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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 2}
\centerline{September 17, 2009}
\vskip 10pt
\noindent
2.10.  (a)  The sample space is the set $\{A,B,AB,O\}$.  

(b)  The probabilities of the simple events are 
$P(A) = .41$, $P(B) = .1$, $P(AB) = .04$, $P(O) = .45$.  

(c)  $P(\{A,AB\}) = P(A) + P(AB) = .41 + .04 = .45$.
\vskip 10pt
\noindent
2.11  (a) Since $E_i$, $i = 1,2,3,4,5$ make up the whole sample
space, $1 = P(E_1) + P(E_2) + P(E_3) + P(E_4) + P(E_5) = .15 + .15 + .4 + 3P(E_5)$.
Then $P(E_5) = .1$ and $P(E_4) = .2$.  

(b)  Similarly, $P(E_1) = .3$ and $P(E_2) = .1$, so
$1 = .3 + .1 + 3P(E_3)$.  Hence $P(E_3) = P(E_4) = P(E_5) = .2$.
\vskip 10pt
\noindent
2.13.  (a) The simple events are the four possible responses, say
$S = \{VL,SL,U,O\}$.  

(b)  Not equally likely, $P(VL) = .24$, $P(SL) = .24$, $P(U) = .4$, 
$P(O) = .12$.  

(c)  $P({SL,VL}) = P(SL) + P(VL) = .24 + .24 = .48$.  
\vskip 10pt
\noindent
2.14.  (a)  The probability that a random adult needs glasses is
$.58$ (add the numbers in the first row).

(b)  The probability that a random adult needs glasses but 
does not use them is $.14$.

(c)  The probability that a random adult uses glasses for
reading (whether or not they are needed) is $.46$ (add numbers
in first column).
\vskip 10pt
\noindent
2.19. (a) There are $9$ points in the sample space:
$$
S = \{(V_1,V_1),(V_1,V_2),(V_1,V_3),(V_2,V_1),(V_2,V_2),(V_2,V_3),(V_3,V_1),
(V_3,V_2),(V_3,V_3)\}.
$$

(b)  Assuming the selections are all random, it makes sense to 
assign the probability $1/9$ to each simple event.

(c)  There are three simple events in $A$, so 
$P(A) = 3/9 = 1/3$.  There are $5$ simple events in $B$
($B= \{(V_2,x),(x,V_2) \mid x \in \{V_1,V_2,V_3\}$).
So $P(B) = 5/9$.  Then $P(A \cap B)  = 1/9$ since
$A \cap B$ includes only $(V_2,V_2)$.  Finally 
$A\cup B$ includes all the pairs except $(V_1,V_3)$
and $(V_3,V_1)$, so $P(A\cup B) = 7/9$. 
\vskip 10pt
\noindent
2.20.  (a)  With no information, and with the choice
of curtain made randomly, $P(G) = P(D_1) = P(D_2) = 1/3$.

(b)  (i)  The probability is $P(G) = 1/3$.
(ii)  She will get the dud prize in this case, since the first
choice was the good prize.  (iii)  If she initially chooses
one of the duds, the other dud is shown, and she switches, then
she wins the good prize.  (iv)  This probability is $2/3$.  To 
see this, think
of the three possible ways this can go -- if she chooses the good
prize (probability $1/3$), the host shows one dud, and she chooses
the other, which is also a dud.  If she chooses one dud (probability
$2/3$), the host shows the other dud, and she switches to the good
prize! 
 (v)  The strategy that maximizes
the contestant's chance of winning the good prize is to
\emph{always switch to the other curtain}.   
\vskip 10pt
\noindent
2.28.  Let the majority candidates be $M1, M2, M3$, and the minority 
candidate $m$.  (a)  $S = \{M1M2, M1M3, M2M3, M1m, M2m, M3m\}$.

(b)  If the selection is really random each of the six simple
events in $S$ should be equally probable so each has probability $1/6$.

(c)  The event $A$ where $m$ is included in the two candidates
chosen is $A = \{M1m, M2m, M3m\}$, so $P(A) = 3/6 = 1/2$. 
\vskip 10pt
\noindent
2.29.  Let the men be $m1,m2,m3,m4$ and the women $w1,w2$.
(a)  The experiment consists of selecting two individuals 
independently and at random out of the 6.  A sample point
will consist of a set (unordered pair) like $m3w1$.

