MATH1715 Tutee Revision Questions

MATH1715 Tutee Revision Questions
Question: 10/1/2011
A blood test for hepatitis is 90% effective in detecting the disease; however, it yields a false
positive result for 1% of healthy people tested. The disease rate in the general population
is 1 in 10000. A patient is sent for a blood test because he has developed jaundice (that
is, yellowness of the skin and the white of the eye). The physician knows that this type of
patient will have hepatitis with probability 0.5. If this patient gets a positive result on his
blood test, what is the probability that he has hepatitis?
Response: This is a Bayes’s theorem question. You require P{Hepatitis|Positive}. You
are told that P{Positive|Hepatitis} = 0.90 and P{Positive|No hepatitis} = 0.10, and for a
member of the general public P{Hepatitis} = 0.0001 so P{No hepatitis} = 0.9999. However
for this patient, with jaundice, P{Hepatitis} = 0.5 so P{No hepatitis} = 0.5.
Question: 10/1/2011
Assuming that the birthday of a randomly chosen person is equally likely to occur in any
one of the 12 months of the year, what is the probability that at least two people in a
random sample of five will have their birthday in the same month?
Response: You have done this already! Recall Homeworks 1, question 1.6. Remember “a
bus starts with 6 people and stops at 10 different stops”? Here you have five people on
the bus and they can get off at any of 12 different stops. If you know the probability that
no-one gets off at the same stop, then the complementary probability is the probability that
at least two get off at the same stop.
Question: 10/1/2011
Five people play a game of “odd man out” to determine who will pay for the pizza they
ordered. Each flips a coin. If only one person gets heads (or tails) while the other four get
tails (or heads), then he is the odd man and has to pay. Otherwise they flip again. What
is the expected number of tosses needed to determine who will pay?
Response: This is quite a hard two-stage question. To see how to approach the question,
imagine the people flipping coins. You will have a series of “failures” (no “odd-man out” situation)
followed by a “success” (an “odd-man out” situation). For example, F F F F F S.
This sort of pattern should remind you of a geometric distribution, the time until an event
(a “success” S) first occurs. You then need to determine the probability of a “success”, an
“odd-man out” situation occurring.
Let P denote the probability of an odd-man out situation occurring when five people
toss a coin. Let X denote the number of heads out of five tosses so X ∼ Bin(n = 5, p = 1
2 ).
Clearly P = P{X = 1} + P{X = 4}.
Now let Y denote the number of times the players toss their coins until the “odd-man
out” situation occurs. Each group of tosses forms a set of Bernoulli trials with “success”
1

eing the “odd-man out” situation, all trials independent, and the probability of a “success”
being P. Thus Y ∼ geometric(P). You require E[Y ].
For “fun”, you might like to determine E[Y ] as a function of n. Would you play this
game if you went out with a football team (n = 11)? 1 Would you play this game if you were
with your MATH1715 tutees (n ≈ 20)? 2
Question: 10/1/2011
I’m sorry, I know I must be bombarding you with emails. I can’t work out whether this
question is telling me conditional probabilities or not. Very ambiguous.
On a multiple-choice exam with five choices for each question, a student either knows the
answer to a question or chooses an answer at random. If the probability that the student
knows a correct answer is 2/3, what is the probability that an answer that was marked as
correct was not chosen randomly? Thanks for all your help!
Response: First write down what you know.
P{Knows correct answer} = 2/3 so P{Does not knows correct answer} = 1/3.
P{Correct|Knows correct answer} = 1, P{Correct|Does not know correct answer} = 1/5.
We are asked to obtain P{Knows correct answer|Correct}. Clearly we need to use
Bayes’s theorem.
The way the question is put makes it difficult at first glance to determine what is asked.
But notice that the phrase “that was marked as correct” is what has happened, it is known
or given. What the question is asking is what is the probability the answer “was not chosen
randomly”, in other words what is the probability the correct answer was known by the
student, given that it was marked as correct.
