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\centerline{Mathematics 375 -- Probability and Statistics I}
\centerline{Solutions for Problem Set 4}
\centerline{October 3, 2003}
\bigskip
\noindent
3.10.  Using the information given in the table,
$$\eqalign{
E(Y) &= (1)(.4) + (2)(.3) + (3)(.2) + (4)(.1) = 2\cr
E(1/Y) &= (1)(.4) + (.5)(.3) + (.\overline{3})(.2) + (.25)(.1) \doteq .6417\cr
E(Y^2 - 1) &= (1 - 1)(.4)+(4 - 1)(.3) + (9 - 1)(.2) + (16 - 1)(.1) = 4\cr
V(Y) &= E(Y^2) - (E(Y))^2 = 5 - 4 = 1\cr}$$
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\noindent
3.17.  There are 8 jacks or queens, 8 kings or aces, and $52 - 16 = 36$
other cards in a standard 52-card deck.  The card is drawn at 
random so the probability of getting a jack or a queen is 
${8\over 52}$, and so forth.  Hence the expected value of 
the winnings in one game is
$$E(W) = (15){8\over 52} + (5){8\over 52} + (-4){36\over 52} = {4\over 13} 
\doteq .31$$
(In other words, on average you ``expect'' to win 31 cents on each game.
``Sweet!!!'')
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\noindent
3.19.  Consider the random variable $Y = $ number of contracts assigned
to firm I.  Since the contracts are assigned randomly to the 3 firms,
the probability distribution function (pdf) for $Y$
is:
$$\eqalign{p(0) = P(Y = 0) = {2\over 3}\cdot{2\over 3} = {4\over 9}\cr
           p(1) = P(Y = 1) = {2\over 3}\cdot{1 \over 3} + {1\over 3}\cdot{2\over 3} = {4\over 9}\cr
           p(2) = P(Y = 2) = {1\over 3}\cdot{1\over 3} = {1\over 9}\cr}$$
({\it Comment:}  Note that in the language from the
next section, $Y$ is a {\it binomial random variable} with success probability
$p = 1/3$.)
The profit is the random variable $90,000 Y$, so 
$$E(90,000Y) = (90,000){4\over 9} + (180,000){1\over 9} = 60,000.$$
The expected profit for each of the three firms is the same, so 
if one person owns I and II, the expected profit for that person
is $60,000 + 60,000 = 120,000$.
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\noindent
3.21.  Let $L$ be the random variable representing the loss expected
from a single policy.  The given information determines the 
pdf for $L$ (note it says, ignore all but total loss and half loss).
If the value of the policy is 85,000, then 
$$E(L) = 85,000 P({\rm total\ loss}) + 42,500 P({\rm half\ loss}) 
= (85,000)(.001) + (42,500)(.01) = 510.$$
The insurance company should charge \$510 to come out even (on average).
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\noindent
3.27.  Let $Y$ be the number of components that fail in less than
1000 hours.  This is a binomial random variable with
$p = .2$ and $q = 1 - p = .8$.  
For a),  $P(Y = 2) = {4\choose 2}(.2)^2(.8)^2 \doteq .1536$.
For b),  since the problem says the system can operate if
any two of the components are operating, the probability that
the system is operating after 1000 hours is 
$P(Y \le 2) = \sum_{k=0}^2 {4\choose k} (.2)^k(.8)^{4-k} \doteq .9728$.
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\noindent
3.28.  The number $Y$ that recover is a binomial random variable
with $p = .8$ and $q = .2$. 
\item{a)}  $P(Y = 14) = {20\choose 14}(.8)^{14}(.2)^6 \doteq .109$.
\item{b)}  $P(10 \le Y \le 20) = \sum_{k=10}^{20} {20\choose k}(.8)^k (.2)^{20-k} \doteq .999$.
\item{c)}  $P(14\le Y \le 18) = \sum_{k=10}^{20} {20\choose k}(.8)^k (.2)^{20-k} \doteq .844$.
\item{d)}  $P(Y \le 16) = \sum_{k=0}^{16} {20\choose k}(.8)^k (.2)^{20-k} \doteq .589$

\noindent
Note: these can be computed by calculator, using Maple, or using the 
Table 1 in Appendix III of the text.  The table gives $P(Y \le k)$
for each $k$, so to use the table for c) for instance you would 
take $P(Y\le 18) - P(Y \le 13) = .931 - .087$.
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\noindent
3.31/ The number of families who qualify is a binomial 
random variable $Y$ with $p = .7$.  For a),
the probability $P(Y = 5) = {5\choose 5}(.7)^5 \doteq .1681$.
