Probabilistic Systems Analysis Exam Help

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1. Suppose that you were foolish enough to save your thesis on only one floppy disk, and that this disk got corrupted. To
make matters worse, you actually have 3 other old corrupted disks lying around, and it is equally likely that any of the 4
disks holds the corrupted remains of your thesis. Before you take all 4 disks to an expensive disk doctor, your friend across
the hall offers to have a look. You know from past experience that the overall probability that your friend will find your
paper on any disk is p. Given that he searches on disk 1 but cannot find your paper, what is the probability that your
thesis is on disk i for i = 1, 2, 3, 4?
2. Bo and Ci are the only two people who will enter the Rover Dog Food Jingle contest. Only one entry is allowed per
contestant, and the judge (Rover) will declare the one winner as soon as he receives a suitably inane entry, which may
be never.
Bo writes inane jingles rapidly but poorly. He has probability 0.7 of submitting his entry first. If Ci has not already won
the contest, Bo’s entry will be declared the winner with probability 0.3. Ci writes slowly, but he has a gift for this sort of
thing. If Bo has not already won the contest by the time of Ci’s entry, Ci will be declared the winner with probability 0.6.
(a) What is the probability that the prize never will be awarded?
(b) What is the probability that Bo will win?
(c) Given that Bo wins, what is the probability that Ci’s entry arrived first?
(d) What is the probability that the first entry wins the contest?
(e) Suppose that the probability that Bo’s entry arrived first were P instead of 0.7. Can you find a value of P for which
“First entry wins” and “Second entry does not win” are independent events?
3. Before leaving to work, Victor checks the weather report before deciding on carrying an umbrella or not. If the forecast
is “rain”, the probability of actually having rain that day is 80%. On the other hand, if the forecast is “no rain” the
probability of actually raining is equal to 10%.
During fall and winter the forecast is 70% of the time “rain” and during summer and spring it is 20%.
Problem

(a) One day, Victor missed the forecast and it rained. What is the probability that the forecast was “rain” if it was
during the winter? What is the probability that the forecast was “rain” if it was during the summer?
(b) The probability of Victor missing the morning forecast is equal to 0.2 on any day in the year. If he misses the
forecast, Victor will flip a fair coin to decide on carrying an umbrella or not. On the day he sees the forecast, if
it says “rain” he will always carry an umbrella, and if it says “no rain”, he will never carry an umbrella. Are the
events “Victor is carrying an umbrella”, and “The forecast is no rain” independent ? Does your answer depend
on the season ?
(c) Victor is carrying an umbrella and it is not raining. What is the probability that he saw the forecast?
4. You are lost in the campus of MIT, where the population is entirely composed of brilliant students and absent-minded
professors. The students comprise two-thirds of the population, 3 and any one student gives a correct answer to a
request for directions with probability . 4 (Assume answers to repeated questions are independent, even if the question
and the person asked are the same.) If you ask a professor for directions, the answer is always false.
(a) You ask a passer-by whether the exit from campus is East or West. The answer is East. What is the probability this
is correct?
(b) You ask the same person again, and receive the same reply. Show that the probability that this second reply is
correct is
(c) You ask the same person again, and receive the same reply. What is the probability that this third reply is correct?
(d) You ask for the fourth time, and receive the answer East again. Show that the probability it is correct is
(e) Show that, had the fourth answer been West instead, the probability that East is nev ertheless correct is
Your friend, Ima Nerd, happens to be in the same position as you are, only she has reason to believe a-priori that,
with probability , East is the correct answer.

(f) Show that whatever answer is first received, Ima continues to believe that East is correct with probability .
(g) Show that if the first two replies are the same (that is, either WW or EE), Ima continues to believe that East is
correct with probability .
(h) Show that after three like answers, Ima will calculate as follows (in the obvious notation):
G1† . Alice and Bob love to challenge each other to coin tossing contests. On one particular day, Alice brings 2n +
1 fair coins, and lets Bob toss n + 1 coins, while she tosses the remaining n coins. Show that the probability that
after all the coins have been tossed Bob will have gotten more heads that Alice is 1/2.