Question: 10/1/2011
An elevator in a building starts with n people and stops at N floors, with n < N. If each
passenger is equally likely to get off at any floor, independently of the others, what is the
probability that at least two passengers get off at the same floor?
Response: You have done this already! Recall Homeworks 1, question 1.6. Remember “a
bus starts with 6 people and stops at 10 different stops”? At least two get off at the same
floor is the complementary event to no-one gets off at the same floor.
Question: 10/1/2011
Suppose that a random variable X has an exponential distribution with parameter λ = 2
and let Y = e −X . What is the set of possible values of the random variable Y , and how
do I find the cumulative distribution function FY (y) = P{Y ≤ y}. How would I obtain the
probability density function fY (y) and the mean E[Y ]? If Y1 is the first digit in a decimal
expansion of Y , what is the probability that Y1 = 6?
Response: I strongly doubt any such transformation question will come up!
1 Only if I wanted to waste 15 minutes, assuming each collective toss takes about ten seconds.
2 Only if we could keep awake for three consecutive days!
2

If X ∼ exponential(λ = 2), then recall that P{X ≤ x} = 1 − e −2x so P{X > x} = e −2x .
Then
FY (y) = P{Y ≤ y} = P � e −X ≤ y � = P{X ≥ − log y} = e 2log y = y 2 .
If x values satisfy x > 0, then 0 < e−x < 1 so the range of y is 0 < y < 1.
For the probability density function and mean, recall that
fY (y) = dFY
�
(y)
, E[Y ] = y fY (y)dy.
dy
By first digit of Y I assume is meant the first digit after the decimal point. Clearly
Y1 = 6 if 0.6 ≤ y < 0.7. Thus P{Y1 = 6} = P{0.6 < Y ≤ 0.7} = FY (0.7) − FY (0.6) 3 .
Question: 10/1/2011
Ten people sit down randomly in a row of 10 chairs. What is the probability that Sam will
be seated on a row end with Asif beside him?
Response: Draw a picture! How many ways can 10 people sit in 10 seats? Here n(S) = 10!,
where S denotes the sample space of possible seatings.
You now need to determine how many ways can Sam and Asif sit next to each other.
Ask Sam to sit down, but not in an end seat. Now have Asif sit next to him. How many
ways can Sam and Asif sit next to each other? Now sit the other eight people down. In how
many ways can the 10 people sit with Sam and Asif next to each other?
Now you need to add on the cases with Sam sat at an end seat. In how many ways can
Asif sit next to him? And now have the other eight people sit down. In how many ways
can they sit down?
(You have already included cases with Asif being in an end seat in the first part above.)
Question: 10/1/2011
Five students are randomly picked from a class of 20 students to form a singing group.
There are 8 girls and 12 boys in the class. What is the probability that the singing group
will contain exactly two girls and three boys?
Response: This is a hypergeometric probability.
In how many ways can you choose two girls from 8? In how many ways can you choose
three boys from 12? Multiplying gives the number of ways n(E) of choosing two girls from
8 and three boys from 12.
For n(S), you need to know the number of ways of choosing any five children from 20.
Then P{E} = n(E)/n(S).
Question: 7/1/2011
We toss a coin repeatedly until we see each side of the coin at least once. What is the mean
number of tosses required?
3 Do not worry that I write the cumulative distribution function as right-continuous. Since Y is a
continuous random variable P{Y ≤ y} ≡ P{Y < y}.
3
y

Response: Doodle! Draw a picture illustrating some examples! For example, HT, HHT,
HHHT, HHHHT, TH, TTH, TTTH, and so on. It looks like a series of Bernoulli trials
which stop when something different (a success) occurs. This should lead you to consider a
geometric distribution, the distribution of the time to the first success. The main difference
between this and the usual geometric distribution is that a “success” can be either H or T
and is determined by the first toss. You need an initial toss to determine what the “success”
is going to be!
Let X denote the number of tosses required to see both sides once. We require E[X].
