\magnification=\magstep1
\centerline{Mathematics 375--Probability and Statistics I}
\centerline{Solutions for Problem Set 3}
\centerline{September 23, 2005}
\bigskip
\noindent
A) 2)  For large numbers of runs (say $\ge 10000$), 
the fraction ${wins\over total games}$ should be roughly $.493$.
3)  a) The probability of winning immediately is the probability
of rolling a total of 7 or 11, which is ${6\over 36} + {2\over 36} = {2\over 9}$.
The probability of losing immediately is the probability of 
rolling a 2, 3, or 12, which is ${1\over 36} + {2\over 36} + {1\over 36} = 
{1\over 9}$.  b) Let $A$ be the event that you win, $F4$ be the
event that you roll a $4$ on the first roll.  The conditional
probability $P(A|F4)$ can be analyzed like this:  $A \cap F4$ is the 
union of ``subevents'' $A_k$:
$$\cases{A_2 = 4 & is rolled on 2nd roll,\cr
         A_3 = 4 & is rolled on 3rd roll and second roll was not a 7,\cr
         A_4 = 4 & is rolled on 4th roll and second and third rolls are not 7's,\cr
          \vdots & \cr}$$
These are mutually exclusive (no overlaps), but any number
of rolls can happen before the next 4 comes up.  Hence
$$P(A\cap F4) = \sum_{k=2}^\infty P(A_k) = \sum_{k=2}^\infty \left({3\over 36}\right)
\left({27\over 36}\right)^{k-2}$$
(the ${3\over 36}$ is the probability of rolling a 4; the 
${27\over 36}$ is the probability of rolling anything other than 
a 4 or a 7).  

Then reasoning similarly, let $F5,F6,F8,F9,F10$ be the 
events corresponding to the different first rolls in this case.
$$\eqalign{P(A\cap F10) &= P(A\cap F4)\cr
           P(A\cap F5) &= P(A\cap F9) = \sum_{k=2}^\infty P(A_k) = \sum_{k=2}^\infty \left({4\over 36}\right)
\left({26\over 36}\right)^{k-2}\cr
P(A\cap F6) &= P(A\cap F8) = \sum_{k=2}^\infty P(A_k) = \sum_{k=2}^\infty \left({5\over 36}\right)
\left({25\over 36}\right)^{k-2}\cr}$$
Then by the Law of Total Probability
$$\eqalign{
P(A) &= P({\rm win\ immediately}) + P(A|F4)P(F4) + P(A|F5)P(F5) + P(A|F6)P(F6)\cr
&\quad  + P(A|F8)P(F8) + 
P(A|F9)P(F9) + P(A|F10)P(F10)\cr
&= {2\over 9} + 2\sum_{k=2}^\infty \left({3\over 36}\right)^2
\left({27\over 36}\right)^{k-2} + 
2\sum_{k=2}^\infty \left({4\over 36}\right)^2
\left({26\over 36}\right)^{k-2}\cr
&\quad + 2 \sum_{k=2}^\infty \left({5\over 36}\right)^2
\left({25\over 36}\right)^{k-2}\cr
&={244\over 495} \doteq .49283\cr}$$
(The sums are {\it geometric series} and can be summed by the usual
formula:
$$\sum_{\ell=0}^\infty cr^\ell = {c\over 1-r}.)$$

\bigskip
2.65/  $P(A) < P(A|B) = {P(A\cap B)\over P(B)}$ implies that
$P(A)P(B) < P(A\cap B)$ (note we are given that $P(B) > 0$). 
Hence $P(B) < {P(A\cap B)\over P(A)} = P(B|A)$.  

\bigskip
\noindent
2.67/  We use the characterization $A,B$ independent
$\Leftrightarrow P(A\cap B) = P(A)P(B)$.
a)  Since $A = (A\cap B) \cup (A\cap \overline{B})$
and $(A \cap B) \cap (A\cap \overline{B}) = \emptyset$,
we have
$P(A) = P(A\cap B) + P(A\cap \overline{B})$.  Hence
$$\eqalign{P(A\cap\overline{B}) 
&=P(A) - P(A\cap B)\cr
&=P(A) - P(A)P(B)\quad (A,B\ {\rm independent})\cr
&=P(A)(1 - P(B))\cr
&=P(A)P(\overline{B})\cr}$$
This shows $A$ and $\overline{B}$ are independent.

b)  The answer is yes:  Similarly, from 
$P(\overline{A} \cap \overline{B}) + P(A\cap\overline{B}) = P(\overline{B})$,
$$\eqalign{P(\overline{A}\cap\overline{B}) 
&=P(\overline{B}) - P(A\cap \overline{B})\cr
&=P(\overline{B}) - P(A)P(\overline{B})\quad {\rm\ by\ part\ a}\cr
&=P(\overline{B})(1 - P(A))\cr
&=P(\overline{B})P(\overline{A})\cr}$$
This shows $\overline{A}$ and $\overline{B}$ are independent.