(b)  The sample space consists of 
\begin{align*} 
S &= \{m1m2,m1m3,m1m4,m1w1,m1w2,m2m3,m2m4,m2w1,m2mw2,m3m4,\\
   & \quad m3w1,m3w2,m4w1,m4w2,w1w2\}.
\end{align*}

(c)  The probability is $1/15$.  
\vskip 10pt
\noindent
2.30.  (a)  A sample point is an ordered listing of the 3 wines,
like $312$ (i.e. wine 3 is best, followed by wine 1 next, and wine 2 worst).

(b)  The sample space is $S = \{123,132,213,231,312,321\}$.

(c)  Say the superior wine is $1$ and the second best is $2$.  The
cases where the ``expert'' ranks $1$ no worse than $2$ are
$123,132,312$.  So the probability is $3/6 = 1/2$.  
\vskip 10pt
\noindent 
2.35.  Since the flights are on successive days, there are no time conflicts
and all pairs are possible.  By the $mn$ rule there are
$6\times 7 = 42$ pairs.
\vskip 10pt
\noindent
2.37. (a) There are $6! = 720$ different orderings of the six cities.

(b)  Intuitively, the answer should be $1/2$ because there should
be an equal chance of visiting Denver first or of visiting
San Francisco first if an itinerary is chosen randomly.  To show
this rigorously, note that the orderings in which Denver (D) is visited 
before San Francisco (S) are of $\binom{6}{2}$ types:  
$DSxyzw, DxSyzw, \cdots, xyzwDS$ (corresponding to the different
possible placements of $D,S$ among the 6 cities.  
For each of these types, the other four cities can be visited
in any of the $4!$ possible orders.  So there are 
$\binom{6}{2}\cdot 4!$ such itineraries.  Then 
$P($D before S$) = \frac{\binom{6}{2}\cdot 4!}{6!} = \frac{1}{2}.$
\vskip 10pt
\noindent
2.43.  The set of 9 taxis is partitioned into three disjoint groups
of sizes $3,5,1$ depending on which airport they go to.  
This is exactly the situation of Theorem 2.3, and 
the number of ways the groups can be selected is 
$\binom{9}{3}\cdot \binom{6}{5}$, which equals the multinomial coefficient
$$\begin{pmatrix} &9& \\
                   3&5&1
\end{pmatrix} = \frac{9!}{3! 5! 1!} = 504.$$
\noindent
2.44.  (a)  If the one taxi that needs repairs is dispatched
to airport C, there are still $\binom{8}{3} = \frac{8!}{3!5!} = 56$ ways the other
taxis can be dispatched to the other two airports.  So the probability
of this case is
$$\frac{56}{504} = \frac{1}{9}.$$
(This can also be seen intuitively by thinking of picking the one
taxi to go to airport C and asking what is the probability that
that one is the bad one.)

(b)  Similar to part (a).  Take out the three bad ones, 
so there are 6 taxis left; 2 going to airport A and 4 to 
airport B.  The probability is 
$$
\frac{\binom{6}{2}}{504} = \frac{5}{168}.
$$
\noindent
2.54.  There are $\binom{8}{4}$ ways to pick four people out of the
whole group.  There are $\binom{3}{2}\cdot \binom{5}{2}$
ways to pick groups that have exactly 2 undergraduates.
So the probability of this case is 
$$
\frac{\binom{3}{2}\cdot\binom{5}{2}}{\binom{8}{4}} = \frac{3}{7}.
$$
\noindent
2.64.  There are $6! = 720$ possible orders of the $6$ distinct
numbers $1,2,3,4,5,6$, and $6^6 = 46656$  possible 
outcomes for the six tosses of the dice, all equally likely.  
Hence the probability
is 
$$
6!\cdot \left(\frac{1}{6}\right)^6 = \frac{720}{46656} = 5/324.
$$
\noindent
2.65.  Now there are $5! = 120$ different orders of the $5$
distinct numbers $1,2,3,4,5$.  The difference between this
situation and the one in 2.64 is that it is now twice as 
likely that a $5$ will be rolled.  So the probability is
$$
5!\cdot \left(\frac{1}{6}\right)^4 \frac{2}{6} = \frac{5}{162}.
$$
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