We toss the coin once and see side A say. Let A now denote a Failure, and the other side
B a Success. The probability of a Success is p = 0.5. Tosses are independent. A toss either
gives side A, a failure, or side B, a success. What is the distribution of the number of
(subsequent) tosses Y until the first Success? (Here Y takes values 1, 2, 3, ....) Clearly
X = Y + 1, so E[X] can be found.
An alternative approach is to determine the distribution of X directly and then calculate
E[X] = �
x P{X = x} .
Thus
P{X = 1} = 0
P{X = 2} = P{HT or TH} = 0.25 + 0.25 = 0.5
x
P{X = 3} = P{HHT or TTH} = 0.125 + 0.125 = 0.25
P{X = 4} = 0.125
and so on. Writing Y = X − 1, then Y has the usual geometric distribution
P{Y = 1} = P{X = 2} = 0.5
P{Y = 2} = P{X = 3} = 0.5 2
P{Y = 3} = P{X = 4} = 0.5 3
and so on. Since E[Y ] = 1/0.5 = 2, then E[Y ] = E[X − 1] = E[X] − 1 gives E[X] = 3.
A simple R program would let you check my answer.
x=numeric(100000) # Initialise the counts of x to zero.
for (i in 1:100000){ # Do 100000 simulations.
u=rbinom(1,1,0.5) # Generate a random Bin(n=1,p=0.5) value.
# u=0 gives a Tail and u=1 gives a Head.
# Now generate tosses 2,3,4,5,.... until we get something not equal to u.
x[i]=1 # For simulation i initialise the count of the number of tosses (1 so far!).
v=-1 # v denotes subsequent tosses, set to -1 to start with.
# This is just to give R a value v for the next line.
while (v != u){ # While v does not equal u keep on tossing the coin.
v=rbinom(1,1,0.5) # Generate subsequent tosses, v=0 is a T and v=1 is a H say.
x[i]=x[i]+1
}
} # End the 100000 simulations.
mean(x) # Mean of x.
4

Three runs gave means 2.99884, 3.00008, 3.00283.
Question: 7/1/2011
Suppose that the waiting time in a queue has an exponential distribution with mean 20
minutes. What is the probability that standing in the queue will take no more that 20
minutes?
It keeps coming out with the answer 1 as I used (1 − e −20.20 ).
Response: Waiting time X ∼ exponential(µ = 20). We require
P{X ≤ 20} =
� 20
0
1
20 exp
�
− x
� � �
dx = − exp −
20
x
��20 = 1 − e
20 0
−1 .
The exponential distribution can be parameterised as having mean µ or parameter λ,
with µ = 1/λ.
Question: 6/1/2011
A contractor has 8 suppliers from which to purchase electrical supplies. He will select 3 of
these at random and ask each supplier to submit a project bid. If your firm is one of the 8
suppliers, what is the probability that you will get the opportunity to bid on the project?
Response: Use the hypergeometric distribution! He selects three firms from a pool of eight.
Seven are “other” firms and one is “your” firm. What is the probability the sample has two
“other” and one “your”?
Question: 6/1/2011
(I got the answer for this question by counting each possible outcome but I’m sure that
there is a more efficient method to find the answer) Four people are chosen at random from
six couples. What is the probability that two men and two women are selected?
Response: Use the hypergeometric distribution! You have six men and six women and are
choosing four people from 12. You want the probability of two men and two women.
Question: 6/1/2011
An integer is selected at random from the set 1, 2, ..., 1000000. What is the probability
that it contains the digit 5? I keep getting 0.1 which isn’t even an option? Thank you.
Response: Of n(S) = 1000000 values, how many contain a 5, event E? Then P{E} =
n(E)/n(S). The problem here is finding n(E)! Not really a Statistics question!
You are essentially asked how many of the integers contains a 5. Count them by listing
them! But you need to avoid counting the same ones more than once. Let x denote any
integer 0, 1, 2, 3, ..., 9. Let e denote any integer excluding 5.
