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\centerline{Mathematics 375 -- Probability and Statistics I}
\centerline{Problem Set 2 -- Solutions}
\centerline{September 16, 2005}
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\noindent
2.20/  Let the majority candidates be $M1, M2, M3$, and the minority 
candidate $m$.  a)  $S = \{M1M2, M1M3,M1m, M2M3, M2m, M3m\}$.
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 $P(A) = 3/6 = 1/2$. 

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\noindent
2.23/ a)  There are ${6\choose 2} = 15$ pairs of systems
so the sample space contains 15 simple events.  There are
${4\choose 2} = 6$ pairs of non-defective systems.  So
$15 - 6 = 9$ pairs contain at least one defective system.  
$A = $ at least one defective $\Rightarrow P(A) = 9/15 = 3/5$.
There is just one pair where both systems are defective, so
$B = $ both defective $\Rightarrow P(B) = 1/15$.

b)  With 4 defective systems out of the 6, 
$A = $ at least one defective $\Rightarrow P(A) = 14/15$, 
$B = $ both defective $\Rightarrow P(B) = 6/15 = 2/5$.

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\noindent
2.26/ Call the stations 1,2,3.  Each of the three patients
chooses one station.  So we can represent the choices as follows.
For instance, $(3,1,1)$ might represent the situation where
the first patient chooses station 3, and the second and third
both choose station 1.
The sample space $S$ is in 1-1 correspondence 
with $\{1,2,3\}^3$ = all ordered triples with entries 1,2,3, 
so $|S| = 3^3 = 27$, and each is equally probable.
(Note there is no requirement that the
patients must choose 3 different stations.  In fact, that is
the case studied next in part a.)
a)  $A = \{123,132,213,321,231,321\}$, so b) $P(A) = 6/27 = 2/9$.

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\noindent
2.27/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.

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\noindent
2.29/ 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 ${6\choose 2}$ 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 
${6\choose 2}\cdot 4!$ such itineraries.  Then 
$P($D before S$) = {{6\choose 2}\cdot 4!\over 6!} = {1\over 2}.$

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\noindent
2.31/ a) $6\cdot 6 = 36$.  b) There are $6$ cases where the 
sum is 7: $1+6, 2+5, 3+4, 4+3, 5+2, 6+1$, so probability of a 7 is
$6/36 = 1/6$.

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\noindent
2.34/ Since the jobs are different, the number is
the number of {\it permutations} of 10 things 3 at a time:
$10\cdot 9 \cdot 8 = 720$.  


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\noindent
2.35/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 
${9\choose 3}\cdot {6\choose 5}$, which equals the multinomial coefficient
$$\pmatrix{ &9& \cr
          3&5&1\cr} = {9!\over 3! 5! 1!} = 504$$

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\noindent
2.37/ The number is the multinomial coefficient:
$${17\choose 2\ 5\ 10} = {10!\over 2!5!10!} = 408408.$$
(This can also be done by expanding using the binomial theorem
in two stages: 
$$(x + (y + z))^{17} = \cdots + {17\choose 2} x^2(y+z)^{15} + \cdots
= \cdots + {17\choose 2} {15\choose 5} x^2y^5z^{10} + \cdots$$
so the coefficient is ${17\choose 2} {15\choose 5} = {17\choose 2\ 5\ 10}$.)

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\noindent
2.38/ The coefficients are ${8\choose 5} = {8\choose 3} = 56$.


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\noindent
2.40/ The likely winning numbers were $444,446,464,644,466,646,664,666$, 
so $8$ of them.

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\noindent
2.49/ If all sets of 5 of the 8 people on the board were
equally likely, then the chance of having the vote break down
strictly along gender lines is $1/{8\choose 5} = 1/56$.  

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\noindent
2.57/ For all parts we use the formula $P(X|Y) = {P(X\cap Y)\over P(Y)}$
from the definition of the conditional probability.

\noindent
a)  $P(A|B) = {.1\over .3} = {1\over 3}$.

\noindent
b)  $P(B|A) = {.1\over .5} = {1\over 5}$.

\noindent
c)  For $P(A|A\cup B)$, we need to know $P(A \cup B)$.  By the 
sum rule, since we know $P(A) = .5$, $P(B) = .3$, and $P(A\cap B) = .1$,
we have $P(A\cup B) = .5 + .3 - .1 = .7$.  Moreover, by usual
``set algebra'', $A \cap (A \cup B) = A$.  Hence
$P(A| A\cup B) = {.5\over .7} = {5\over 7}$.

\noindent
d)  $P(A|A\cap B) = 
{P(A \cap (A\cap B))\over P(A\cap B)} = {P(A\cap B)\over A\cap B} = 1$.
{\it Comment:}  The same reasoning shows $P(X|Y) = 1$ whenever
$Y \subset X$, which is certainly what we would hope!

\noindent
e) $P(A\cap B| A\cup B) = {P(A\cap B)\over P(A\cup B)} = {1\over 7}$.
(Note: Here we used the set algebra fact that 
$(A\cap B) \cap (A \cup B) = A\cap B$.

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\noindent
2.60/ Let $S$ denote the event customer is satisfied and $A$
denote the event that plumber $A$ was the one that did the work.
We have $P(S) = .9$, so $P(\overline{S}) = .1$, and $P(A) = .4$.
We also know that half of the complaints dealt with $A$, which 
says $P(A|\overline{S}) = .5 \Rightarrow P(A\cap \overline{S}) = .5 \cdot .1 = .05$.   a)  By the definition
of conditional probabilities, 
$P(\overline{S}| A) = {P(\overline{S}\cap A)\over P(A)} = .05/.4 = .125$.
b)  $P(S|A) = {P(S\cap A)\over P(A)} = .35/.4 = .875$.
(Note that $P(S \cap A) = .35$ because $P(S \cap A) + P(\overline{S} \cap A) = 
P(A) = .4$.) 
\bye
