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1.Fred is giving out samples of dog food. He makes calls door to door, but he leaves a sample (one can) only
on those calls for which the door is answered and a dog is in residence. On any call the probability of the
door being answered is 3/4, and the probability that any household has a dog is 2/3. Assume that the
events “Door answered” and “A dog lives here” are independent and also that the outcomes of all calls are
independent.
(a) Determine the probability that Fred gives away his first sample on his third call.
(b)Given that he has given away exactly four samples on his first eight calls, determine the conditional
probability that Fred will give away his fifth sample on his eleventh call.
(c) Determine the probability that he gives away his second sample on his fifth call.
(d)Given that he did not give away his second sample on his second call, determine the conditional
probability that he will leave his second sample on his fifth call.
(e)We will say that Fred “needs a new supply” immediately after the call on which he gives away his last
can. If he starts out with two cans, determine the probability that he completes at least five calls before he
needs a new supply.
(f)If he starts out with exactly m cans, determine the expected value and variance of Dm, the number of
homes with dogs which he passes up (because of no answer) before he needs a new supply.
2.Each night, the probability of a robbery attempt at the local warehouse is 1/5. A robbery attempt is
successful with probability 3/4, independent of other nights. After any particular successful robbery, the
robber celebrates by taking off either the next 2 or 4 nights (with equal probability), during which time
there will be no robbery attempts. After that, the robber returns to his original routine.
Probabilistic Systems Analysis
Problem
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(a)Let K be the number of robbery attempts up to (and including) the first successful robbery. Find the
PMF of K.
(b)Let D be the number of days until (and including) the second successful robbery, in cluding the days of
celebration after the first robbery. Find the PMF of D, or its transform (whichever you find more
convenient).
During a successful robbery, the robber steals a random number of candy bars, which is 1, 2, or 3, with
equal probabilities. This number is independent for each successful robbery and independent of everything
else. No candy bars are stolen in unsuccessful robberies.
(c)Let S be the number of candy bars collected in two successful robberies. Find the PMF of S.
(d)Let Q be the number of candy bars collected in ten robbery attempts (whether successful or not). Find
the PMF of Q, or its transform, whichever is easier. Find the expectation and the variance of Q.
3.You are visiting the rainforest, but unfortunately your insect repellent has run out.
(a)As a result, at each second, a mosquito lands on your neck with probability 0.2.
i. What’s the PMF for the time until the first mosquito lands on you?
ii.What’s the expected time until the first mosquito lands on you?
iii.If you weren’t bitten for the first ten seconds, what would be the expected time until the first
mosquito lands on you (from time t=10) ?
(b)Instead, imagine the rainforest had only one mosquito, which arrived in the following way: the time of
arrival is exponentially distributed with λ = 0.2.
i. What’s the expected time until the first mosquito lands on you?
ii.If you weren’t bitten for the first ten seconds, what would be the expected time until the first
mosquito lands on you (from time t=10) ?
4.A particular medical operation proves fatal in 1% of the cases. Find an approximation to the probability that
there will be at least 2 fatalities in 200 operations.

5.Arrivals of certain events at points in time are known to constitute a Poisson process, but it is not
known which of two possible values of λ, the average arrival rate, describes the process. Our a priori
estimate is that λ = 2 or λ = 4 with equal probability. We observe the process for t units of time and
observe exactly k arrivals. Given this information, determine the conditional probability that λ = 2.
Check to see whether or not your answer is reasonable for some simple limiting values for k and t.
G1† . A wombat in the San Diego zoo spends the day walking from a burrow to a food tray, eating, walking
back to the burrow, resting and repeating the cycle. The amount of time to walk from secs. The wombat,
with 1 3 the burrow to the tray (and also from the tray to the burrow) is 20 probability , will momentarily
stand still (for a negligibly small time) during a walk to or from the tray, with all times being equally likely
(and independently of what happened in the past). A photographer arrives at a random time and will take a
picture at the first time the wombat will stand still. What is the expected value of the length of time the
pohotographer has to wait to snap the wombat’s picture ?

1. A successful call occurs with probability
(a) Fred will give away his first sample on the third call if the first two calls are failures and the third is a success.
Since the trials are independent, the probability of this sequence of events is simply
(b) The event of interest requires failures on the ninth and tenth trials and a success on the eleventh trial. For a
Bernoulli process, the outcomes of these three trials are independent of the results of any other trials and again
our answer is
(c) We desire the probability that L2, the second-order interarrival time is equal to five trials. We know that pL2(l)
is a Pascal PMF, and we have
(d) Here we require the conditional probability that the experimental value of L2 is equal to 5, given that it is
greater than 2.
Solutions

(e) The probability that Fred will complete at least five calls before he needs a new supply is equal to the
probability that the experimental value of L2 is greater than or equal to 5.
(f) Let discrete random variable F represent the number of failures before Fred runs out of samples on his mth
successful call. Since Lm is the number of trials up to and including the mth success, we have F = Lm −m. Given that
Fred makes Lm calls before he needs a new supply, we can regard each of the F unsuccessful calls as trials in
another Bernoulli process with parameter r, where r is the probability of a success (a disappointed dog) obtained
by
We define X to be a Bernoulli random variable with parameter r. Then, the number of dogs passed up before Fred
runs out, Dm, is equal to the sum of F Bernoulli random variables each with parameter r = 1 3 , where F is a
random variable. In other words,
Note that Dm is a sum of a random number of independent random variables. Further, F is independent of the
Xi’s since the Xi’s are defined in the conditional universe where the door is not answered, in which case,
whether there is a dog or not does not affect the probability of that trial being a failed trial or not. From our
results in class, we can calculate its expectation and variance by

2. Define the following events:
Event R: Robber attempts robbery.
Event S: Robbery is successful.
Define the following random variables:
x: The number of days up to and including the first successful robbery.
b: The number of candy bars stolen during a successful robbery.
c: The number of days of rest after a successful robbery.