1. Let A be the event that your friend searches disk 1 and finds nothing, and let Bi be the event that your thesis is on
disk i. The sample space is described below.
Note that B1, B2, B3, and B4 partition the sample space, so applying Bayes’ rule, we have
Solution

We see that since P(B) is not equal to 1 for any value of P, there is no value of P for which P(A)P(B)= P(AB). Thus
events A and B are never independent.
3. (a) The tree representation during the winter can be drawn as the following:

Let A be the event that the forecast was “Rain”,
Let B be the event that it rained, Let p be the probability that the forecast says “Rain”,
If it is in the winter, p = 0.7,
(b) Let C be the event that Victor is carrying an umbrella. Let D be the event that the forecast is no rain. The tree
diagram in this case is:

Therefore, P(C)= P(C | D) if and only if p = 0. However, p can only be 0.7 or 0.2, which implies the event C and D
can never be independent, and this result does not depend on the season.
(c) Let us first find the probability of rainning if Victor missed the forecast.
P(actually rains | missed forecast) = (0.8)p + (0.1)(1 − p)=0.1+0.7p.
Then, we can extend the tree in part b) as follows,
Therefore, given that Victor is carrying an umbrella and it is not raining, we are looking at the two shaded cases.

4. Without prior bias on whether the exit of campus lies East or West, the exact answers of the passerby are not as
important as whether a string of answers is similar or not. Let Rr denote the event that we receive r similar answers and
T denote the event that these repeated answers are truthful. Let S denote the event that the questioned passerby is a
student. Note that, because a professor always gives a false answer, T ∩ Sc = ∅ and thus P(T ∩ Sc) = 0. Therefore,
where the stated independence of a passerby’s successive answers implies P(T ∩Rr|S)= Applying the Total
Probability Theorem and again making use of independence, we also 3 4 . deduce
(a) Applying the above formulas for r = 1, we have P(R1) = 1 and thus

(e) As soon as we receive a dissimilar answer from the same passerby, we know that this passerby is a student; a
professor will always give the same (false) answer. Let D denote the event of receiving the first dissimilar answer. Given D
on the fourth answer, either the student has provided three truthful answers followed by one untruthful answer,
In parts (a) - (d), notice the decreasing trend in the probability of the passer-by being truthful as the number of
similar answers grows. Intuitively, our confidence that the passerby is a professor grows as the sequence of similar
answers gets longer, because we know a professor will always give the same (false) answer while a student has a
chance to answer either way. However, as part (e) demonstrates, the first indication that the passerby is a student
will boost our confidence that the previous string of similar answers are truthful, because any single answer by the
student has a 3-to-1 chance of being a truthful one.
For the remainder of this problem, let E and W represent the events that a passerby provides East and West,
respectively, as an answer and let TE represent the event that East is the correct answer. We are told Ima’s a-priori
bias is P(TE)= .
(f) Using Bayes’s Rule and all the arguments used in parts (a) - (e), we have

In particular, we have used that P(E) = P(E|TE)P(TE)+P(E|Tc E)P(Tc E) (and similarly for P(W))
(g) Likewise, given two consecutive and similar answers from the same passerby, we have
(h) Finally, given three consecutive and similar answers from the same passerby,
Notice that the E, EE and EEE answers to parts (f) - (h) match the answers to parts (a)-(c) when = 1 2 , or when Ima’s
prior bias does not favor either possibility.

G1†. Through the first n coins, Alice and Bob are equally likely to have flipped the same number of heads as the other
(since they are using fair coins and each flip is independent of the other flips). Given this, when Bob flips his last coin
but Alice doesn’t flip a coin, Bob has a 1/2 chance of getting a head and thus having more heads than Alice.
Let’s look at it another way. First define the following events:
1) A = the number of heads Alice tossed 2) B = the number of heads Bob tossed Now suppose both Bob and Alice
toss n coins. A sample space of interst is shown below where the shaded area represents Alice having more heads
than Bob and the unshaded and uncrossed area represents Bob having more heads than Alice.

Well, each box has the same weight and there are the same number of boxes on each side of the diagonal.
Consequently the above equation holds, meaning Alice and Bob are equally likely to have flipped more heads than
the other. Furthermore, Alice and Bob are equally likely to have flipped the same number of heads, again since
each box with an x in it has the same probabilistic weight.
Now Bob picks up the last coin. Given that both Alice and Bob are equally likely to have the same number of
heads, the event Bob having more heads than Alice boils down to Bob getting a head on the last coin flip. Since
this coin is fair and the flip is independent of past flips, this probability is simply 1/2.
An alternative solution is shown as follows,
Let B be the event that Bob tossed more heads,
let X be the event that after each has tossed II of their coins, Bob has more heads than Alice,
let Y be the event that under the same conditions, Alice has more heads than Bob,
and let Z be the event that they have the same number of heads.
Since the coins are fair, we have P(X) = P(Y), and also P(Z) = 1— P(X) — P(Y). Further-
more, we see that
Now we have, using the theorem of total probability,

as required.

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