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\centerline{Mathematics 375 -- Probability Theory}
\centerline{Solutions for Problem Set 3}
\centerline{September 23, 2011}
\vskip 10pt
\noindent
2.85.  Since $S = B \cup \overline{B}$ and $B \cap \overline{B} = \emptyset$, 
we have $A = (A \cap B) \cup (A\cap \overline{B})$ and 
$P(A) = P(A\cap B) + P(A\cap\overline{B})$.  
Since $A$ and $B$ are independent, $P(A\cap B) = P(A)P(B)$.  Substituting
for $P(A)$, we obtain:
$$
P(A\cap B) = (P(A\cap B) + P(A\cap\overline{B}))P(B),
$$
so after algebraic rearrangement
$$
P(A\cap B)(1 - P(B)) = P(A\cap\overline{B})P(B).
$$
Since $P(A\cap B)= P(A)P(B)$ again, this simplifies to 
$$
P(A)P(\overline{B}) = P(A\cap\overline{B}),
$$
which shows that $A$ and $\overline{B}$ are independent.  
The answer is yes for the second question.  From 
$P(\overline{A} \cap \overline{B}) + P(A\cap\overline{B}) = P(\overline{B})$,
\begin{align*}
P(\overline{A}\cap\overline{B}) 
&=P(\overline{B}) - P(A\cap \overline{B})\\
&=P(\overline{B}) - P(A)P(\overline{B})\quad \text{ by the first part }\\
&=P(\overline{B})(1 - P(A))\\
&=P(\overline{B})P(\overline{A})
\end{align*}
This shows $\overline{A}$ and $\overline{B}$ are independent.
\vskip 10pt
\noindent
2.86.  (a) No it is not possible.  That would say
$P(A \cup B) = .8 + .7 - .1 = 1.4$.  But $P(A\cup B) \le 1$.  (b)  The
smallest value is $P(A\cap B) = .5$.  (c) No it is not possible,
$P(B) = .7$ and $A\cap B \subseteq B$, so $P(A\cap B) \le P(B)$.  
(d)  The largest value is $P(A\cap B) = .7$.   
\vskip 10pt
\noindent
2.90. (a) Let $A$ be the event that the sky diver is injured
 on the first jump, and $B$ the event that she is injured on the second
jump.  Under the independence assumption 
$$P(A\cap B) = P(A)P(B) = \frac{1}{50} \cdot \frac{1}{50} = \frac{1}{2500}.$$
Hence the probability that she is injured (on at least one jump) is:
$$P(A\cup B) = \frac{1}{50} + \frac{1}{50} - \frac{1}{2500} = \frac{99}{2500}.$$
(b) No, your friend has it wrong.  Note that the probability of injury after
two jumps is already slightly less than $\frac{2}{50}$. After 50 jumps, the probability
that you have been injured in at least one jump can be computed like this.  The event
we want is the complement of the event indicating that all jumps occurred without
injury.  So the probability is
$$1 - \left(\frac{49}{50}\right)^{50} \doteq .636$$
(that is you have about a $64\%$ chance of being injured at least once).  
\vskip 10pt
\noindent
2.91.  If events $A$ and $B$ are mutually exclusive, then $P(A\cap B) = 0$.
If $P(A) = .4$ and $P(B) = .7$, then $P(A \cap B) \ge .1$ (as in 2.86 above).
So $A$ and $B$ are not mutually exclusive in this case.  If $P(A) = .4$ and
$P(B) = .3$, then $P(A \cap B) = 0$ is possible (but not always the case, of course!).
\vskip 10pt
\noindent
2.92.  (a)  Let $A_1,A_2,A_3$ be the events that the three
employees are telling the truth but the lie detector 
test indicates that each is lying.  Because of independence,
the probability that all three events occur is 
$$P(A_1 \cap A_2 \cap A_3) =  (.05)^3 = .000125$$
(b)  The probability that at least one occurs is the complementary probability
for the event that the lie detector never gives an incorrect result.  This is 
$1 - (.95)^3 \doteq .143$.  This can also be computed by the additive rule
for unions of more than two events:  By the formula from page 58 in the text,
$$P(A_1 \cup A_2 \cup A_3) = 3(.05) - 3(.05)^2 + (.05)^3 \doteq .143.$$
Either way, there is about a $14\%$ chance that the lie detector test 
is wrong at least once.  
\vskip 10pt
\noindent
2.94.  (a)  This is $P(A \cup B) = .95 + .90 - .88 = .97$.  (The combination
is more reliable than either separately.)  (b)  The probability 
that the smoke is undetected is $P(\overline{A\cup B}) = 1 - .97 = .03$.   
\vskip 10pt
\noindent
2.95.  (a)  From the additive law (Theorem 2.6), $.4 = .2 + .3 - P(A\cap B)$,
so $P(A\cap B) = .1$.  

