\magnification=\magstep1
\centerline{Mathematics 375--Probability and Statistics I}
\centerline{Solutions for Problem Set 3}
\centerline{September 26, 2003}
\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:
$$\sum_{\ell=0}^\infty cr^\ell = {c\over 1-r}.)$$

The extra credit part is entirely similar, 
but you must use the new probabilities for
each dice roll.  The formula 
from the Law of Total Probability is still valid, but now
$P(F7) = {4\over 64} + {2\over 16} = {3\over 16}$, $P(F11) = {1\over 32}$
$P(F4) = P(F10) = {2\over 32} + {1\over 64} = {5\over 64}$, and so on.
The result for the winning probability is ${7581\over 14960} \doteq .507$.
In other words, ``Buggsy is going to be very happy''(!)

\bigskip
2.61/  $P(A) = .4$ is the sum of the entries on the first row (but not the 
total, of course!),
$P(B) = .37$ is the sum of the entries in the first column,
$P(A\cap B) = .1$ is the entry in the upper left, 
$P(A\cup B) = .67$ is the sum of the three entries excluding 
lower right (and excluding totals), $P(\overline{A}) = 1 - P(A) = .6$, 
$P(\overline{A \cup B}) = 1 - P(A\cup B) = .67$, $P(\overline{A\cap B}) = 
1 - P(A\cap B) = .9$, $P(A|B) = P(A\cap B)/P(B) = .1/.37 \doteq .2703$,
and $P(B|A) = .1/.4 = .25$.

\bigskip
\noindent
2.63/  The events are mutally exclusive, so $P(A\cap B) = 0$.
Hence if $P(B) > 0$, then $0 = P(A\cap B) \ne P(A)P(B)$.  So 
the events are not necessarily independent.

\bigskip
\noindent
2.64/  Since $A \subset B$, we have $P(A\cap B) = P(A)$.
Hence if $P(B) < 1$, it cannot be true that $P(A\cap B) = P(A)P(B)$.
So the events are not necessarily independent.

\bigskip
\noindent
2.66/  The general rule for triple unions is derived from the sum
rule in the text:
$$\eqalign{
P(A_1\cup A_2\cup A_3) &= P(A_1) + P(A_2) + P(A_3) - 
P(A_1\cap A_2) - P(A_1\cap A_3) - P(A_2\cap A_3)\cr
&\qquad + P(A_1\cap A_2\cap A_3)}$$
We are given that $P(A_2\cap A_3) = 0$, and that implies 
$P(A_1\cap A_2\cap A_3) = 0$ since $A_1\cap A_2\cap A_3 \subset A_2 \cap A_3$.
So the last two terms above are both zero.  Furthermore, we are given
$P(A_1\cap A_3) = P(A_1\cap A_2)$, so 
$$P(A_1\cup A_2\cup A_3) = P(A_1) + P(A_2) + P(A_3) - 
2P(A_1\cap A_2),$$
which is what we had to show.

\bigskip
\noindent
2.68/  Let $F_i$ be the event that the $i$th person is 
telling the truth, but the lie detector indicates that he/she
is lying.  Part a) asks for $P(F_1 \cap F_2 \cap F_3) = (.05)^3 = .000125$, since
the $F_i$ are independent events.  Part b) asks for 
$P(F_1\cup F_2\cup F_3)$, which we can compute using the formula
from 2.66:
$$P(F_1\cup F_2\cup F_3) = 3(.05) - 3(.0025) + .000125 = .142625.$$
({\it Alternate Method:}  We could also derive the answer for b)
by using the complement rule: 
$$P(F_1\cup F_2\cup F_3) = 1 - P(\overline{F_1} \cap \overline{F_2} \cap 
\overline{F_3}) = 1 - (.95)^3 = .142625.$$
(Note: $\overline{F_i}$ is the event: ``The person is telling the truth
and the lie detector agrees.'')

\bigskip
\noindent
2.69/ Let $H$ represent a hit and $M$ a miss on one attempt, and 
let $L,R$ represent which hand is used.  The given information
says that $P(H|R) = .7$ (so $P(M|R) = 1 - .7 = .3$) and 
$P(H|L) = .4$ (so $P(M|L) = .6$).  The winning sequences of
outcomes in the three attempts are $HHM$, $MHH$, and $HHH$.
Starting on the right hand and using the independence,
$P(HHM) = P(H|R)P(H|L)P(M|R) = (.7)(.4)(.3)$, 
$P(MHH) = P(M|R)P(H|L)P(H|R) = (.3)(.4)(.7)$, and
$P(HHH) = P(H|R)P(H|L)P(H|R)$, which $= (.7)(.4)(.7)$.  The
sum gives the total probability of winning: $.364$.

\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
2.80/  Using the DeMorgan and sum laws we have
$$\eqalign{1 - P(A\cap B) &=P(\overline{A\cap B}) \cr
                      &= P(\overline{A} \cup \overline{B})\cr
                      &= P(\overline{A}) + P(\overline{B}) - P(\overline{A}\cap \overline{B})\cr
                      &= P(\overline{A}) + P(\overline{B}) - P(\overline{A \cup B}).\cr}$$
So rearranging,
$$P(A\cap B) = 1 - P(\overline{A}) - P(\overline{B}) + P(\overline{A\cup B})
\ge  1 - P(\overline{A}) - P(\overline{B}),$$
(since $P(\overline{A\cup B}) \ge 0$), which is what we had to show.