Note that x is a geometric random variable with parameter 3 20 (the probability of a successful robbery on a
given night). Also, it is given that b is uniform over {1, 2, 3} with probability 1 each, and that c is uniform over {2,
4} with probability 1 each. 2
(a) Observe that since the probability that any given robbery attempt succeeds is 3 4 , the 3 random variable k is
geometric with parameter p = 4 . Thus,
(b) We will derive first the PMF and then the transform of d, the number of days up to and including the second
successful robbery.
The PMF of d can be easily found by conditioning on c, the number of days the robber rests after the first
successful robbery (which only takes on values 2 or 4):
There must be at least one day before the rest period, since it follows a successful robbery; similarly, there
must be at least one day after the rest period. Thus we can view the coefficient in the preceding formula as the
number of ways to choose the beginning of a four-day period in a block of do days. Then we multiply the
probability of the first success and the probability of do − 4 failures and finally the probability of the second
success at trial do. Similarly, we have:

Alternately, we can solve this problem using transforms. Note that d is the sum of three independent
random variables: x1, the number of days until the first successful robbery; c, the number of days of
rest after the first success; and x2, the number of days from the end of the rest period until the
second successful robbery. Since d = x1 + c + x2, and x1, x2, and c are mutually independent, we find
that the transform of d is Md(s) = [Mx(s)]2Mc(s).
Now, x1 and x2 are (independent) geometric random variables with parameter 3 (the 20 probability of
a successful robbery). Again, c is equally likely to be 2 or 4. Thus we conclude that
(c) Given a successful robbery, the PMF for y is py(yo) = 1 for yo = 1, 2, 3, and py(yo) = 0 3 otherwise. The total
number of candybars collected in 2 successful robberies is s = y1 +y2, where y1 and y2 are independent and
identically distributed as py(yo). Therefore, the PMF for s is

(d) Observe that since the probability of a robbery attempt being successful is 3 4 , with the number of candy bars
taken in a successful attempt equally likely to be 1, 2, or 3, we can view each attempt as resulting in b candy bars,
with the following PMF for b:
Now, let t, u, and v be the number of attempts that result in 1, 2, and 3 candy bars, respectively. Then if a total
of q candy bars are stolen during the ten robbery attempts, 0 ≤ v ≤ q 3 , 0 ≤ u ≤ q−3v , and 0 ≤ t ≤ q − 3v − 2u,
with exactly 10 − t − u − v attempts 2 failing. Thus the PMF for q is

3. (a) i. Since it is implied that each second is independent from each other (i.e. if a mosquito landed on your
neck last second, it doesn’t affect the likelihood of one landing this or next second), the PMF for the time until
the first mosquito lands is simply a geometric random variable. In this random variable, the “success” event is
when a mosquito lands with probability 0.2, with the “failure” event being 0.8.
ii. The expected time of the geometric PMF with parameter is , so the expected p time until the first mosquito
lands on you is . 1 2 = 5 seconds.
iii. The scenario in which mosquitoes independently land on you each second can be modeled as a Bernoulli
process, with each Bernoulli trial being the event that a mosquito lands on you on a given second. Because the
Bernoulli process is memoryless, it doesn’t matter whether or not you were bitten for the first one, ten, or two
hundred seconds. Thus, the expected time from T = 10 is identical to the answer in the previous part, . 1 2 = 5
seconds
(b) i. Because the PDF that models the time until the first mosquito arrives is exponential, the expected time until
it lands is . 1 2 = 5 seconds.
ii. It has been previously shown that the exponential PDF exhibits the memorylessness property. In other words,
looking at an exponential PDF from some future time �= 0 will still yield an exponential PDF with the same
parameter. Thus, the expected time from T = 10 is identical to the answer in the previous part, . 1 2 = 5 seconds.
4. We could find an exact value by using the binomial probability mass function. A reasonable, and much more
efficient method is to use the Poisson approximation to the binomial, which tells us that for a binomial random
variable with parameters n and p, we have:

where λ = np. The desired probability is
5. We have a Poisson process with an average arrival rate λ which is equally likely to be either 2 or 4. Thus,
We observe the process for t time units and observe k arrivals. The conditional probability that λ = 2 is, by
definition
Now,
we know that −2t (2t)ke 1 P(λ = 2 and k arrivals in time t) = P(k arrivals in time t | λ = 2) · P(λ = 2) = ·
Similarly,
Thus,
To check whether this answer is reasonable, suppose t is large and k = 2t (observed arrival rate equals 2). Then,
P(λ = 2 | k arrivals in time t) approaches 1 as t goes to ∞. Similarly, if t is large and k = 4t (observed arrival rate
equals 4), then, P(λ = 2 | k arrivals in time t) approaches 0 as t goes to ∞

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