\magnification=\magstep1
\centerline{Mathematics 375 -- Probability and Statistics I}
\centerline{Problem Set 2 -- Solutions}
\centerline{September 19, 2003}
\bigskip
\noindent
2.19/  a) Let $A = \{R,L,S\}$ be the set of possible behaviors of a 
single car.  The sample space for two cars is the Cartesian product
$$S = A \times A = \{(R,R),(R,L),(R,S),(L,R),(L,L),(L,S),(S,R),(S,L),(S,S)\}$$
By the $mn$ rule, $|S| = 3\cdot 3 = 9$.

\noindent
b)  Since the behaviors are equally likely for each car, each sample
point should be assigned $P(E_i) = {1\over 9}$, and hence
the probability that at least one car turns left is 
$$P(\{(R,L),(L,R),(S,L),(L,S),(L,L)\}) = {5\over 9}.$$

\noindent
c)  Similiarly the probability that at most one car turns is 
$$P(\{(S,L),(S,R),(R,S),(L,S),(S,S)\}) = {5\over 9}.$$

\bigskip
\noindent
2.24/  a)  Call the two types of consoles $1,2$.  The sample space
is $\{1,2\}^4$, so there are 16 points in the sample space:
$$\eqalign{
S = \{&(1,1,1,1),(1,1,1,2),(1,1,2,1),(1,2,1,1),(2,1,1,1),(1,1,2,2),\cr
      &(1,2,1,2),(1,2,2,1),(2,1,1,2),(2,1,2,1),(2,2,1,1),(1,2,2,2),\cr
      &(2,1,2,2),(2,2,1,2),(2,2,2,1),(2,2,2,2)\}\cr}$$

\noindent
b)  Since we are given that the two types of consoles are in equal demand,
each sample point should have same probability, $P(E_i) = {1\over 16}$.

\noindent
c)  $A = \{(1,1,1,1),(2,2,2,2)\}$ so $P(A) = {2\over 16} = {1\over 8}$.

\bigskip
\noindent
2.25/ a)  For each family, let outcome 1 represent income $\le \$35353$, and
let outcome 2 represent income $> \$35353$.  Then the sample
space is represented by the same set as given in 2.24 above.  

\noindent
b) 
$$
\eqalign{A = &\{(1,1,2,2),(1,2,1,2),(1,2,2,1),(2,1,1,2),(2,1,2,1),(2,2,1,1),\cr
             &(1,2,2,2),(2,1,2,2),(2,2,1,2),(2,2,2,1),(2,2,2,2)\}\cr}$$  
$$B = \{(1,1,2,2),(1,2,1,2),(1,2,2,1),(2,1,1,2),(2,1,2,1),(2,2,1,1)\}$$
$$C = \{(1,2,2,2),(2,1,2,2),(2,2,1,2),(2,2,2,1)\}$$

\noindent
c) By the definition of the median, half the families in the
population would have 1 and the other half 2.  Hence 
$P(E_i) = {1\over 16}$ for each of the 16 simple events.
Hence $P(A) = {11\over 16}$, $P(B) = {3\over 8}$, and $P(C) = {1\over 4}$.

\bigskip
\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.

\bigskip
\noindent
2.28/This is the number of permutations of three objects $P_3^3 = 3! = 6$.

\bigskip
\noindent
2.30/Using the $mn$ rule repeatedly, we have $4\cdot 3\cdot 4\cdot 5 = 240$
possible meals (each appetizer can be ordered with every combination
of salad, entree, and dessert). 

\bigskip
\noindent
2.33/Using the $mn$ rule, there are $9\times 10^6 = 9000000$ such numbers.

\bigskip
\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 the multinomial coefficient
$$\pmatrix{ &9& \cr
          3&5&1\cr} = {9!\over 3! 5! 1!} = 504$$

\bigskip          
\noindent
2.36/ a)  We want to count the sample points in the event 
``taxi needing repair goes to C''.
There is no choice for the taxi needing repair -- it goes to C.
But there are $\pmatrix{ 8&\cr
                         3&5\cr} = 56$ ways the other $8$ taxis can be
divided into two groups of sizes 3 and 5.  Hence the probability of this
event is ${56\over 504} = {1\over 9}$.

b)  The three taxis needing repair each go to a separate airport, so 
there are $6 = P_3^3$ ways in which those three taxis can be assigned
to the three airports $A,B,C$.  There are $\pmatrix{ 6&\cr
                                                    2&4\cr} = 15$
ways to assign the other taxis to airports (note the one taxi for 
airport $C$ is already accounted for).  The probability of this
event is ${6\cdot 15\over 504} = {5\over 28}$.

