problem set

Problems and results for the tenth week
Mathematics A3 for Civil Engineering students
1. Steve has not prepared for his exam, where he has to answer 10 yes or no questions. A small
part of the material dawns on him, and so he can give the correct answer to each question
with probability 60%. What is the probability he will pass if that needs at least 8 correct
answers?
2. Each student on a test has to answer 20 yes or no questions. Assume that independently
for each question, a student knows the correct answer with probability 0.7, believes that he
knows the correct answer, but he is wrong with probability 0.1, and doesn’t know the answer
with probability 0.2. In this case he answers yes or no with probability 1 1 − . What is the
2 2
probability that he will answer at least 19 questions correctly?
3. On a multiple-choice exam with 3 possible answers for each of the 5 questions, what is the
probability that a student would get 4 or more correct answers just by guessing?
4. A communications channel transmits the digits 0 and 1. However, due to static, the digit
transmitted is incorrectly received with probability 0.2. Suppose that we want to transmit an
important message consisting of one binary digit. To reduce the chance of error, we transmit
00000 instead of 0 and 11111 instead of 1. If the receiver of the message uses “majority”
decoding, what is the probability that the message will be wrong when decoded?
5. A wolverine starts from the origin of the integer line. At each step he moves one unit to the
right with probability 1/2, and to the left with probability 1/2, independently of his previous
steps. After 20 moves,
(a) What is the probability that he is at position 0?
(b) What is the probability that he is at position 1?
(c) What is the probability that he is at position (-2)?
(d) What is the probability that he is at position (-2), if he was at (-3) one step ago?
6. I have two coins, a fair one and a biased one, but I cannot distinguish them. The biased coin
comes up heads with probability 3/4. I pick one of the two coins from my pocket, the fair
one with probability 1/2 and the biased one with probability 1/2. Then I flip the chosen coin
30 times, and I find that it came up heads 25 times. What is the probability that I chose the
biased coin?
7. When coin A is flipped, it lands heads with probability 0.4; when coins B is flipped, it land
heads with probability 0.7. One of these coins is randomly chosen and flipped 10 times.
(a) Are these coin flips independent of each other?
(b) What is the probability that exactly 7 of the 10 flips land on heads?
(c) Given that the first of these ten flips lands heads, what is the probability that we are
using coin A?
(d) Given that the first of these ten flips lands heads, what is the probability that exactly 7
of the 10 flips land on heads?
8. It is known that diskettes produced by a certain company will be defective with probability
0.01, independently of each other. The company sells the diskettes in packages of size 10
and offers a money-back guarantee that at most 1 of the 10 diskettes in the package will be
defective.
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(a) What is the probability that a box contain more than one defective diskettes?
(b) If someone buys 3 packages, what is the probability that he or she will return exactly 1
of them?
9. A newsboy purchases papers at HUF 100 and sells them at HUF 150. However, he is not
allowed to return unsold papers. If his daily demand is a binomial random variable with
n = 10 and p = 1/3, approximately how many papers should he purchase so as to maximize
his expected profit?
10. Approximately 80000 marriages took place in a country last year. Estimate the probability
that for at least one of these couples
(a) both partners were born on April 30;
(b) both partners celebrate their birthday on the same day of the year.
State your assumptions.
11. There are 200 typos, randomly distributed, in a book of 400 pages. What is the probability
that on page 13 there are more than one typos?
12. How many chocolate chips should there be in a muffin on average if we want at least one
chocolate chip in any given muffin with probability at least 0.99?
13. The Run With Us Movement organized a foot-race at the Danube Bend. Unfortunately, the
track passed through an area infected with ticks. After the race, 300 contestants found one
tick, 75 found two ticks on their bodies. Based on this information, approximate the number
of contestants in this race.
14. Consider a roulette wheel consisting of 38 numbers – 1 through 36, 0, and double 0. If Smith
always bets that the outcome will be one of the numbers 1 through 12, what is the probability
that
(a) Smith will loose his first 5 bets;
(b) his first win will occur on his fourth bet?
15. An urn contains 4 white and 4 black balls. We randomly choose 4 balls. If 2 of them are white
and 2 are black, we stop. If not, we replace the balls in the urn and again randomly select 4
balls. This continues until exactly 2 of the 4 chosen are white. What is the probability that
we shall make exactly n selections?
16. We repeatedly roll a die until we roll a 6. What is the expected number of times we roll that
die? And if we roll two dice a time until we see a 6 on at least one of them?
17. We repeatedly roll a die until we see the same number twice in a row. What is the expected
number of rolls we make?
18. Out of our 100 keys, only one opens the door in front of us. In the dark we don’t see the keys
we already tried, and so we might pick and try any given key more than once. What is the
probability that we open the door by at most 50 trials? And what if we put away the ones
we already tried?
