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123.tex

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+\item{2.} In a batch of 30 resistors, 10 are defective.
+\itemitem{(a)} If 10 resistors are randomly selected from the batch,
+what is the probability that exactly 3 are defective?
+\itemitem{(b)} If 8 resistors are randomly selected from the batch,
+what is the probability that at most 2 are defective?
+\itemitem{(c)} If 5 resistors are randomly selected from the batch,
+what is the probability that at least two are defective?
+
+\bigskip
+Solution: Here the random variable is the number of
+successes in a sample.
+Hence, use the hypergeometric distribution with $N=30$ and $D=10$.
+We have
+$$p(x)={{}_DC_x\cdot{}_{N-D}C_{n-x}\over {}_NC_n}=
+{{}_{10}C_x\cdot{}_{20}C_{n-x}\over {}_{30}C_n}$$
+
+\bigskip
+{\bf (a) If 10 resistors are randomly selected from the batch,
+what is the probability that exactly 3 are defective?}
+
+\bigskip
+Solution: $n=10$, $x=3$. Check with Excel {\tt =HYPGEOMDIST(3,10,10,30).}
+$$p(3)={{}_{10}C_3\cdot{}_{20}C_{7}\over {}_{30}C_{10}}=0.3096$$
+
+\bigskip
+{\bf (b) If 8 resistors are randomly selected from the batch,
+what is the probability that at most 2 are defective?}
+
+\bigskip
+Solution: $n=8$, $P(X\le2)=p(0)+p(1)+p(2)=0.4520$.
+
+\bigskip
+{\bf (c) If 5 resistors are randomly selected from the batch,
+what is the probability that at least two are defective?}
+
+\bigskip
+Solution: $n=5$, $P(X\ge2)=1-p(0)-p(1)=0.5512$.
+
+
+\end

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+\item{3.} A certain experiment is to be performed until a successful result
+is obtained.
+The trials are independent and the probability of success on a single trial
+is 0.25.
+The cost of performing the experiment is \$25{,}000; however, if a failure
+results, it costs \$5{,}000 to ``set-up'' for the next trial.
+
+\bigskip
+{\bf (a) What is the expected cost of the project?}
+
+\bigskip
+Solution: Here we are interested in the number of trials before a success
+so it must be a geometric distribution.
+The key word above is ``expected'' which means ``average'' or mean.
+From lecture 35 slide 10 we have
+$$\mu={1-p\over p}$$
+In this case we have
+$$\mu={1-0.25\over0.25}=3$$
+So on average the experiment will fail 3 times.
+That means the experiment must be done 4 times
+so the expected cost is $4\times25{,}000+3\times5{,}000=115{,}000$
+dollars.
+
+\bigskip
+{\bf (b) Suppose the experimenter has a maximum of \$500{,}000, what is the
+probability that the experimental work would cost more than this amount?}
+
+\bigskip
+Solution: First we need to find $x$ such that
+$$25{,}000(x+1)+5{,}000x=500{,}000$$
+We have
+$$x={475{,}000\over30{,}000}=15.83$$
+Since $x$ is discrete, 16 failures or
+more will break the budget.
+Now we need to compute $P(X\ge16)$.
+Fortunately, lecture 35 slide 16 gives us a distribution function
+that makes it easy.
+$$P(X>15)=(1-0.25)^{16}=0.0100$$
+
+\end

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+\beginsection Negative Binomial (Pascal) Distribution
+
+\bigskip
+\item{4.} A military commander wishes to destroy an enemy bridge.
+Each flight of planes he sends out has a probability of 0.8 of
+scoring a direct hit on the bridge.
+It takes four direct hits to completely destroy the bridge.
+If he can mount seven assaults before the bridge is
+tactically unimportant, what is the probability that the
+bridge will be destroyed?
+
+\bigskip
+Solution: We need 4 successes so this looks like a
+Pascal distribution.
+For example, $p(5)$ is the probability of
+success on the 5th flight.
+With $p=0.8$ and $k=4$ we have
+$$p(x)={}_{x-1}C_{k-1}p^k(1-p)^{x-k}=
+{}_{x-1}C_{3}(0.8)^4(0.2)^{x-4}$$
+The probability of destroying the bridge in 7 flights
+or less is
+$$P(X\le7)=\sum_{x=4}^7p(x)=0.9667$$
+
+\end

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+\item{5.} The number of red blood cells per square unit
+visible under a microscope follows a Poisson distribution
+with average number 4.
+
+\bigskip
+{\bf (a) Find the probability of more than 5 blood cells
+visible from a square unit.}
+
+\bigskip
+Solution: $\lambda=4$
+$$P(X>5)=1-\sum_{x=0}^5{4^xe^{-4}\over x!}=0.2149$$
+Check with Excel {\tt =1-POISSON(5,4,TRUE)}
+
+\bigskip
+{\bf (b) Find the probability that exactly 4 blood
+cells are visible from two square units.}
+
+\bigskip
+Solution: The trick here is to use the independence
+property. (See lecture 37, slide 1.)
+In other words, the average for 2 square units is
+twice the average of 1 square unit,
+hence $\lambda=8$.
+$$p(4)={8^4e^{-8}\over4!}=0.0573$$
+Check with Excel {\tt =POISSON(4,8,FALSE)}
+
+\bigskip
+{\bf (c) Find the probability that less than 2 blood cells
+are visible from a half square unit.}
+
+\bigskip
+Solution: As above, $\lambda=2$.
+$$P(X<2)=p(0)+p(1)=0.4060$$
+
+\bigskip
+{\bf (d) Find the probability that exactly 6 blood cells are visible from a square unit.}
+
+\bigskip
+Solution: $\lambda=4$
+$$p(6)={4^6e^{-4}\over6!}=0.1042$$
+
+
+\end

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+\bigskip
+\item{6.} An urn contains 10 black balls and 5 white balls.
+A ball is drawn at random from the urn and replaced by a
+black ball.
+What is the expected number of draws needed so that
+all of the balls in the urn are black?
+
+\bigskip
+Solution: The problem here is that the trials are
+not independent.
+So let us just think for a moment about replacing the first
+white ball.
+We see that it is going take a number of trials
+until a white ball is drawn, so it is like a
+geometric distribution with $p=5/15=1/3$.
+We have
+$$\mu={1-p\over p}=2$$
+So on average 2 draws are required {\it before} drawing the first
+white ball.
+Then we need to add 1 draw for the white ball itself.
+After that, the probability of drawing a white ball
+changes to $p=4/15$ and now we see the pattern.
+$$\mu=\sum_{n=5,4,3,2,1}\left({1-n/15\over n/15}+1\right)
+=\sum{15\over n}=3+3.75+5+7.5+15=34.25$$
+
+\end