5

Pattern Examples Number of values
xxxxxx5 05, 15, 25, 35, ..., 999995 100000
xxxxx5e 05e, 15e, 25e, 35e, ..., 99995e 10000 × 9 = 90000
xxxx5ee 05ee, 15ee, 25ee, ..., 9995ee 1000 × 9 × 9 = 81000
xxx5eee 05eee, 15eee, 25eee, 35eee, ..., 995eee 100 × 9 × 9 × 9 = 72900
· · · · · · · · ·
I get the answer n(E) = 468559 so P{E} = 0.469.
How can you “see” your way to the solution? Try a simpler problem! (Though I do
think the question quite hard!) Suppose the question was about the numbers 1,2, ...,1000.
You could almost write all these down. Write these as
000 001 ... 009
010 011 ... 019
... ... ... ...
090 991 ... 999
You can see the values making up event E are then:
• 005, 015, 025, ..., 995. There are 100 values here.
• 050, 051, 052, ..., 059, 150, 151, 152, ..., 159, 250, 251, 252, ..., 259, all the way to
950, 951, ..., 959. There are 10 × 1 × 9 = 90 values here.
• 500, 501, 502, ..., 504, 506, ..., 509, 510, 511, 512, ..., 599. There are 1 ×9×9 = 81
values here.
So the answer here is (100 + 90 + 81)/1000 = 0.1981.
Question: 6/1/2011
Let X1, X2 and X3 be independent random variables with zero mean and variance σ 2 . Find
the correlation coefficient between Y1 = 2X1 + X2 and Y2 = X2 − 2X3.
Response: You need Var[aX1 +bX2] = a 2 Var[X1]+b 2 Var[X2]+2abcov(X1, X2) where here
cov(X1, X2) = 0 as X1 and X2 are independent. Also,
cov(aX1+bX2, cXi+dXj) = ac cov(X1, Xi)+ad cov(X1, Xj)+bc cov(X2, Xi)+bd cov(X2, Xj).
But note that you will do all of this in MATH1725!
Question: 6/1/2011
A retailer offers a replacement policy for a new dishwasher that will pay the full product
cost of £300 if it breaks down within a year. The retailer’s statistics show that about 4%
of dishwashers of that brand will break down within a year. The administrative cost per
policy is £3 at the outset plus an extra £10 if a claim has been made. What should be
the policy premium (i.e., the amount the customer is charged) in order that the retailer’s
expected net profit per policy be £5?
6

Response: If the machine breaks down in a year the firm will pay out £(300+3+10) = £313
which occurs with probability 0.04. If the machine does not break down in a year, with
probability 0.96, the firm will pay £3. The firm needs to charge £((313×0.04)+(3×0.96)+
5) = £20.4.
Question: 6/1/2011
An event E occurs at least once in four independent trials with probability 0.76. What is
the probability of its occurrence in one trial?
Response: Let X be the number of times that E occurs in four trials. Clearly X ∼ Bin(n =
4, p). We are told that P{X ≥ 1} = 0.76 so that P{X = 0} = (1 − p) 4 = 0.24. Thus you
can obtain p.
Question: 3/1/2011
I am sorry to bother you but I just had a couple of probability questions that I just cannot
get my head around how to attempt and was hoping you could maybe push me in the right
direction.
A computer company provides an insurance policy for one of its systems. If the system
fails during the first year the policy pays £3000. The benefit decreases by £1000 each year
until it reaches zero. If the system has not failed at the beginning of a year, the probability
that it fails during that year is 0.1. How much should the company charge for the insurance
policy so that, on average, its net profit per policy is £100?
Response: The probability the system fails in year 1 is 0.1 and the firm pays out £3000.
The probability it fails in year 2 is 0.9 × 0.1 (it does not fail in year 1 but 4 fails in year
2) and the firm pays out £2000. Similarly for year 3 the system fails with probability
0.9 × 0.9 × 0.1 (assuming independence of failure in each year). For years 4 onwards we
either use the formula for a geometric distribution being greater than 3, or use the fact that
the failure probabilities must sum to unity. We thus have the following information for the
payout by the firm.