For b),  $P(Y \ge 4) = {5\choose 4}(.7)^4(.3) + {5\choose 5}(.7)^5 
\doteq .5282$.
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\noindent
3.36/  The number of radar sets $Y$ that detect the missile
is a binomial random variable with $p = .9$.  The probability
that exactly 4 sets detect the missile
is $P(Y = 4) = {5\choose 4}(.9)^4(.1) \doteq .32805$.  
The probability that at least one does is
$P(Y \ge 1) = 1 - P(Y = 0) = 1 - (.1)^5 = .99999$.
For b), the general pattern is that with $n$ radar
sets, $P(Y \ge 1) = 1 - (.1)^n$.  This is $\ge .999$ for 
all $n \ge 3$.  So (only) 3 radar sets are necessary.
(Of course, you don't want to be in the .001 case where
the missile goes undetected!!!)
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\noindent
3.39/  We follow the pattern from the proof of the 
formula for the expected value of $Y(Y-1)$:
$$\eqalign{
E(Y(Y-1)(Y-2)) &= \sum_{y=0}^n y(y-1)(y-2) {n\choose y}p^y q^{n-y}\cr
&= \sum_{y=0}^n y(y-1)(y-2) {n!\over y(y-1)(y-2)(y-3)! (n-y)!}p^y q^{n-y}\cr
&= n(n-1)(n-2)p^3\sum_{y=3}^n {(n-3)!\over (y-3)! ((n-3)-(y-3))!}p^{y-3} q^{(n-3)-(y-3)}\cr
&= n(n-1)(n-2)p^3\sum_{z=0}^{n-3} {(n-3)!\over z! ((n-3)-z)!}p^z q^{(n-3)-z}\cr
&= n(n-1)(n-2)p^3 \cr}$$
(where we wrote $z = y - 3$ to reindex the sum, and then used the fact
that the sum was sum of the pdf-values for a binomial random variable
with $n-3$ trials and success probability $p$.
By the linearity of the expectation:
$$n(n-1)(n-2)p^3 = E(Y^3 - 3Y^2 + 2Y) = E(Y^3) - 3E(Y^2) + 2E(Y)$$
Since we already know $E(Y^2) = n(n-1)p^2 + np$ (from derivation
of formula for $V(Y)$ for a binomial random variable) and $E(Y) = np$, 
we get
$$E(Y^3) = n(n-1)(n-2)p^3 + 3n(n-1)p^2 + np.$$ 
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\noindent
3.51/ Let $Y$ be the number of the interview on which the 
first one with advanced training is found.  This is a geometric
random variable with $p = .3$.  Hence
$P(Y = 5) = (.7)^4(.3) \doteq .07203$.
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\noindent
3.52/  Continuing from 3.51, we have 
$E(Y) = 1/.3 = 10/3 = 3.\overline{3}$ by the formula we 
proved in class for the expected value of a geometric random
variable.
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\noindent
3.55/  For a geometric random variable $Y$, 
$$P(Y > a) = \sum_{y=a+1}^\infty q^{y-1} p = {pq^a \over 1 - q} = q^a$$
(where we used the geometric series sum formula -- this is a 
geometric series with first term $q^ap$ and ratio $q$).  
Similarly, for part b) using the definition of conditional probability
and part a),
$$P(Y> a + b| Y > a) = P(Y>a+b)/P(Y > a) = q^{a+b}/q^a = q^b.$$
{\it Comment:} This is the same as $P(Y > b)$.  In other words,
for geometric random variables, the probability that
$Y > a + b$ given that $Y > a$ (i.e. the probability that 
the first success occurs after trial $a + b$, given that
the first success occurs after trial $a$) is the same
as the probability that the first success occurs after trial $b$.
Geometric random variables thus are ``memoryless'' in 
the sense that starting from the $a$-th trial (with no previous
successes), the probability
that the first success occurs more than $b$ trials later is the same 
as if we started from the beginning with the $0$th trial.  
The previous non-successes don't affect the probability 
of later successes -- no ``memory''.
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\noindent
3.57/ $Y$ is the number on which the first audit
with substantial errors is found -- geometric with $p = .9$.
For a), $P(Y = 3) = (.1)^2(.9) = .009$.  For b),
$P(Y\ge 3) = P(Y > 2) = (.1)^2 = .01$ by problem 55.
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\noindent
3.58/  Use the formulas for geometric random variables
from the text Theorem 3.8 or class notes:  
$E(Y) = 1/.9 = 10/9$.  $\sigma = \sqrt{(1-.9)/(.9)^2} \doteq .35$.
\bye