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19. If X is a geometric random variable, show analytically that
(1) P{X = n + k | X > n} = P{X = k}.
Give a verbal argument using the interpretation of a geometric random variable as to why
the preceding equation is true.
20. Mr. Bilk takes the tram to work each day, but he has no monthly pass nor ticket. Every day
the ticket controller gets on the tram with probability 0.2, and then catches Mr. Bilk with
probability 0.95. (Every day the ticket controller decides independently whether to get on
Mr. Bilk’s tram or not.)
(a) What is the probability that Mr. Bilk has a “lucky week” that is, he won’t get fined on
the five working days of the week?
(b) What is the probability that he will get fined exactly twice on the five working days of
the week?
(c) Given that Mr. Bilk had a lucky week, what is the probability that there was a ticket
controller on his tram on each of the five working days?
(d) What is the probability that he will get fined on Thursday the first time?
21. The suicide rate in a certain country is 1 suicide per 100000 inhabitants per month.
(a) Find the probability that in a city of 400000 inhabitants within this country, there will
be 8 or more suicides in a given month.
(b) What is the probability that there will be at least 2 months during the year that will
have 8 or more suicides?
(c) Counting the present month as month number 1, what is the probability that the first
month to have 8 or more suicides will be month number i, i ≥ 1?
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1.00
0.75
0.50
0.25
0.00
Answers
1. Let X be the number of correct answers given. This is a binomial random variable with
parameters n = 10, p = 0.6. The answer is
P{X ≥ 8} = P{X = 8} + P{X = 9} + P{X = 10}
10
= · 0.6
8
8 · 0.4 2
10
+ · 0.6
9
9 · 0.4 1
10
+ · 0.6
10
10 · 0.4 0 ≃ 0.167.
3.
5
·
4
1
4 ·
3
2
1 +
3
5
·
5
1
5 ·
3
2
0 ≃ 0.045.
3
4. The information is wrongly decoded if at least three bits are incorrectly received. The probability
of this is
5
· 0.2
3
3 · 0.8 2
5
+ · 0.2
4
4 · 0.8 1
5
+ · 0.2
5
5 · 0.8 0 ≃ 0.058.
5.(a) The wolverine will be at 0 if and only if exactly 10 out of his 20 steps were left moves, and
10 were right moves. The probability of this is 20 1 10
1 10
· · ≃ 0.176.
10 2 2
(b) He can only be at even positions after an even number of steps, hence the answer is zero.
(c) He will be at (-2) if and only if he made 11 steps to the left and 9 steps to the right. The
probability of this is 20 1 11
1 9
· · ≃ 0.160.
11 2 2
(d) Given that he is at (-3) before the last step, he will move one to the right, to (-2), with
probability 1
2 .
6. Let X be the number of heads flipped, and {B} is the event that we use the biased coin. Then
P{B | X = 25} =
=
P{X = 25 | B} · P{B}
P{X = 25 | B} · P{B} + P{X = 25 | Bc } · P{Bc }
30 3 25
1 5 1
· · · 25 4 4 2
30 3 25
1 5 1
· · · 25 4 4 2 + 30 1 25
1 5 =
1
· · · 25 2 2 2
325 325 ≃ 0.9987.
+ 230 In general, having flipped heads n times (0 ≤ n ≤ 30) out of 30, the probability in question is
P{B |X = n}
P{B | X = n} =
3 n
3 n + 2 30.
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
4
n

On the graph of this conditional probability we explore a sharp transition at around n = 19.
7. Let X be the number of heads flipped, H the event that the first flip lands heads, A and B the
events of picking the corresponding coins.
(b)
(c)
P{A | H} = P{HA}
P{H} =
P{X = 7} = P{X = 7 | A} · P{A} + P{X = 7 | B} · P{B}
10
= · 0.4
7
7 · 0.6 3 · 1
2 +
10
· 0.7
7
7 · 0.3 3 · 1
≃ 0.155.
2
P{H | A} · P{A}
P{H | A} · P{A} + P{H | B} · P{B} =
0.4 · 1
2
0.4 · 1
2
+ 0.7 · 1
2
= 4
11
≃ 0.364.
(a) Flipping heads first will make it likely that we are using coin B, which in turn will increase
the chances for flipping heads the second time. Hence the flips are not independent. Let’s
see a computation for this:
From the denominator of the previous display, we have P{H} = 0.55 that is, any given flip
will land heads with probability 55%. However, we already know that the first flip landing
heads will modify the probabilities of our coins in accordance with part (c): P{A | H} =
4/11, P{B | H} = 7/11. Hence the probability that the second flip comes heads (let us
denote this event by H2), given that the first flip landed heads, is
P{H2 | H} = P{H2 | A} · P{A | H} + P{H2 | B} · P{B | H} = 0.4 · 4 7 13
+ 0.7 · =
11 11 22
≃ 0.591.