Year 1 2 3 ≥ 4
Payout Y 3000 2000 1000 0
Probability 0.1 0.09 0.081 0.729
The mean payout is E[Y ] = £561 (can you prove this?) so the firm should charge £661 to
make £100 per policy.
Question: 3/1/2011
An urn contains five chips, four marked Failure and one marked Success. Two players take
turns drawing a single chip from the urn, without replacement. The player who is first to
draw Success wins the game. What is the probability that the player starting the game will
be the winner? (This one seems very simple but I can’t seem to reach the right answer)
4 Think of “but” here as meaning “and yet”
7

Response: This is very similar to the question asked on 16/12/2010. The answer can be
obtained using a tree diagram or as indicated on 16/10/2010. Thus let Ar denote player
A’s draw on stage r, r = 1, 3, 5, and Bs denote player B’s draw on stage s, s = 2, 4.
P{A wins} = P{A1 = S} + P{(A1 = F) ∩ (B2 = F) ∩ (A3 = S)} + · · ·
where P{A1 = S} = 1
, and 5
and so on.
P{(A1 = F) ∩ (B2 = F) ∩ (A3 = S)} = 4 3 1
× ×
5 4 3 ,
Question: 3/1/2011
A stick of length L is broken into two parts at a random point chosen with uniform distribution.
What is the expected length of the shortest part?
Response: This is quite hard.
Let the stick be broken at a point x so the two halves have length x and L − x for
0 < x < L. Without loss of generality we can suppose that the shortest length is x for
0 < x < 1L.
(If this is the longer length, then use the other piece which has length
2
uniformly distributed between 0 and 1
L.) This shorter piece has a uniform distribution
2
between 0 and 1L
so the probability density function is f(x) = 2/L. (This is the key part
2
to understand!)
The mean length is
E[X] =
� 1
2 L
0
xf(x)dx =
� 1
2 L
0
2x
dx =
L
� x 2
L
�1
2 L
0
= 1
4 L.
Question: 3/1/2011
Could you help me with this questions please about correlation coefficients.
Let X be a random variable with mean 1 and variance 2. Let Y = −2X + 3. What is
the correlation coefficient between X and Y ?
Response: You can see that X and Y are linearly related. The negative slope −2 shows
that as X increases, then Y decreases. The correlation corr(X, Y ) = −1.
More formally
Thus
Var[Y ] = (−2) 2 Var[X] = 8, cov(X, Y ) = cov(X, −2X + 3) = −2Var[X] = −4.
corr(X, Y ) =
Question: 1/1/2011
Please help me with the following question.
cov(X, Y )
� Var[X]Var[Y ] = −4
√ 2 × 8 = −1.
8

Let X be a random variable with mean 2 and variance 3. Let Y be a random variable
with mean −5 and variance 1. Assuming that X and Y are independent, what is the variance
of Z = 4X − Y ?
Response:
Var[aX + bY + c] = a 2 Var[X] + b 2 Var[Y ] + 2ab cov(X, Y )
where a, b, and c are constants. Here you are told that X and Y are independent, so that
cov(X, Y ) = 0.
Question: 1/1/2011
Please help me answer the following question from the 2010 examination.
The probability Susan cycles to work given that it is raining is 0.7. The probability that
Susan cycles to work given that it is not raining is 0.95. Tomorrow there is a 60% chance
of rain. What is the probability that Susan cycles to work tomorrow?
Response: Let C denote the event she cycles to work and R the event it is raining. We are
given the following information: P{C|R} = 0.7, P{C|R c } = 0.95, and P{R} = 0.60. We
require the probability P{C}.
Using the total probability rule
P{C} = P{C ∩ R} + P{C ∩ R c } (1)
P{C} = P{C|R}P{R} + P{C|R c }P{R c } .
Use a Venn diagram to prove to yourself that equation (1) is true.
Notice that if the question had instead been “what is the probability it is raining if
Susan has cycled to work” and had not given P{R}, then we would use Bayes’s theorem to
determine P{R|C}.
