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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 3}
\centerline{September 24, 2009}
\vskip 10pt
\noindent
\vskip 10pt
2.71.  (a)  $P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{1}{3}$.

(b) $P(B|A) = \frac{P(A\cap B)}{P(A)} = \frac{1}{5}$

(c)  We have $P(A\cup B) = P(A) + P(B) - P(A\cap B) = .7$.
Therefore, $P(A|A\cup B) = \frac{P(A)}{P(A\cup B)} = \frac{5}{7}$.

(d) $P(A|A\cap B) = 1$, since $(A\cap B)\cap A = A\cap B$.
 
(e) $P(A\cap B|A\cup B) = \frac{1}{7}$.
\vskip 10pt
\noindent
2.72.  (a)  $P(A\cap M) = .24$ while $P(A) = .6$ and $P(M) = .4$.
Since $P(A\cap M) = P(A)P(M)$, the events $A, M$ \emph{are independent}
(see Definition 2.10).

(b)  $P(\overline{A} \cap F) = .24$ while $P(\overline{A}) = .4$
and $P(F) = .6$.  Since $P(\overline{A} \cap F) = P(\overline{A})P(F)$,
these events are also independent. 
\vskip 10pt
\noindent
2.74. (a)  $P(A\cap D) = .2$, while $P(A) = .61$ and $P(D) = .3$ 
since $.2 \ne (.61)(.3)$, $A$ and $D$ are \emph{not independent}.

(b)  $P(B \cap D) = .09 = P(B)P(D) = (.3)(.3)$.  So $B$ and $D$
\emph{are independent}.

(c)  $P(C \cap D) = .01$, but  $P(C)P(D) = (.09)(.3) = .027$.
So $C$ and $D$ are \emph{not independent}.
\vskip 10pt
\noindent
2.75. (a) Let $A$ be the event that the first two cards dealt are spades,
so $P(A) = \frac{13}{52}\cdot\frac{12}{51} = \frac{1}{17}$.  Next, let
$B$ be the event that cards $3,4,5$ are also spades.  We want
$$
P(B|A) = \frac{P(B\cap A)}{P(A)} = 
\frac{\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}\cdot\frac{9}{48}}{\frac{13}{52}\cdot\frac{12}{51}} = \frac{33}{3920} \doteq .008.
$$

(b) Similarly, let $C$ be the event that the first three cards are
spades, and $D$ be the event that cards $4,5$ are spades.
$$
P(D|C) = \frac{P(D\cap C)}{P(C)} = 
\frac{\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}\cdot\frac{9}{48}}{\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}} = \frac{15}{392} \doteq .04.
$$

(c)  Similarly, let $F$ be the event that the first four cards are spades,
and $F$ be the envent that card $5$ is a spade.
$$P(F|E) = \frac{P(F\cap E)}{P(E)} = 
\frac{\frac{13}{52}\cdot\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{10}{49}\cdot\frac{9}{48}}{\frac{13}{52}\cdot\frac{12}{51}\cdot \frac{11}{50}\cdot\frac{10}{49}} = \frac{9}{48} \doteq
.19.
$$

\noindent
2.81.  We are given that $P(A),P(B) > 0$ and $P(A) < P(A|B)$.
By the definition of conditional probability, this last statement says
$P(A) < \frac{P(A\cap B)}{P(B)}$, so since $P(B) > 0$, $P(A)P(B) < P(A\cap B)$.
Now divide this statement through by $P(A)$:  $P(B) < \frac{P(A\cap B)}{P(A)}$.
But this says $P(B) < P(B|A)$, which is what we wanted to show.
\vskip 10pt
\noindent
2.84.  ``At least one $A_i$'' means $A_1 \cup A_2 \cup A_3$.  The general 
formula for the probability of a 3-fold union is given in the text on 
page 58.  In the terms on the right, we know $P(A_2 \cap A_3) = 0$,
and it follows that $P(A_1 \cap A_2 \cap A_3) = 0$, since 
$A_1 \cap A_2 \cap A_3 \subset A_2 \cap A_3$.  Moreover, we are 
given that $P(A_1 \cap A_2) = P(A_1 \cap A_3)$.  The required formula
follows.
\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
2.88.  (a)  It is possible for $P(A\cap B) = .1$ if $P(A\cup B) = .8$. 

(b)  The smallest possible value for $P(A\cap B)$ is $0$ (i.e. $A$
and $B$ can be disjoint).

(c)  It is \emph{not possible} for $P(A \cap B) = .7$, since
$A \cap B \subset A$ and $P(A) = .6$.  

(d)  The largest possible value for $P(A\cap B)$ is $.3$ (i.e.
$B$ can be contained in $A$, so $A \cap B = B$ is possible.)
\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
{\it Additional Problem}
\vskip 10pt
(2)  For large numbers of runs (say $\ge 10000$), 
the fraction $\frac{\text{wins}}{\text{total games}}$ should be roughly $.493$.

(3)  a) The probability of winning immediately is the probability
of rolling a total of 7 or 11, which is $\frac{6}{36} + \frac{2}{36} = 
\frac{2}{9}$.
The probability of losing immediately is the probability of 
rolling a 2, 3, or 12, which is $\frac{1}{36} + \frac{2}{36} + \frac{1}{36} = 
\frac{1}{9}$.  

(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'' $A_k$:
$$\begin{cases} A_2 = 4 & \text{ is rolled on 2nd roll,}\\
         A_3 = 4 & \text{ is rolled on 3rd roll and second roll was not a 7,}\\
         A_4 = 4 & \text{ is rolled on 4th roll and second and third rolls are not 7's,}\\
          & \vdots  \end{cases} $$
These are mutually exclusive (no overlaps), but any number
of rolls can happen before the next 4 comes up.  Hence
$$P(A\cap F4) = \sum_{k=2}^\infty P(A_k) = \sum_{k=2}^\infty \left(\frac{3}{36}\right)
\left(\frac{27}{36}\right)^{k-2}$$
(the $\frac{3}{36}$ is the probability of rolling a 4; the 
$\frac{2}{36}$ 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.
\begin{align*}
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(\frac{4}{36}\right)
\left(\frac{26}{36}\right)^{k-2}\\
P(A\cap F6) &= P(A\cap F8) = \sum_{k=2}^\infty P(A_k) = \sum_{k=2}^\infty \left(\frac{5}{36}\right)
\left(\frac{25}{36}\right)^{k-2}
\end{align*}
Then by the Law of Total Probability
\begin{align*}
P(A) &= P({\rm win\ immediately}) + P(A|F4)P(F4) + P(A|F5)P(F5) + P(A|F6)P(F6)\\
&\quad  + P(A|F8)P(F8) + 
P(A|F9)P(F9) + P(A|F10)P(F10)\\
&= \frac{2}{9} + 2\sum_{k=2}^\infty \left(\frac{3}{36}\right)^2
\left(\frac{27}{36}\right)^{k-2} + 
2\sum_{k=2}^\infty \left(\frac{4}{36}\right)^2
\left(\frac{26}{36}\right)^{k-2} + 2 \sum_{k=2}^\infty \left(\frac{5}{36}\right)^2
\left(\frac{25}{36}\right)^{k-2}\\
&=\frac{244}{495} \doteq .49283
\end{align*}
(The sums are {\it geometric series} and can be summed by the usual
formula:
$$\sum_{\ell=0}^\infty cr^\ell = \frac{c}{1-r}.)$$
\end{document}