\bigskip
\noindent
2.84/ Using 2.80 twice:
$$\eqalign{
P(A\cap B \cap C) &= P((A\cap B) \cap C) \cr
                  &\ge 1 - P(\overline{A\cap B}) - P(\overline{C})\cr
                  &=P(A\cap B) - P(\overline{C})\cr
                  &\ge 1 - P(\overline{A}) - P(\overline{B}) - P(\overline{C})\cr}$$
which is what we had to show.
{\it Comment:}  The same argument extends to any finite number of sets:
$$P(\cap_{i=1}^n A_i) \ge 1 - \sum_{i=1}^n P(\overline{A_i}).$$

\bigskip
\noindent
2.87/  Consider following events: $M = $ person sees magazine ad,
$T =$ person sees TV ad, $S = M\cup T$ person sees ad in at least
one place, and $B = $ person buys.  The given information is first that
$P(M) = .02$, $P(T) = .2$, $P(M\cap T) = .01$, so by the sum 
rule we also know $P(S) = P(M \cup T) = .02 + .2 - .01 = .21$.
We are also given $P(B|S) = .\overline{3}$, and $P(B|\overline{S}) = .1$.
By the Law of Total Probability,
$$P(B) = P(B|S)P(S) + P(B|\overline{S})P(\overline{S}) = (.\overline{3})(.21) + 
(.1)(1 - .21) = .149$$

\bigskip
\noindent
2.93/  Let $P$ be the event that a car passes inspection, and 
let $T_i$ be the event that team $i$ is chosen to do the inspection.
The given information says $P(P|T_1) = 1$ and $P(P|T_2) = 0$.  
The sample space for the four cars is the set of ordered 4-tuples
in $\{T_1,T_2\}^4$ (indicating which team is selected for each car).
a)  The probability that three will be rejected is 
$P(E)$ where $E = \{(T_1,T_2,T_2,T_2),(T_2,T_1,T_2,T_2),(T_2,T_2,T_1,T_2),
(T_2,T_2,T_2,T_1)\}$.  So $P(E) = 4/16 = 1/4$.
For b), we have $P(\{T_1,T_1,T_1,T_1\}) = 1/16$.

\bigskip
\noindent
2.99/  Let $D$ be the event: person has the disease, and let
$T$ be the event: test is positive (indicates disease is present).
We are given $P(T|D) = .9$ and $P(\overline{T}|\overline{D}) = .9$
(so $P(T|\overline{D}) = .1$ -- the probability that the test will
be positive even though the patient does not have the disease -- or
``false positive'' test results).  We also know $P(D) = .01$.
For a random person, we can find $P(D|T)$ using Bayes' Rule:
$$P(D|T) = {P(T|D)P(D) \over P(T|D)P(D) + P(T|\overline{D})P(\overline{D}) }
= {(.9)(.01) \over (.9)(.01) + (.1)(.99)} = .083.$$
{\it Comment}:  This should be an {\bf absolutely shocking result}.
It says a ``90\% reliable'' test in this sense is virtually worthless.
The probability that the disease is present, given that the test
was positive is only $.083$(!)  The ``moral'' is that tests have 
to be much more reliable than this to give useful information(!)

\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,
$$\eqalign{
P(A) &= P(A|C)P(C) + P(A|\overline{C})P(\overline{C}) \cr
     &> P(B|C)P(C) + P(B|\overline{C})P(\overline{C}) \cr
     &= P(B).\cr}$$
     
\bigskip
\noindent
2.107/We have $A\cup B = (A\Delta B) \cup (A\cap B)$, and 
$(A\Delta B)\cap (A\cap B) = \emptyset$.  Hence by the Sum Rule
$$P(A\cup B) = P(A) + P(B) - P(A\cap B) = P(A \Delta B) + P(A\cap B)$$
Rearranging yields:
$$P(A \Delta B) = P(A) + P(B) - 2P(A\cap B).$$

\bigskip
\noindent
3.1/ Given $P(A) = .4$, $P(B) = .5$, and $P(\overline{A\cup B}) = .2$.
Then by the Sum Rule $P(A\cup B) = .5 + .4 - P(A\cap B) = 1 - .2 = .8$
Hence $P(A \cap B) = .1$.  Therefore the probability distribution
for the random variable $Y = $ number of impurities found is
$$\eqalign{p(0) &= P(Y = 0) = P(\overline{A\cup B}) = .2\cr
           p(1) &= P(Y = 1) = P(A \Delta B) = .7\cr
           p(2) &= P(Y = 2) = P(A\cap B) = .1\cr}$$
           (all other values 0).
The value for $p(1)$ is computed using the formula from 2.107, or
by taking $1 - (.2 + .1)$.

\bigskip
\noindent
3.3/ The random variable $Y$ is the number of the test on which the
second defective is found.  We can find the probability distribution
by considering the different possible sequences in which the 2 
OK and 2 defective components could be found:
$$\eqalign{
DDOO &\Rightarrow Y = 2\cr
DODO &\Rightarrow Y = 3\cr
DOOD &\Rightarrow Y = 4\cr
ODDO &\Rightarrow Y = 3\cr
ODOD &\Rightarrow Y = 4\cr
OODD &\Rightarrow Y = 4\cr}$$
Since the components are selected randomly any sequence has
probability $1/6$, and the distribution is:
$$\eqalign{p(2) &= P(Y = 2) = 1/6\cr
           p(3) &= P(Y = 3) = 2/6 = 1/3\cr
           p(4) &= P(Y = 4) = 3/6 = 1/2}$$
           (all other values 0).
 \bye

  