Question: 1/1/2011
A survey showed that 0.5% of 20000 homes have suffered from damp problems in the past
10 years. A second survey investigated whether 1000 other homes have suffered from damp
problems in the past 10 years. Use an appropriate approximation to estimate the probability
that more than two homes have suffered from damp problems in the second survey.
Response: The first survey provides an estimate for the probability a home has suffered
from damp over the last 10 years. This probability is p = 0.005.
Clearly homes either suffer from damp or they do not, the probability p a home suffers
from damp can be assumed a constant, and it can further be assumed that homes are
independent of each other. 5
For the second survey let X denote the number of homes suffering from damp over the
last 10 years. Clearly X ∼ Bin(n = 1000, p = 0.005). We require P{X > 2}. Since n is
large and p is small, then approximately X ≈ Poisson(µ = np = 5).
5 The latter two assumptions are perhaps not strictly true for all homes. Can you think why?
9

Question: 29/12/2010
I am not quite sure how to obtain the answer for the following question.
The demand for a certain weekly magazine at a news stand is a random variable with
probability mass function pX(x) given by
x 3 4 5 6
pX(x) 0.20 0.44 0.32 0.04
The magazine sells for £1.50, and the cost to the owner is 50 pence per copy. Unsold copies
cannot be returned to the publisher.
(i) What is the expected weekly demand for this magazine?
(ii) If the owner orders three copies of the magazine, his net profit will be £3.00. What will
be his expected profit if he orders four copies?
(iii) How many copies should be ordered per week in order to maximise the expected profit?
Response: Here X denotes the number of copies demanded per week.
(i) E[X] = �
xpX(x) = µ say.
x
(ii) Let N denote the number of copies he orders. To see how to tackle the question, look
at some specific cases.
If N = 3, then he pays 3 × 50p = £1.50. He sells them all if X ≥ 3, which occurs with
probability unity. He receives 3 × 1.50 = £4.50, so his net profit is £3 as specified.
If N = 4, then he pays 4 ×50p = £2.00. There are two cases to now consider. If X = 3,
with probability 0.2, he sells 3, making net profit P3. If X ≥ 4, with probability 0.8, he sells
all 4, making net profit P4. His net profit P then satisfies E[P] = (0.2 × P3) + (0.8 × P4).
(iii) Repeat the calculations for N = 5 and N = 6. For example, for N = 5 you can
determine what he has spent buying the copies. For each of the cases X = 3, X = 4, X ≥ 5
you can determine his net profit and the associated probability. Thus you can determine
the mean profit if N = 5. Similarly for N = 6.
You could also consider the cases N = 0, 1, 2 and N = 7, 8, 9, . . .. In the latter case it
is easy to see that his costs increase while his return is capped by the maximum demand
X = 6; thus his net profit must certainly be monotonic decreasing for all N > 6. For N = 0
his net profit is zero (a safe risk-free place to be!). For 0 ≤ N ≤ 3 it can be seen that his
profit must be monotonic increasing as he always sells at least three copies. The maximum
must therefore lie between N = 3 and N = 6.
Question: 26/12/2010
Is it a general rule that Var[aX + bY ] = a Var[X] + b 2 Var[Y ] + 2ab cov(X, Y )?
Response: Yes.
Question: 26/12/2010
We toss a coin repeatedly until we see each side of the coin at least once. What is the mean
number of tosses required?
Response: Let X denote the number of tosses until we see each side at least once. What
is the distribution of X? What is E[X]?
10

Hint: Recall kite flying and counting Sundays until a kite flies.
Question: 26/12/2010
An insurance company writes a policy to the effect that an amount of £1000 must be paid
if a certain event A occurs within a year. Probability of A to occur within a year is 0.02.
What should be the premium c, the amount in pounds the customer is charged when buying
the policy, in order that the companys expected profit per policy be 50% of c.
Response: Let company’s profit per policy be X.
If A occurs, with probability 0.02, the company receives c and pays out £1000. Net gain
is c − 1000.