Therefore, the first flip landing on heads modified the probability of the second flip coming
up heads from 0.55 to 0.591; the flips are not independent.
Rather than independence, we have the so-called conditional independence property here:
given that we use coin A, flips are independent. Similarly, given that we use coin B, flips are
independent. Without those conditions, however, we do not have independence.
(d) If we already know that the first flip lands heads, that will modify the probabilities of our
coins in accordance with part (c): P{A | H} = 4/11, P{B | H} = 7/11. After the first flip we
have 9 more flips to make, out of which we want Y = 6 to be heads. The probability of this
is
P{Y = 6} = P{Y = 6 | A} · P{A | H} + P{Y = 6 | B} · P{B | H}
9
= · 0.4
6
6 · 0.6 3 · 4
11 +
9
· 0.7
6
6 · 0.3 3 · 7
≃ 0.197.
11
8.(a) Let X be the number of defective diskettes in a package. Then X is a binomial random
variable with parameters n = 10 and p = 0.01. The package can be returned if X > 1. The
probability of this is
10
P{X > 1} = 1−P{X = 0}−P{X = 1} = 1− ·0.01
0
0 ·0.99 10
10
− ·0.01
1
1 ·0.99 9 ≃ 0.00427.
(b) Packages can be returned independently of each other with the above probability. Hence the
answer is
3
· 0.00427
1
1 · [1 − 0.00427] 2 ≃ 0.0127.
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9. Let us denote the number of purchased papers by m, the number of papers sold by X, and the
demand by Y . Then
Y, if Y < m,
X =
m, if Y ≥ m.
Since Y is binomial with parameters n = 10 and p = 1/3, X will have mass function
⎧
10
1
i
2
10−i ⎪⎨
P{Y = i} = · ·
, if i < m,
i 3 3
p(i) = P{X = i} =
10
10
1
j
2
10−j ⎪⎩
P{Y ≥ m} = · · , if i = m.
j 3 3
j=m
The cost of buying papers is 100 · m forints, the revenue is 150 · X forints, and so the newsboy’s
expected profit is
N = E(150 · X − 100 · m) = 150 · E(X) − 100 · m.
To determine E(X), we use the above mass function:
m m−1
E(X) = i · p(i) = i ·
i=0
i=0
10
·
i
1
i ·
3
2
10−i + m ·
3
10
j=m
10
·
j
1
j ·
3
2
10−j .
3
For each given m, this formula gives a concrete number, which can be plugged in the display above
to compute the expected profit (with a calculator or a computer algebra system). Results are:
m 0 1 2 3 4 5 6 7 8 9 10
N 0 47.4 81.79 86.92 53.03 -15 -103.52 -200.57 -300.06 -400 -500
The greatest expected profit is realized when 3 papers are bought.
10. We assume that spouses were independently born with equal chance on any day of the year
(we do not deal with leap years).
(a) The probability that both partners of a given couple were born on April 30 is p = 1/3652 ≃
7.51 · 10−6 . The number X of such couples is binomial with parameters n = 80 000 and the
above p. The probability in question is
80 000
P{X ≥ 1} = 1 − P{X = 0} = 1 − ·
0
1
3652 0
· 1 − 1
3652 80 000
≃ 0.451.
Our formulas simplify a lot after realizing that the data given place the problem in the setting
of the Poisson approximation: n is large, p is small, and their product is λ = np =
80 000/365 2 ≃ 0.6. Hence the distribution of X can be approximated by a Poisson distribution:
P{X ≥ 1} = 1 − P{X = 0} ≃ 1 − e −0.6 ≃ 0.451,
which agrees to at least three digits with the above binomial probability.
(b) The probability that partners of a given couple have the same birthday is 1/365. The number
Y of such couples is now binomial with parameters n = 80 000 and p = 1/365. It is still true
that n >> λ = np, hence the Poisson approximation is still valid. The above probabilities in
this case are:
80 000
1
0
P{Y ≥ 1} = 1 − P{Y = 0} = 1 − · · 1 −
0 365
1
80 000
≃ 1 − 4.8 · 10
365
−96 ,
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that is practically = 1. With the Poisson approximation λ = np = 80 000/365 ≃ 219.2, and
that is practically = 1.
P{Y ≥ 1} = 1 − P{Y = 0} = 1 − e −219.2 ≃ 1 − 6.49 · 10 −96 ,
13. The number X of ticks found in a contestant is Poisson distributed, with an unknown λ
parameter. Assuming that n contestants were racing, we approximate P{X = 1} ≃ 300/n and
P{X = 2} ≃ 75/n based on the data given. With the Poisson mass function, we now have to solve
the system of equations
λ1 1! · e−λ ≃ 300
n
λ2 2! · e−λ ≃ 75
n .
