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\centerline{Mathematics 375 -- Probability Theory}
\centerline{Solutions for Problem Set 4}
\centerline{September 30, 2011}
\vskip 10pt
\noindent
3.12.  We have  $E(Y) = (1)(.4) + (2)(.3) + (3)(.2) + (4)(.1) = 2$.
Similarly $E(1/Y) \doteq (1)(.4) + (.5)(.3) + (.333)(.2) + (.25)(.1) \doteq .642$.
Then $E(Y^2-1) = (0)(.4) + (3)(.3) + (8)(.2) + (15)(.1) = 4$.  Finally
$V(Y) = E(Y^2) - (E(Y))^2 = 5 - 2^2 = 1$.  
\vskip 10pt
\noindent
3.14 (a)  $E(Y) = 7.9$.  (b) $V(Y) = E(Y^2) - (E(Y))^2 = 67.14 - (7.9)^2 = 4.73$, so 
$\sigma(Y) = \sqrt{4.73} \doteq 2.17$.  (c)  The interval $\mu \pm 2\sigma$ is
$7.9 \pm 4.35 = [3.55,12.25]$. The probability that the patent life is
in this range is $1 - .03 - .01 = .96$.  (This is quite good agreement with
the empirical rule, of course.)
\vskip 10pt
\noindent
3.17.  Recall from PS 3 that the probability mass function for $Y = $ number
of empty bowls was 
\begin{align*}
P(Y = 0) &= \frac{6}{27} = \frac{2}{9}\\
P(Y = 1) &= \frac{18}{27} = \frac{2}{3}\\
P(Y = 2) &= \frac{3}{27} = \frac{1}{9}.
\end{align*}
So $E(Y) = (0)(2/9) + (1)(2/3) + 2(1/9) = 8/9$
and $V(Y) = E(Y^2) - E(Y)^2 = 10/9 - 64/81 = 26/81$ and $\sigma \doteq .567$.  The
interval $\mu \pm 2\sigma$ is approximately $[-.24,2.02]$.  \emph{All} values of $Y$ are in 
this interval, so the probability = 1.   
\vskip 10pt
\noindent
3.18.  Recall from PS 3 that the probability function for $Y = $
number of cells in second generation was
\begin{align*}
P(Y = 0) &= .1 + (.9)(.1)^2 = .109\\
P(Y = 2) &= (.9)((.1)(.9) + (.9)(.1)) = .162\\
P(Y = 4) &= (.9)^3 = .729.   
\end{align*}    
So the mean number of cells in generation 2 is $(0)(.109) + (2)(.162) + (4)(.729) = 3.24$.
\vskip 10pt
\noindent
3.19.  If $C$ is the premium charged for the policy, then the company's
profit is $C - 15$ if the event $A$ does not occur, and $C - 15 - 1000$
if it does.  Since the probability of $A$ is $.02$ and the probability 
of $A$ not occurring is $.98$, the expected profit is
$E(P) = (.98)(C - 15) + (.02)(C - 15 - 1000)$.  To make this equal $50$,
$C$ should solve 
$$50 = C - 15 - 20,\ {\rm so}\  P = 85. $$
(Note that we need to use the probabilities $.98$ and $.02$ here, not
the percentages.)
\vskip 10pt
\noindent
3.31.  (a) Whether or not the mean of $W = 2Y$ is larger than the mean of $Y$
will depend on whether $\mu = E(Y)$ is positive or negative.  If $E(Y) > 0$, 
then the mean of $W$ will be larger; if $E(Y) < 0$, then it will be 
smaller.    
(b)  By Theorem 3.4, $E(W) = 2E(Y)$.  This does agree with (a).  (c)  The
variance of $W$ should also be larger since the values of $W$ will be
more spread out.  (d)  $V(W) = E(W^2) - (E(W))^2 = E(4Y^2) - (2E(Y))^2$.
This equals $4(E(Y^2) - (E(Y))^2) = 4V(Y)$.  
\vskip 10pt
\noindent
3.33.  (a)  By Theorem 3.5, $E(aY + b) = E(aY) + E(b)$.  Then since
$a,b$ are constant, Theorems 3.4 and 3.3 imply this equals $aE(Y) + b$.  
(b)  We have by part (a) and Theorems 3.3 - 3.5:
\begin{align*}
V(aY + b) &= E((aY + b)^2) - (E(aY+b))^2\\
&= E(a^2 Y^2 + 2 a b Y + b^2 - (a E(Y) + b)^2\\ 
&= a^2 E(Y^2) + 2 a b E(Y) + b^2 - a^2 (E(Y))^2 - 2 a b E(Y) - b^2\\
&= a^2(E(Y^2) - (E(Y))^2) \\
&= a^2 V(Y).
\end{align*}
\vskip 10pt
\noindent
3.37.  Of these quantities, (c) and (e) could be approximated by binomial
random variables.  $n = 100$ for both, but $p$ cannot be determined in 
either case from the given information.  Note that (a) is not a random
variable at all since the description does not say ``in the sample.''
The (total) number of students who took the SAT that year is a constant.  (b) and (d)
are random variables but they do not match the number of successes in a binomial
experiments since there are not just two possible outcomes for the score or
the time taken for each of the students in the sample.  
\vskip 10pt 
\noindent
3.39.  (a)  $Y = $ the number of components
that last longer than 1000 hours has a binomial
distribution.   $P(Y = 2) = \binom{4}{2}(.8)^2(.2)^2 \doteq .1536$.

