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\centerline{Mathematics 375 -- Probability and Statistics 1}
\centerline{Solutions for Problem Set 5}
\centerline{October 15, 2009}
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\noindent
3.39.  (a)  $Y = $ the number of components
that last longer than 1000 hours has a binomial
distribution.   $P(Y = 2) = \binom{4}{2}(.8)^2(.2)^2 \doteq .1536$.

(b)  The subsystem will operate if $2,3$, or $4$ of the 
components last longer than 1000 hours.  So this 
probability is 
$P(Y\ge 2) = \binom{4}{2}(.8)^2(.2)^2 + \binom{4}{3}(.8)^3(.2) + \binom{4}{4}(.8)^4 \doteq .973$.  
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\noindent
3.40.  These are all binomial probabilities. Let $Y$ be the number
who recover out of the 20.
(a)  $P(Y = 14) = \binom{14}{20} (.8)^{14}(.2)^6 \doteq .11$.

(b)  $P(Y \ge 10) = \displaystyle{\sum_{y = 10}^{20} \binom{20}{y}(.8)^y(.2)^{20-y}}\doteq 
.9994$.  

(c)  $P(14 \le Y \le 18) = 
\displaystyle{\sum_{y=14}^{18} \binom{20}{y}(.8)^y(.2)^{20-y}}\doteq .844$.

(d) $P(0 \le Y \le 16) = 
\displaystyle{\sum_{y=0}^{16} \binom{20}{y}(.8)^y(.2)^{20-y}}\doteq .589$.
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\noindent
3.43.  Let $Y$ be the number who qualify for the favorable rates (out of the 5),
a binomial random variable. (a) $P(Y = 5) = (.7)^5 \doteq .168$.

(b)  $P(Y \ge 4) = 
\displaystyle{\sum_{y=4}^5 \binom{5}{y} (.7)^y (.3)^{5-y}} \doteq .528$.  
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\noindent
3.44.  Let $Y$ be the number of successful operations (binomial).
(a)  $P(Y = 5) = (.8)^5 \doteq .328$.

(b)  $P(Y = 4) = \binom{5}{4}(.6)^4(.4)\doteq .259.$

(c)  $P(Y < 2) = 
\displaystyle{\sum_{y=0}^1 \binom{5}{y} (.3)^y (.7)^{5-y}} \doteq .528$.
(Note that this is the same as the answer for 3.43 (b).  Do you see why?  For
a formal statement of the pattern here, see Exercise 3.54.)
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\noindent
3.53.  $Y = $ the number of children (out of 3) that develop 
Tay-Sachs has a binomial distribution.  (a)  $P(Y = 3) = (.25)^3 \doteq .0156$.

(b)  $P(Y = 1) = \binom{3}{1}(.25)(.75)^2 \doteq .422$.

(c)  This conditional probability is $\frac{(.75)^2(.25)}{(.75)^2} = .25$.
(This follows because of the independence.)  
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\noindent
3.59.  Let $Y$ be the number of good motors, so $10 - Y$
is the number of defectives.  The problem is asking for the 
expected value of $G = 1000 - 200(10 - Y) = 200Y - 1000$.  
Since $Y$ is binomial with $n = 10$ and $p = .92$, we have
$E(G) = E(200Y - 10000) = 200E(Y) - 1000 = (200)(9.2) - 1000 = 840$
(dollars).
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\noindent
3.62.  (a)  To justify multiplying $p_1p_2p_3$ to get the probability
of detecting a wing crack, you need to assume the individual events
are independent.

(b)  $p = (.9)(.8)(.5) = .36$.  Then the probability desired is
$P(Y \ge 1)= 
\displaystyle{\sum_{y=1}^3 \binom{3}{y} (.36)^y (.64)^{3-y}} \doteq .738$.
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\noindent
3.67.  This is geometric with $p = .3$.  So $P(Y = 5) = (.7)^4(.3) \doteq .072$.
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\noindent
3.69.  $Y$ is a geometric random variable with $p = .59$, so 
$P(Y = y) = (.41)^{y-1} (.59)$.  
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