ECON250B-2003
Assignment 2
Solutions
1. The table below is an approximate tally of the 1912 Titanic disaster.
C
W
M
1st
2nd
3rd
1st
2nd
3rd
1st
2nd
3rd
Died
0
0
50
5
15
80
100
150
400
800
Survived
5
25
30
130
80
60
50
20
80
480
5
25
80
135
95
140
150
170
480
Calculate the following probabilities and indicate whether they are joint, marginal or
conditional probabilities. Calculate the following probabilities and indicate whether they
are joint, marginal or conditional probabilities. The probability that a randomly chosen
passenger:
(a) is a woman;
(b) is either a child or a woman given that s/he travelled 3rd class;
(c) travelled 3rd class and survived.
Are the events:
(d) “Survived” and “3rd / C” independent?
(e) “Child” and “Died” mutually exclusive? Are there any mutually exclusive events?
Sol’ns
(a) P(W) = 37/128
(b) P(C  W  3rd) = 11/35
(c) P(3rd  S) = 17/128
(d) P(S  3rd  C) = 3/8 = P(S) thus independent
(e) P(C  D) = 5/128  0 thus not mutually exclusive but, for example, D and C-1st are as
P(D  C-1st) = 0
2. A manufacturer purchases a particular component from three suppliers Hall, Schuller
and Crawford, respectively 30%, 20% and 50% of the total. The manufacturer, having
long-standing links with these suppliers, knows that 3% of the components coming from
Hall, 5% from Schuller and 4% from Crawford are defective. All items received, being
identical, are stored in the same storage area prior to assembly. What’s the probability
that a randomly chosen component from storage is:
(a) defective and manufactured by Crawford?
(b) defective?
(c) manufactured by Crawford if it was found to be defective?
Sol’ns
(a) P(D  H) = P(H) P(D  H) = .9 %
P(D  S) = P(S) P(D  S) = 1.0 %
P(D  C) = P(C) P(D  C) = 2.0 %
Therefore: P(D  C) = 2.0 %
(b) P(D) = P(D  H) + P(D  S) + P(D  C) = 3.9 %
(c) P(C  D) = P(D  C) / P(D) = .020 / .039 = .513
3. A student couple are taking a statistics course. There are only 4 students in the course.
The professor regularly awards, potentially, 2 bonus points every class. Any such point is
earned upon correctly answering a snap and nearly trivial question. Each student is
equally likely to be selected to answer and, once selected, s/he gets it correctly with a
probability 90%. What is the probability that:
(a) both points were earned by this couple?
(b) at least one of the two points was earned by someone else?
(c) the same person earned both points?
(d) the same person earned both points if the couple earned both points?
(e) In which of the above cases, your answer depended on your sampling assumption?
Sol’ns
Note that there are 16 simple events, each with probability 1/16.
(a) 4 (1/16) (.9)2 = .2025
(b) 12 (1/16) (.9)2 = .6075
(c) 4 (1/16)(.9)2 = .2025
(d) .5
(e) The person drawn for the first point is no longer eligible to be drawn for the second
(sampling without replacement) thus there are 12 simple events. In (a), the probability
is (2/12)(.9)2  13.5%; in (b), (10/12)(.9)2 = 67.5%; in (c) and (d), zero.
4. Four different mathematics books, six different physics books and two different
chemistry books are to be arranged on a shelf. How many different arrangements are
possible if:
(a) the books in each particular subject must all stand together?
(b) only the mathematics books must stand together?
Sol’ns
(a) Math books can be arranged amongst themselves in P44 = 4! Ways, physics books in
6! ways and chemistry books 2! ways. Similarly, the three groups can be arranged in
3! ways. Thus: 4!6!2!3! = 207,360
(b) If one considered the math books as one, there will be 9 books to be arranged: 9!
whereas the four math books can be arranged in 4! ways. Thus: 9!4! = 8,709,120
5. Out of 5 mathematicians and 7 physicists, a committee consisting of 2 mathematiciians
and 3 physicists is to be formed. In how many ways can this be done if:
(a) any mathematician and any physicist can be included?
(b) one particular physicist must be on the committee?
(c) Two particular mathematicians can not be on the committee?
Sol’ns
(a) There are C25 ways to choose two mathematicians out of five and, similarly, C37 ways
to choose three physicists out of seven. Thus: C25 C37 = 350
(b) C25 C26 = 150
(c) C23 C37 = 105
6. Show that, for two events A and B with P(A) > 0 and P(B) > 0:
(a) “Mutually exclusive” implies “Not independent” but “Not independent” does not
imply “Mutually exclusive”;
(b) Pxn is never smaller than Cxn.
Sol’ns
(a) Mutual exclusivity requires P(A  B) = 0. Now check
P(A  B) = P(A  B) / P(B) = 0 / P(B) = 0 whereas P(A) > 0. Thus
P(A  B) < P(A) hence not independent.
(b) Pxn / Cxn = [n! / (n – x)!] / [n! / x!(n – x)!] = x! thus Pxn > Cxn if x > 1.