(b)  The subsystem will operate if $2,3$, or $4$ of the 
components last longer than 1000 hours.  So this 
probability is 
$P(Y\ge 2) = \binom{4}{2}(.8)^2(.2)^2 + \binom{4}{3}(.8)^3(.2) + \binom{4}{4}(.8)^4 \doteq .973$.  
\vskip 10pt 
\noindent  
3.40.  (a) $P(Y = 8) = \binom{20}{14}(.8)^{14}(.2)^6 \doteq .109$.
\vskip 10pt\noindent
(b) 
$$P(Y \ge 10) = \sum_{y=10}^{20} \binom{20}{y}(.8)^y(.2)^{20-y} \doteq .999.$$  
(c)
$$P(14 \le Y \le 18) = \sum_{y=14}^{18} \binom{20}{y}(.8)^y(.2)^{20-y} \doteq .844.$$ 
(d) 
$$P(Y \le 16) = \sum_{y=0}^{16} \binom{20}{y}(.8)^y(.2)^{20-y} \doteq .589.$$ 

\noindent
3.41.  The probability is 
$$P(Y \ge 10) = \sum_{y=10}^{15} \binom{15}{y} (.2)^y (.8)^{15-y} \doteq .0001.$$
\noindent
3.42.  (a)  If one of the incorrect answers can be eliminated each time, then
the probability of finding the correct one by guessing goes up 
to $p = .25$.  So the probability of getting at least 10 correct answers
is now
$$P(Y \ge 10) = \sum_{y=10}^{15} \binom{15}{y} (.25)^y (.75)^{15-y} \doteq .0008.$$
Similarly if two incorrect answers can be eliminated each time,
$$P(Y \ge 10) = \sum_{y=10}^{15} \binom{15}{y} (.333)^y (.667)^{15-y} \doteq .008.$$
(In other words, this much-discussed ``test-taking strategy'' is really useful only
when you can use it sparingly.   \emph{Really knowing the material--so guessing
is unnecessary most of the time--is a much superior strategy!!})
\vskip 10pt
\noindent
3.46.  The probability histogram for $p = .1$ should have boxes of heights
$$.590490, .32805, 0.07290, 0.00810, 0.00045, 0.000010$$ 
(the first box extends from $-1/2$ to $1/2$, etc..
This one is skewed to the left.  With $p = .5$, the boxes have heights
$$0.03125, .15625, .3125, .3125, .15625, 0.03125$$
This one is symmetrical.  Finally with $p = .9$ the heights are
the same as for $.1$ but in reverse order.  This one is skewed right.
\vskip 10pt
\noindent
3.48.  (a) The number of sets that detect the incoming missile is binomial
with $n = 5$ and $p = .9$.  The probability that 
exactly $4$ sets detect it is $\binom{5}{4}(.9)^4(.1) = .32805$.
The probability that at least one detects it is
$$1 - P(Y = 0) = 1 - (.00001) = .99999.$$
(b)  ``Detecting'' should mean that at least one set detects
the incoming missile.  The result of part (a) shows that $n = 5$
is certainly big enough.  But in fact if $n = 3$, 
$$1 - P(Y = 0) = 1 - \binom{3}{0}(.1)^3 = .999.$$
So $n = 3$ is large enough.  
\vskip 10pt
\noindent
3.64.  We have $P = P(Y = y_0) = \binom{n}{y_0}p^{y_0} (1 - p)^{n-y_0}$, where
we think of this as a function of the \emph{unknown value} $p$.  Given that
$y_0$ successes were observed, the question is to determine a best estimate
for $p$.    

This function of $p$ will be maximized at a point where $\frac{dP}{dp} = 0$.  
In office hours, I suggested maximizing $\ln(P)$ instead because 
that simplifies the algebra.  It is also possible to differentiate $P$
directly, of course.  We have
by the product rule:
$$ 
\frac{dP}{dp} = \binom{n}{y_0}\left(y_0 p^{y_0 - 1} (1 - p)^{n-y_0} 
+ (n-y_0)p^{y_0}(1 - p)^{n-y_0-1}\right).$$
Set this equal to zero and divide through by $\binom{n}{y_0} p^{y_0-1}(1 - p)^{n-y_0-1}$
yields the equation:
$$0 = y_0(1-p) + (n-y_0)p$$
Solving for $p$ we find $p = \frac{y_0}{n}$.  This essentially has to be 
a local (and global) maximum since it is the only critical point.  Note
that the probability is always non-negative, but it equals zero when $p = 0, 1$.
Hence on the interval $0 \le p \le 1$ the critical point has to be a local maximum.   
\vskip 10pt
\noindent
3.65.  (a)  We want to think of $Y$ as binomial random
variable, so using 3.33 (a),
$$E\left(\frac{Y}{n}\right) = \frac{1}{n} E(Y) = \frac{1}{n}\cdot np = p.$$
(b) Similarly, using the result of 3.33 (b) 
$$V\left(\frac{Y}{n}\right) = \frac{1}{n^2} V(Y) = \frac{npq}{n^2} = \frac{pq}{n}. 
$$
From this we see that as $n \to \infty$, the variance of $\frac{Y}{n}$  \emph{goes to zero}.  

\end{document}
