\magnification=\magstep1
\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions For Whole Class Discussion Problems -- The ``Event-Composition Method''}
\centerline{September 16, 2011}
\bigskip
\noindent
{\it Discussion Questions}
\bigskip
\noindent
A)  Given: $P(A) = .2$, $P(B) = .3$, $P(A\cup B) = .4$.  Determine:
\bigskip
\item{1)} $P(\overline{A} \cup \overline{B})$.
\item{}{\it Solution:}  By the additive law, we have
$$P(\overline{A} \cup \overline{B}) = 
P(\overline{A}) + P(\overline{B}) - P(\overline{A} \cap \overline{B}).$$
Since $P(A) = .2$, $P(\overline{A}) = 1 - .2 = .8$.  Similarly, $P(\overline{B}) = 1 - .3 = .7$.
Finally, $\overline{A} \cap \overline{B} = \overline{A\cup B}$ by one of the DeMorgan
Laws from set theory.  So $P(\overline{A} \cap \overline{B}) = 1 - .4 = .6$.  Putting everything
together, 
$$P(\overline{A} \cup \overline{B}) = .8 + .7 - .6 = .9$$
\item{2)} $P(\overline{A} | B)$.  
\item{}{\it Solution:} By the definition of conditional probability,
$$P(\overline{A} | B) = {P(\overline{A} \cap B) \over P(B)}$$
We know $P(B) = .3$ from the given information.  In the course of solving part A, 
we saw $P(\overline{A} \cap \overline{B}) = .6$ and $P(\overline{A}) = .8$.  Since 
$$\overline{A} = (\overline{A} \cap B) \cup (\overline{A} \cap \overline{B}),$$  
and the two intersections in the parentheses don't intersect each other, we
have 
$$.8 = P(\overline{A} \cap B) + .6,\  {\rm hence}\  P(\overline{A} \cap B) = .2$$  Then
$$P(\overline{A} | B) = {.2\over .3} \doteq .67.$$
\bigskip
\noindent
B)  The circles marked 1,2,3,4 in the diagrams on the sheet given in class
represent electrical relays that operate
independently and function properly with probability $p = .85$.
Which design yields the higher probability that
current will flow when the relays are activated?
\bigskip
{\it Solution:}  Let $1,2,3,4$ represent the events that
the corresponding relays function correctly.  Layout I allows the current to
flow if either 1 and 2 function correctly, or 3 and 4 do, or both sets of two do.
So the probability we want is 
$P((1 \cap 2) \cup (3 \cap 4))$.  By the additive law and independence, this is
$$\eqalign{P(1\cap 2) + P(3 \cap 4) - P(1\cap 2\cap 3\cap 4) &= (.85)^2 + (.85)^2 - (.85)^4\cr
&\doteq .923.}$$
Similarly, Layout II allows the current to flow if
any one of the 4 relays functions correctly.  The current {\it does not} flow only if
all four relays fail, so we can compute the probability we want as 
$$1 - P(\overline{1}\cap\overline{2}\cap\overline{3}\cap\overline{4}) = 1 - (.15)^4 \doteq .9995.$$
As you should have seen intuitively, Layout II gives a significantly higher probability.
\bigskip
\noindent
C)  An accident victim will die unless he receives type 
$A+$ blood (strictly) before 8 minutes elapse.  He will be saved if
he does get the blood transfusion.  
Potential donors and a reusable blood typing kit are available, 
but it takes 2 minutes to determine each possible donor's blood type, 
and only $40\%$ of them have type $A+$ blood. 
What is the probability that the victim will be saved
if only one donor's blood can be typed at a time?  (Follow-up:  Are you making 
an assumption in your solution?)
\bigskip
\noindent
{\it Solution:}  There are just three chances for an $A+$ donor
to be found since each typing takes 2 minutes and the blood has to 
be given to the victim strictly before 8 minutes have passed.  
The chance of finding a good donor on the first test is $.4$.  When that test fails,
there is a $(.6)(.4)$ chance of success on the second test.  Similarly, 
when the first two tests fail, there is a $(.6)^2(.4)$ chance of success on
the last test before time runs out.  The total probability of finding 
a donor and saving the victim is 
$$(.4) + (.4)(.6) + (.4)(.6)^2 = .784$$
More formally, say $F$ is the event that a match for the blood
is found on or before the third try.  Then
$$F = F_1 \cup F_2 \cup F_3$$
where $F_i = $ {\it first match found on try} $i$, for $i = 1,2,3$.  Then
$F_i\cap F_j = \emptyset$ if $i\ne j$, so 
$$P(F) = P(F_1) + P(F_2) + P(F_3).$$
As before $P(F_1) = .4$.  But then $F_2 = $ match found on the second person
but not on the first.   So $P(F_2)$ is the probability that 
the first person tested does not have type $A+$ blood, and the 
second person does have type $A+$ blood.  The value $(.4)(.6)$ is OK here
(only) as long as the two people's blood types are {\it independent}.
Similarly for the $(.4)(.6)^2$ in the last term.     
\bye
