can determine the probability of the various values of $X_{t+1}$ knowing only the state $X_{t}$ (and without knowing the previous states).

          2. For example the long-run probability of the state $(i,j,k)$ is, with MATH MATH

          3. The probability that record $j$ is in position $k=1,2,3$ is, with MATH MATH respectively. The expected cost of accessing a record in the long run is MATH Substitute MATH so MATH and MATH and (q8.28) is 1.7214.

          4. If they are in random order, the expected costMATH If they are ordered in terms of decreasing $p_{j},$ expected cost is MATH

        Chapter 9:

        9.1 MATH for MATH

        9.2 (a) MATH
        (b) Find $c$ such that $c^{3}-3c+1.9=0$. This gives $c=.811$

        9.3 (a) 1/2, 1/24 (b) 0.2828 (c) 0.2043

        9.4 $f(y)=1;~~0<y<1$

        9.5 (a) $\alpha>-1$ (b) 0.5MATH (c) MATH

        9.6 (a) $(1-e^{-2})^{3}$ (b) $e^{-.4}$

        9.7 MATH

        9.8 (a) .0668, .2417, .3829, .2417, .0668 (b) .0062 (c) .0771

        9.9 (a) .5 (b) $\mu\geq2.023$

        9.10 .4134

        9.11 (a) .3868 (b) .6083 (c) 6.94

        9.12 (a) .0062 (b) .9927

        9.13 (a) .2327, .1841 (b) .8212, .8665; Guess if $p_{i}=0.45$, don't guess if $p_{i}=0.55$

        9.14 6.092 cents

        9.15 574

        9.16 (a) 7.6478, 88.7630 (b) 764.78, 8876.30, people within pooled samples are independent and each pooled sample is independent of each other pooled sample. (c) 0.3520

        9.17 .5969

        9.18 (a) .6728 (b) 250,088

        9.19 (a) MATH (b) MATH (using table) for $n=20,50,100$ The more you play, the smaller your chance of winning. (c) 1264.51
        With probability .99 the casino's profit is at least $1264.51.

        9.20 (a) $X$ is approximately MATH (b) (i)MATH (ii) MATH

        9.21 (a) (i) .202 (ii) .106 (b) .045, .048

        9.22 (a) False positive probabilities are .048, .091, .023 for (i), (ii), (iii). False negative probabilities are .023, .091, .048.

          1. Let $Y=$total change over day. Given $N=n,$ $Y$ has a Normal$(0,n\sigma^{2})$ distribution and therefore MATH Not a MGF in this course at least. The mean is MATH and the variance is MATH

            1. $\exp(t+t^{2})$

            2. $\exp(2t+2t^{2})$

            3. $\exp(nt+nt^{2})$

            4. $\exp(t^{2})$










          Summary of Distributions




          Discrete
          Notation and
          Parameters
          Probability function
          $f(x)$
          		Mean Variance
          Moment generating
          function $M_{X}(t)$
          MATH $\binom{n}{x}$ $p^{x}q^{n-x}$ $np$ $npq$ $(pe^{t}+q)^{n}$
          Bernoulli$(p)$
          $x=0,1$
          $0<p<1,q=1-p$
          		$p^{x}(1-p)^{1-x}$ $p$ $p(1-p)$ $(pe^{t}+q)$
          Negative Binomial$(k,p)$
          $x=0,1,...$
          $0<p<1,q=1-p$
          		MATH $\frac{kq}{p}$ $\frac{kq}{p^{2}}$ MATH
          Geometric$(p)$
          $x=0,1,...$
          $0<p<1,q=1-p$
          		$pq^{x}$ $\frac{q}{p}$ $\frac{q}{p^{2}}$ MATH
          Hypergeometric$(N,r,n)$
          $x=0,1,...$min($r,n)$
          $r<N,n<N$
          		MATH $\frac{nr}{N}$ MATH intractible
          Poisson$(\lambda)$ MATH $\lambda$ $\lambda$ MATH
          Continuous
          p.d.f.
          $f(x)$
          		Mean Variance
          Moment generating
          function $M_{X}(t)$
          Uniform$(a,b)$
          $a<x<b$
          		$\frac{1}{b-a}$ $\frac{a+b}{2}$ MATH MATH
          Exponential$(\theta)$
          $0<x,$ $0<\theta$
          		MATH $\theta$ $\theta^{2}$ MATH
          Normal$(\mu,\sigma^{2})$
          MATH
          MATH $\ \sigma^{2}>0$
          		MATH $\mu$ $\sigma^{2}$ MATH

          Probabilities For the Standard Normal Distribution $N(0,1)$