\bigskip
\noindent
2.70/  a)  $P(A\cup B) = P(A) + P(B) - P(A\cap B) = .95 + .90 - .88 = .97$.
b)  $P(\overline{A\cup B}) = 1 - .97 = .03$.


\bigskip
\noindent
2.73/  Let $C_i$ be the event: relay $i$ activates correctly.
We have $P(C_i) = .9$, and the $C_i$ are independent.  For a),
the probability that at least one open path from $A$ to $B$ exists
is $P(C_1\cup C_2 \cup C_3) = 3\times .9 - 3\times .81 + .729 = .999$
using the formula from 2.66 above.
For b) we want $P(C_1| C_1\cup C_2 \cup C_3) = P(C_1)/P(C_1\cup C_2 \cup C_3)$,
using the definition of the conditional probability.  This 
gives $.9/.999 \doteq .9009$.

\bigskip
\noindent
2.79/ a)  From problem 2.40, there were 8 likely winning numbers
out of the 1000 possible combinations of 3 digits $0,1,\ldots, 9$.
a) Assuming that the drawings in CT and PA were independent
(note the ``all indications CT was due to pure chance'' in problem!),
$P($666 in CT $|$ 666 in PA$) = P($666 in CT$) = 1/1000$.
b)  Again by independence, 
$P($666 in CT and 666 in PA$)(1/8)(1/1000) = 1/8000$. 

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\noindent
2.88/ Let $D_i$ be the event that the $i$th radar set
detects the plane.  $P(D_i) = .98$ and $P(\overline{D_i}) = .02$.
a)  Since $D_i$ are independent $P(\overline{D_i})$ are
also independent (see 2.67b above).  Hence
$P(\overline{D_1}\cap \overline{D_2}\cap \overline{D_3}) = (.02)^3 = .000008$.
b) The probability that all 3 radar sets do 
detect the plane is $P(D_1\cap D_2\cap D_3) = (.98)^3 = .941192$
\bigskip
\noindent
2.91/  Because of independence $(.75)^4$.

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\noindent
2.101/ By Bayes' Rule,
$$P(I_1 | H) = {(.9)(.01)\over (.9)(.01) + (.95)(.005) + (.75)(.02)}
\doteq .313$$

\bigskip
\noindent
2.102/a)  Since $P(A|B) = P(A|\overline{B})$, by the Law of Total Probability,
$$\eqalign{
P(A) &= P(A|B)P(B) + P(A|\overline{B})P(\overline{B}) \cr
     &= P(A|B)(P(B) + P(\overline{B}))\cr
     &=P(A|B)\cr}$$
Then $A$ and $B$ are independent according to Definition 2.10.
For b), given $P(A|C) > P(B|C)$ and $P(A|\overline{C}) > P(B|\overline{C})$,
by the Law of Total Probability,
$$\eqalign{
P(A) &= P(A|C)P(C) + P(A|\overline{C})P(\overline{C}) \cr
     &> P(B|C)P(C) + P(B|\overline{C})P(\overline{C}) \cr
     &= P(B).\cr}$$
     
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\noindent
2.103/ Let $N$ denote a negative reaction, $\overline{N}$ a positive one, 
$M$ denote a male, $\overline{M}$ a female (that means
no mixed reactions, no hermaphrodites or neuters :) ).  The given information
in the problem is: $P(N|\overline{M}) = .3$, $P(N|M) = .6$, which
implies $P(\overline{N}|\overline{M}) = .7$, $P(\overline{N}|M) = .4$.
Finally $P(M) = .25$ and $P(\overline{M}) = .75$.  Hence by Bayes' Rule,
$P(M|N) = {(.6)(.25) \over (.6)(.25) + (.3)(.75)} = .4$.  
 \bye

  