Dividing the second equation by the first one we have λ/2 ≃ 1/4, that is λ ≃ 1/2. Hence by the
first equation n ≃ 300 · eλ /λ ≃ 989.
14. Smith will win each of his bets with probability p = 12/38 = 6/19. Hence
(a) (1 − p) 5 =
13 5,
19
(b) (1 − p) 3 · p =
13 3 6 · 19 19 .
15. The probability of drawing exactly two white balls at a given draw is p = 4 4 8 18
· / = 2 2 4 35 .
The probability that we make n draws is, according to the geometric distribution, (1 − p) n−1 · p =
17 n−1 · 18/35 n .
16. Rolling one die, the number of rolls we make is geometric with parameter p = 1/6. The
expected number of rolls is 1/p = 6.
Rolling two dice, we see at least one 6 if and only if we do not roll something else on both dice,
that is, with probability p = 1 −
5 2
= 11/36. The number of rolls is now geometric with this
6
parameter, the expectation is now 1/p = 36/11.
17. Except for the first roll, each time independently we roll the same as the previous time with
probability 1/6. Hence the number of rolls we make after the first one is geometric with parameter
p = 1/6, and with expectation 6. Together with the first roll, we are expected to roll 7 times.
18. We assume that keys are tried independently and by giving equal chance to each key. Then
each time we will succeed with probability 1/100. The first 50 trials will all fail, if 50 times we have
a wrong key in our hands. The probability of this is
99 50,
and the answer is the probability of
100
the complement event, 1 −
99 50
≃ 0.395.
100
If we put away the keys we already tried, then we open the door by at most 50 trials if the right
key was among the first 50 in the random order in which we tried the keys. Since the right key can
be at any position of that order with equal chance, the probability is now 50/100 = 0.5.
19. The problem clearly asks for k > 0, as in other cases both sides of the equality are zero. For
k > 0
P{X = n + k and X > n}
P{X = n + k | X > n} = =
P{X > n}
P{X = n + k}
P{X > n}
= (1 − p)n+k−1 · p
(1 − p) n
= (1 − p) k−1 · p = P{X = k}.
Here we used the geometric mass function, and the fact that P{X > n}, the probability that the
first n trials all fail, is (1 − p) n .
7

The verbal argument can be the following: assume that none of the first n experiments succeeded.
This is precisely the condition on the left hand-side. On the left we see the conditional probability,
under this condition, that the first experiment to succeed, counted from the n + 1st trial, will be
the kth one. By independence, the condition does not matter, and the conditional probability will
agree with the unconditional probability of seeing the first success at the kth trial if we just restart
the experiments. This is the probability we see on the right hand-side.
Equation (1) expresses the memoryless property of the geometric distribution: the condition that
time for success has not come yet will not give any information about the number of further
experiments needed in order to see the first success. This random number is the same as if the
experiments would have been restarted.
20. Mr. Bilk will be fined independently with probability p = 0.2 · 0.95 = 0.19. In the five days the
number of fines will be binomial with the above parameter p and n = 5. Hence
(a) P{X = 0} = 5 0 5 · 0.19 · 0.81 ≃ 0.349.
0
(b) P{X = 2} = 5 2 3 · 0.19 · 0.81 ≃ 0.192.
2
(c) Let E be the event that on each of the five days the ticket controller was on the tram, and F
the event that Mr. Bilk had a lucky week. Then P{EF } = P{F | E} · P{E} is the probability
that on each day tickets were inspected, but Mr. Bilk escaped fines all five times. P{F } has
already been computed in part (a):
P{E | F } =
P{F | E} · P{E}
P{F }
≃ 0.055 · 0.2 5
0.349 ≃ 2.87 · 10−10 .
(d) On the first three days Mr. Bilk was not fined, and on the fourth day he finally got fined.
The probability of this is (1 − p) 3 · p = 0.81 3 · 0.19 ≃ 0.101.
21. We assume that inhabitants will commit suicide independently, with the same chance in each
period of the year.
(a) In the city of 400000 habitants we expect 4 suicides per month, hence the number X of
suicides is Poisson distributed with parameter λ = 4.
p : = P{X ≥ 8} = 1 −
7
P{X = j} = 1 −
j=0
7
j=0
4 j
j! · e−4 ≃ 0.0511.
(b) The number Y of such months of the year is binomial with the above p and n = 12 parameters.
P{Y ≥ 2} = 1 − P{Y = 0} − P{Y = 1}
12
= 1 − · 0.0511
0
0 · (1 − 0.0511) 12 −
12
· 0.0511
1
1 · (1 − 0.0511) 11 ≃ 0.123.
(c) The number Z of the first such month is a geometric random variable with the above parameter
p.
P{Z = i} = (1 − p) i−1 · p = (1 − 0.0511) i−1 · 0.0511.
8