\bigskip
\noindent
2.41 For all four parts, the sample space is the set of all possible
ways to assign the three prizes among the 50 lottery tickets.
Hence the sample space has size ${50 \choose 3}$.  For the events given,
we want to think about choosing the three prizewinners
in the specified ways.

\noindent a)  There are ${4\choose 3}{46\choose 0}$ ways to select 
the prizewinners if the officers win them all.  The probability of 
this event is ${4\choose 3}{46\choose 0}/{50\choose 3}$.

\noindent b)  If exactly two officers win prizes there are 
${4 \choose 2}$ possibilities for the officers who win, and 
${46\choose 1}$ possibilities for the other prizewinner.
Hence the probability of this event is ${4\choose 2}{46\choose 1}/{50\choose 3}$.

\noindent c)  Similar to b):  ${4\choose 1}{46\choose 2}/{50 \choose 3}$

\noindent d)  Similar to b) and c):  ${4\choose 0}{46\choose 3}/{50 \choose 3}$.

\noindent
{\it Comment:}  It should be clear that the four events in the four
parts of the problem account for the {\it whole sample space}.  Do our 
formulas agree with that?  That is, do the answers
from parts a), b), c), d) sum to $1$?  Or, in other words, is 
$${4\choose 3}{46\choose 0} + {4\choose 2}{46\choose 1} + 
{4\choose 1}{46\choose 2} + {4\choose 0}{46\choose 3} = {50\choose 3}?\leqno(1)$$
The answer is {\it yes}(!), and we can see it like this.  By the polynomial
interpretation of the binomial coefficients, ${50\choose 3}$ is the 
coefficient of $x^3$ in the expansion of $(1+x)^{50}$.  But
$$\eqalign{
(1+x)^{50} &= (1+x)^4(1+x)^{46}\cr
&=({4\choose 0} + {4\choose 1}x + {4\choose 2}x^2 + {4\choose 3}x^3 + \cdots )\cdot\cr
&\quad ({46\choose 0} + {46\choose 1}x + {46\choose 2}x^2 + {46\choose 3}x^3 + \cdots )\cr}$$
Collecting all $x^3$ terms in the product gives exactly the left side 
of (1) above.  This is a special case of a general binomial coefficient
identity sometimes called the {\it Vandermonde convolution formula\/}.

\bigskip
\noindent
2.54/ a)  ${n\choose n} = {n!\over n! 0!} = 1$.  This says there is 
only one way to choose a subset of size $n$ from a set with $n$
elements -- you have to take all of the elements!

\noindent
b)  ${n\choose 0} = {n!\over 0! n!} = 1$.  This says there is 
only one way to choose a subset of size $0$ from a set with $n$
elements -- you have to take the empty subset!

\noindent
c)  ${n \choose r} = {n!\over r! (n-r)!} = {n \choose n-r}$, since
$n - (n - r) = r$.  This says that the number of ways to choose
a subset of size $r$ from a set with $n$ elements is the same
as the number of ways to choose a subset of size $n-r$.  This 
makes sense because for each subset $A$ of size $r$, we have 
the complement $\overline{A}$ which contains exactly $n - r$ elements.
There are as many ways to choose $A$ as there are to choose $\overline{A}$.

\noindent
d)  We have $(1 + x)^n = \sum_{k=0}^n {n \choose k} x^k$ by 
a result proved in class.  This is an equality of polynomials,
so we get another true equality by substituting any value for $x$.
With $x = 1$, we get $2^n = \sum_{k=0}^n {n \choose k}$ which 
is what we had to show.

\bigskip
\noindent
2.56/By Theorem 2.3, the multinomial expansion is:
$$(x_1 + \cdots + x_k)^n = \sum_{n_1+\cdots+n_k = n} 
\pmatrix{ &n& \cr
         n_1&\cdots&n_k\cr} x_1^{n_1} \cdots x_k^{n_k}.$$
Similarly to 2.54d, set $x_1 = \cdots = x_k = 1$, and we
get 
$$k^n = \sum_{n_1+\cdots+n_k = n} \pmatrix{ &n& \cr n_1&\cdots&n_k\cr}$$
as claimed.


\bigskip
\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 
$(A\cap B) \cap (A \cup B) = A\cap B$.

\bigskip
\noindent
2.58/  $A$ and $M$ are independent since 
$P(A\cap M) = {24\over 100} = {6\over 25} = P(A)P(M)$. 
Similiarly $\overline{A}$ and $F$ are independent
since $P(\overline{A}\cap F) = {24\over 100} = P(\overline{A})P(F)$.
{\it Comment}: Intuitively, these independences are evident from the table,
since the fraction of the total males  who passed
equals the fraction of the total females who passed
($24/40 = 3/5 = 36/60$) and similarly for the fractions who failed.
\bye