If A does not occur, with probability 0.98, the company receives c. Net gain is c.
Thus E[X] = 0.02(c−1000)+0.98c. We require to find c such that E[X] = 0.5c. Answer
is c = £40.
Question: 26/12/2010
For a candidate to win an election he needs to win districts I, II and III. Probability of
winning I and III is 0.55, losing II but winning I is 0.34, and losing II and III but not I is
0.15. Find probability the candidate will win all 3 districts?
Response: You require P{I ∩ II ∩ III}. You are given P{I ∩ III} = 0.55, P{I ∩ II c } =
0.34, and P{I ∩ II c ∩ III c } = 0.15.
Draw a Venn diagram! It can be seen that P{I ∩ II c ∩ III} = 0.19. Thus P{I ∩ II ∩ III} =
0.36. Alternatively, notice that
and also
P{I ∩ III} = P{I ∩ II ∩ III} + P{I ∩ II c ∩ III}
P{I ∩ II c } = P{I ∩ II c ∩ III} + P{I ∩ II c ∩ III c } .
Question: 26/12/2010
Two hunters A and B shoot a duck, which is hit by exactly one bullet. If the hunter A hits
the target with probability 0.3 and the second with probability 0.6, what is the probability
that the second hunter killed the duck?
Response: Draw a Venn diagram!
We are told there was one hit. We require
P{B hit|one hit} =
P{B hit ∩ one hit}
.
P{one hit}
Can you show that P{two hits} = 0.18 (assume independence of hits)? Also, P{at least one hit} =
0.72 (union of events), so that P{one hit} = 0.72−0.18 = 0.54 (why?). Also, P{B hit ∩ one hit} =
0.6 − 0.18 = 0.42 (why?).
11

Question: 26/12/2010
The probability that a certain type of inoculation takes effect is 0.995. Using a Poisson
approximation or otherwise, compute the probability that at most 2 out of 400 people given
the inoculation will find that it has not taken effect.
Response: Let X be the number for whom the inoculation does not take effect. For n = 400
people, X ∼ Bin(n = 400, p = 0.005). What is the approximate distribution satisfied by
X? The question requires P{X ≤ 2}.
Question: 26/12/2010
Three factories manufacture 20%, 30% and 50% of computer chips a company sells. If
the fractions of defective chips are 0.4%, 0.3% and 0.2%, respectively, what fraction of the
defective chips come from the third factory?
Response: Let D denote defective, and A, B, C the source factory. Thus we are told, for
example, P{A} = 0.20 and P{D|A} = 0.004.
We require P{C|D}. Thus use Bayes’s theorem.
P{C|D} =
P{D|C} P{C}
P{D|A} P{A} + · · · + P{D|C}P{C} .
Question: 26/12/2010
A box contains 3 blue balls and 5 red balls. Ann and Bob take turns (with Ann going first)
to draw a ball from the box, without replacement, until a blue ball is drawn. What is the
probability that Ann will draw the blue ball first?
Response: Draw a tree diagram is one way. On the first stage Ann either draws blue (with
probability 3/8) or red (with probability 5/8). If Ann draws red on stage one, on the second
stage Bob either draws blue (with probability 3/7) or red (with probability 4/7). If Bob
draws red on stage two, on the third stage Ann either draws blue or red, and so on. You
require all routes which end with Ann drawing blue.
The question can be answered in other ways. For example, let Ar denote Ann’s draw on
stage r, r = 1, 3, 5, 7, and Bs denote Bob’s draw on stage s, s = 2, 4, 6, 8.
P{Ann wins} = P{A1 = blue} + P{(A1 = red) ∩ (B2 = red) ∩ (A3 = blue)} + · · ·
where P{A1 = blue} = 3,
and 8
and so on.