(b)  By DeMorgan's Laws, $P(\overline{A} \cup \overline{B}) = P(\overline{A \cap B})
 = 1 - .1 = .9$.

(c)  Similarly,  $P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B})
 = 1 - .4 = .6$.

(d)  $P(\overline{A}|B) = \frac{P(\overline{A} \cap B)}{P(B)}$.
The numerator can be computed since we know $P(A\cap B) = .1$
and $P(A \cap B) + P(\overline{A}\cap B) = P(B) = .3$.  So the 
probability we want is $\frac{.2}{.3} = \frac{2}{3}$.  
\vskip 10pt
\noindent
2.97.  (a)  The probability of current flowing is 
\begin{align*}
P(1\cup 2\cup 3) &= P(1) + P(2) + P(3) - P(1 \cap 2) - P(1\cap 3) - P(2\cap 3) + 
P(1\cap 2\cap 3)\\
&= .9 + .9 + .9 - .81 - .81 - .81 + .729\\
&= .999
\end{align*}
(Alternate method:  The only way current can fail to flow is 
if all three relays fail, which happens with probability 
$(.1)^3 = .001$.  So the complementary probability is $.999$.)

(b)  If relay 1 functioned properly, then current did flow,
so $P(\text{flow} \cap 1) = P(1) = .9$.  Hence $P(1|\text{flow}) = \frac{.9}{.999} = 
.\overline{900}$.
\vskip 10pt 
\noindent
2.99.  Since $A$ and $B$ are independent, $P(A\cap B) = P(A)P(B)$.  So
\begin{align*}
a &= P(\overline{A} \cap \overline{B})\\
&= P(\overline{A \cup B})\\ 
&= 1 - P(A\cup B)\\
&= 1 - (P(A) + P(B) - P(A)P(B))\\
&= 1 - P(A)(1 - b) - b.
\end{align*}
Solving for $P(A)$, we get 
$$P(A) = \frac{1 - a - b}{1 - b},$$
as desired.
\vskip 10pt
\noindent
2.114.  The given information says that one person is guilty and will 
lie when asked ``did you do it?'' and the other is innocent and tells the truth when 
asked that same question.   In the products in the following, in each case
the first factor corresponds to the innocent person, and the second one corresponds
to the guilty person.   (a) The probability that the lie detector gives
a positive reading for both (i.e.
indicates both suspects are lying) is (because of independence)
$$(.10)(.95) = .095.$$
(b)  Similarly, $(.90)(.95) = .855$.  (c)  $(.10)(.05) = .005$.   (d) Using the Additive Law, $.10 + .95 - (.10)(.95) = .955$.  
\vskip 10pt
\noindent
2.115.  Because of independence of the games, the probability is $(.75)^4 \doteq .32$.
\vskip 10pt
\noindent
2.120.  To me, the most straightforward way to attack these is to list out the 
different sequences in which the 2 defective fridges ($D$'s) and the 4 good ones ($G$'s)
could be found in the testing.  I'm assuming the fridges are not distinguishable here:
\begin{align*}
DDGGGG,&DGDGGG,DGGDGG,DGGGDG,DGGGGD,\\
GDDGGG,&GDGDGG,GDGGDG,GDGGGD,\\
GGDDGG,&GGDGDG,GGDGGD,\\
GGGDDG,&GGGDGD,\\
GGGGDD.&
\end{align*}
These $\binom{6}{2} = 15$ possible sequences would be equally likely,
given that the fridges are chosen at random for the testing.  Of the 15, exactly 3 have
the second $D$ in the 4th position -- 
$$\{DGGDGG,GDGDGG,GGDDGG\}.$$  
So this probability is $\frac{3}{15} = \frac{1}{5}$.  
(b)  The second $D$ is found no later than the 4th test in 6 of the cases --
the three from part (a) and 
$$\{DDGGGG,DGDGGG,GDDGGG\}$$   
So this probability is $\frac{6}{15} = \frac{2}{5}$.  (c)  The event
$A = $ exactly one defective found in first two tests consists of 
$$A = \{DGDGGG,DGGDGG,DGGGDG,DGGGGD,GDGDGG,GDDGGG,GDGGDG,GDGGGD\}$$
Then $B = $ remaining defective found in 3rd or 4th test gives
$$B \cap A = \{DGDGGG,DGGDGG,GDDGGG,GDGDGG\}.$$
So
$$P(B|A) = \frac{P(B\cap A)}{P(A)} = \frac{\frac{4}{15}}{\frac{8}{15}} = \frac{1}{2}.$$ 
\vskip 10pt
\noindent
2.125.  Let $A$ be the event that the test indicates presence of the disease, and
$B$ be the event that the disease is actually present.  With a randomly chosen
person, $P(B) = .01$ and $P(\overline{B}) = .99$.