P{(A1 = red) ∩ (B2 = red) ∩ (A3 = blue)} = 5 4 3
× ×
8 7 6 ,
Question: 26/12/2010
In the Euro millions lottery, a player must choose 5 correct digits from {1, 2, . . ., 50} on the
main board and two “lucky stars” from the nine digits 1-9. The lottery draw yields a “5+2”
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winning combination, that is, five numbers on the main board and two “lucky stars” (the
order in which they appear does not matter). What is the probability that the player will
guess correctly two numbers (out of 5) on the main board and one “lucky star” (out of 2)?
Response: A finite population has N = 50 values made up of R = 5 successes and
N − R = 45 failures. You sample n = 5 values. What is the probability that your sample
contains x = 2 successes and 3 failures? Recall sampling without replacement and the
hypergeometric distribution!
Similarly for choosing the “lucky stars”. Then multiply because of independence of
choosing from the main board and the “lucky stars”.
Question: 21/12/2010
Let X be a random variable with mean 2 and variance 1. Let Y be a random variable with
mean 2 and variance 4. The variance of 4X −3Y is 16. Determine the correlation coefficient
between X and Y . How do I do this?
Response: You are told that Var[X] = 1 and Var[Y ] = 4. Also Var[4X − 3Y ] = 16. Here
Thus cov(X, Y ) = 3
2
Var[4X − 3Y ] = 16Var[X] + 9Var[Y ] − 24cov(X, Y )
= 16 + 36 − 24cov(X, Y ).
and this can be used to obtain corr(X, Y ).
Question: 21/12/2010
A variety of chocolate bars are stored in a box containing 10 bars. On average, 45% of
the chocolate bars will contain nuts. Using the normal approximation with a continuity
correction, evaluate the probability that at least two bars will contain nuts in a randomly
selected small box? How do I do this?
Response: Let X be the number of bars out of 10 which contain nuts. We require
P{X ≥ 2}. But what is the distribution of X?
(i) A bar either contains nuts or it does not.
(ii) The probability a random bar contains nuts is 0.45.
(iii) We can assume all the bars are independent of each other (if a box contains a random
collection of bars this is plausible).
Why is X ∼ Bin(n = 10, p = 0.45)? Assuming n is large and p ≈ 0.5, then a normal
approximation can be used. What normal distribution can be used as an approximation to
the distribution of X?
Hint: What is E[X] and Var[X]?
Question: 21/12/2010
When on a holiday in the Alps, the time X Sarah spends snowboarding each day (in hours)
has a probability function of the form:
�
2 c(x + 4x) if 0 ≤ x ≤ 9,
fX(x) =
0 otherwise.
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I calculate c = 1
. What is the expected amount of time Sarah spends snowboarding on
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a randomly picked day within her holiday? What is the probability that Sarah will spend
over 3 hours snowboarding on a randomly picked day within her holiday?
Response: The first question asks for E[X] which is found using
E[X] =
� 9
0
xfX(x)dx.
The second question is asking for P{X > 3} which is found using
P{X > 3} =
� 9
3
fX(x)dx.
Question: 16/12/2010
Urn 1 contains two white and four red balls, whereas urn 2 contains two white and one
red. A ball is randomly chosen from urn 1 and put into urn 2, and a ball is then randomly
selected from urn 2. What is the probability that the ball selected from urn 2 is white?
Response: Let A1 be the event of selecting the first ball and A2 the event of selecting the
second ball. Clearly P{A1 = w} = 2
6 and P{A1 = r} = 4.
This ball is now placed into urn
6
2.
If this first ball is white, then urn 2 now contains three white and one red ball and so
P{A2 = w|A1 = w} = 3
4 .
If this first ball is red, then urn 2 now contains three white and one red ball and so
P{A2 = w|A1 = r} = 2
4 .
We want P{A2 = w}. This does not require Bayes’s theorem as we are not conditioning
on anything. We use the total probability rule. Either the first ball is white or it is red, so
that
P{A2 = w} = P{A2 = w ∩ A1 = w} + P{A2 = w ∩ A1 = r}.
Thus
P{A2 = w} = P{A2 = w|A1 = w} P{A1 = w} + P{A2 = w|A1 = r}P{A1 = r}.
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