The given information says $P(A\mid B) = .9$ and $P(\overline{A}\mid\overline{B}) = .9$,
so $P(A\mid\overline{B}) = .1$.  But
we are interested in the conditional probability $P(B\mid A)$ -- the chance
that the person has the disease, given that the test was positive.  Using
the partition $B \cup \overline{B} = U$,  Bayes' rule says
\begin{align*}
P(B\mid A) &= \frac{P(A\mid B)P(B)}{P(A\mid B)P(B) + P(A\mid\overline{B})P(\overline{B})}\\
&= \frac{(.9)(.01)}{(.9)(.01) + (.1)(.99)}\\
&\doteq .083.
\end{align*}
Your reaction should be: ``What??? That can't possibly be right.''  This should seem absurdly low, and 
completely contradictory to what the ``$90\%$ reliability'' given information 
would seem to be saying.  But it {\it is correct} and a standard example of how the results
of diagnostic testing can give misleading results.  The problem
is the large size of the second term on the bottom relative to the 
first term.  The term $P(A\mid\overline{B})P(\overline{B}) = P(A\cap\overline{B})$
represents the chance of a \emph{false positive} result -- a case where
the person does not have the disease, but the test comes back positive
anyway.   The large number of false positives swamps the true positive
results that the test is designed to produce.  {\bf Comment:} The pattern seen in this question has major implications concerning quality and affordability of health care!
\vskip 10pt 
\noindent
2.128.  (a)  By LTP, we always have
$$P(A) = P(A\mid B)P(B) + P(A\mid \overline{B})P(\overline{B}).$$
If $P(A\mid \overline{B}) = P(A \mid B)$, then we substitute into the 
second term in the right and collect terms like this:
$$P(A) = P(A\mid B)(P(B) + P(\overline{B})) = P(A\mid B).$$
This is one of the three equivalent forms of the statement ``$A$ and $B$
are independent.'' (See Definition 2.10.)  (b)  By the LTP, 
$$P(A) =   P(A\mid C)P(C) + P(A\mid \overline{C})P(\overline{C}).$$
If $P(A\mid C) > P(B \mid C)$ and $P(A \mid\overline{C}) > P(B \mid\overline{C})$,
then since $P(C), P(\overline{C}) \ge 0$, we have
$$P(A) > P(B\mid C)P(C) + P(B \mid\overline{C})P(\overline{C}) = P(B).$$
\vskip 10pt
\noindent
2.129.  Let $A$ be the event that the randomly chosen person is a woman,
and $B$ be the event that the person is a man.  Let $C$ be the event that
he or she reacts positively to the mysterious ``circumstances'' in the 
question.  We have $P(A) = \frac{15}{20} = .75$ and $P(B) = .25$.
Moreover, $P(C\mid A) = .7$ (so $P(\overline{C}\mid A) = .3$) 
and $P(C\mid B) = .4$, (so $P(\overline{C}\mid B) = .6$).  We want
$P(B\mid \overline{C})$.  By Bayes' rule:
\begin{align*}
P(B\mid \overline{C}) &= \frac{P(\overline{C}\mid B)P(B)}{P(\overline{C}\mid B)P(B) + P(\overline{C}\mid A)P(A)}\\
&= \frac{(.6)(.25)}{(.6)(.25) + (.3)(.75)}\\
&= .4.
\end{align*} 
\vskip 10pt
\noindent
2.134.  Again, we use Bayes' rule.  Let $A,B$ be the events that 
the corresponding teaching methods were used and $F$ be the event that
the person fails the test.  We have $P(F\mid A) = .2$ and $P(F\mid B) = .3$.
Moreover, $P(A) = .7$ and $P(B) = .3$, so $A$ and $B$ form the partition 
of the sample space in this case.
\begin{align*}
P(A\mid F) &= \frac{P(F\mid A)P(A)}{P(F\mid A)P(A) + P(F\mid B)P(B)}\\
&= \frac{(.2)(.7)}{(.2)(.7) + (.1)(.3)}\\
&\doteq .824.
\end{align*}  
\vskip 10pt
\noindent
3.1.  Let $Y = $ number of impurities found in one well.  Since
$20\%$ of the wells had neither, $80\%$ had at least one.  Thus,
by the additive law, the percent that had both was $(40 + 50 - 80)\% = 10\%$.  
Then $P(Y = 0) = .2$, $P(Y = 1) = .7$ and $P(Y = 2) = .1$.   
\vskip 10pt
\noindent
3.4.  There are two open paths only when all three valves open correctly.
This happens with probability $(.8)^3 = .512$.  There are no open
paths when valve 1 fails, and valve 2 or valve 3 or both fail.  This
happens with probability $(.2)(.2 + .2 - (.2)^2) = .072$.  Since only 
0,1, or 2 paths can occur, the probability of exactly one path 
open is then $1 - (.512 + .072) = .416$.  The probability distribution for
$Y = $ number of open paths is $P(Y = 0) = .072$, $P(Y = 1) = .416$, $P(Y = 2) = .512$.     
\vskip 10pt
\noindent
3.5.  Say the correct assignment is $ABC$ in that order.  The possible assignments
the child makes are then 
$ABC, ACB, BAC, BCA, CAB, CBA$.
The first gives all three correct, the next three give one correct, and the 
last two give no correct.  Hence if $Y = $ the exact number correct if one of the 6 
assignments is chosen at random, we have 
$P(Y = 0) = \frac{1}{3}$, $P(Y = 1) = \frac{1}{2}$, $P(Y = 2) = 0$, and
$P(Y = 3) = \frac{1}{6}$.  (Note that it is not possible to get exactly 
2 correct, since then the third would have to be correct too.)  
\vskip 10pt
\noindent
3.7.  The answer in the book corresponds to an interpretation where you 
think of the balls as being \emph{distinguishable} (perhaps differently colored, or
different sizes).  This means that \emph{it matters which balls go into which bowls}, not
just the total numbers that end up in each bowl.
One way to represent the sample space is by giving a collection of 
triples giving the bowl into which the corresponding ball is placed.  For instance
$(1,1,2)$ would indicate balls 1 and 2 go into bowl 1 and ball 3 goes
into bowl 2.   In this case, bowl 3 is empty.  There are $3^3 = 27$ sample 
points obtained this way.  There are 3 ways to have two empty bowls (sample points
$(1,1,1),(2,2,2),(3,3,3)$), 6 ways
to have no empty bowls (sample points $(1,2,3)$ and all permutations), 
and hence 18 ways to have 1 empty bowl.  If $Y = $ number
of empty bowls, this gives
$P(Y = 0) = \frac{6}{27} = \frac{2}{9}$, $P(Y = 1) = \frac{18}{27} = \frac{2}{3}$,
and $P(Y = 2) = \frac{3}{27} = \frac{1}{9}$.  
\vskip 10pt
\noindent
Note:  I also gave full credit for solutions that treated the balls as 
\emph{indistinguishable}.  If you thought about it that way, then
there are only 10 different outcomes, corresponding to the ways of
splitting the 3 balls over the 3 bowls.  Here you could represent
the sample space by ordered triples with one entry for each bowl,
giving the number of balls that end up in that bowl:
$$S = \{(3,0,0),(0,3,0),(0,0,3),(2,1,0),(2,0,1),(1,2,0),(0,2,1),(1,0,2),(0,1,2),(1,1,1)\}$$
Note that if $Y = $ number of empty bowls again, then the probabilities end up being
$P(Y = 2) = 3/10$, $P(Y = 1) = 6/10$, and $P(Y = 0) = 1/10$.  
\vskip 10pt
\noindent
Comment:  It is sometimes difficult to state problems in this subject
in ways that are totally unambiguous!
\vskip 10pt
\noindent
3.8.  The question asks you to compute the probabilities for the numbers
of cells in the second generation after the start.  We
start from a single cell (generation 0, so to speak).  
If that dies then there are no
cells in generations 1 or 2.  If the single cell splits, but both
cells in generation 1 die, then there is another way for generation
2 to have 0 cells:  $P(Y = 0) = .1 + (.9)(.1)^2 = .109$.  If both cells
in generation 1 split then we have 4 cells in generation 2:
$P(Y = 4) = (.9)^3 = .729$.  Finally, if the original cell split
to form 2 cells in generation 1, but only one of those split and
the other one died, there are two cells
in generation 2 and $P(Y = 2) = (.9)((.1)(.9) + (.9)(.1)) = .162$.    
\vskip 10pt
\noindent
3.9.  We are supposed to assume that the auditor finds all the 
incorrect entries that are present.  So this is really just
the same as computing the probabilities of the different 
numbers of errors that could be made in the three entries
that are tested.  If $Y = $ number of
incorrect entries out of the three, $P(Y = 0) = (.95)^3 = .857375$, 
$P(Y = 1) = 3(.05)(.95)^2 = .135375$,
$P(Y = 2) = 3(.05)^2(.95) = .007125$ and $P(Y = 3) = (.05)^3 = .000125$.
(Comment:  This is a {\it binomial} random variable.  We will study
these in detail later in Chapter 3.)
(b)  The probability histogram is similar to the relative frequency
histograms constructed in Chapter 1.
(c)  $P(Y > 1) = P(Y = 2) + P(Y = 3) = .0725$.  
\end{document}
