MA22S6
Assignment 1 - probability
Mike Peardon (mjp@maths.tcd.ie)
January 27, 2012
Answer all four parts. Each part is worth five marks. Hand the homework in by the end of next
Friday’s tutorial (3rd February).
Question 1
A bag contains seven equal-sized balls, five red and two black. Balls are drawn until both black balls
have been taken from the bag. What is the probability five red balls are drawn consecutively, given the
first two balls drawn are red?
Define events A =“five red balls drawn in a row” and B =“two red balls drawn in a row The theorem
of conditional probability tells us
P (A ∩ B )
.
P (A|B ) =
P (B )
In this example, A ∩ B is the same as A (if five reds are drawn, then the first two must have been red)
so we just need to work out the probabilities of A and B . These are
P (B ) =
P (A) =
5 4 10
× =
7 6 21
5 4 3 21
1
× × ×
=
7 6 5 4 3 21
So
P (A|B ) =
1
10
Question 2
Three phone networks ABC , PQR and X Y Z carry all calls in a particular country. The fraction of
calls that fail and the fraction of calls carried by each network is given in the table below. Find the
probability a call will fail.
Carrier
ABC
PQR
XY Z
Failure rate
90%
5%
1%
1
Fraction of traffic
5%
80%
15%
Define the event F =“call fails” and the events E 1,2,3 =“call was with carrier ABC , PQR and X Y Z
respectively”, The first column of the table gives P (F |E 1 ), P (F |E 2 ) and P (F |E 3 ) while the last column
gives P (E 1 ), P (E 2 ) and P (E 3 . Now since E 1 , E 2 and E 3 are disjoint with S = E 1 ∪ E 2 ∪ E 3 , we can use the
additive property to write
P (F ) = P (F ∩ E 1 ) + P (F ∩ E 2 ) + P (F ∩ E 3 )
= P (F |E 1 )P (E 1 ) + P (F |E 2 )P (E 2 ) + P (F |E 3 )P (E 3 )
= 0.9 × 0.05 + 0.05 × 0.8 + 0.01 × 0.15
= 0.0865
Question 3
The probability one of your classmates (chosen at random) finds a 22S6 lecture interesting is 95%.
22S6 lectures make up 20% of their timetable this semester. The probability a classmate finds another
course interesting is 5%. You overhear your classmate saying they have just enjoyed an interesting
lecture. What is the probability they just attended a 22S6 lecture?
Define two events, A =“classmate attends interesting lecture” and B =“classmate attends 22S6
lecture”. The question then gives us P (A|B ) = 0.95, P (A|B c ) = 0.05 and P (B ) = 0.2 and asks us to find
P (B |A), the conditional probability that given a student just attended an interesting lecture, that it
was a 22S6 lecture. We can use Bayes’ theorem to find the solution:
P (B |A) =
P (B )
P (A|B )
P (A)
We’re missing P (A) from the right-hand side, but we can work this out using the theorem of conditional probability:
P (A) = P (A ∩ B ) + P (A ∩ B c )
= P (A|B )P (B ) + P (A|B c )P (B c )
= 0.95 × 0.2 + 0.05 × 0.8
= 0.23
Now substitute into Bayes’ theorem, giving;
P (B |A) =
0.2
P (B )
P (A|B ) =
× 0.95 = 0.826...
P (A)
0.23
Question 4
The probability I set an easy homework question and it is raining today is 13 . The probability I set an
1
easy homework question and it is not raining is 12
. My office has no view outside, so these events
(easy homework; raining outside) are independent. What is the probability my homework questions
are not easy when it is not raining?
2
Define two events, E and R corresponding to “my homework is easy” and “it is raining” respectively. The Venn diagram below helps us work out the probability of R ∩ E c , which then will let us
figure out R c ∩ E c . Define x to be x = P (R ∩ E c ), then this is the area of the region marked in the
diagram. Now if E and R are independent, this means
P (E ∩ R) = P (E ) × P (R)
1 1
1
1
= ( + ) × ( + x)
3
12 3
3
8
so x =
15
E
R
1
12
1
3
x
S
c
c
Now we need the area outside both circles (which is P (E ∩ R )) and we find the answer
P (E c ∩ R c ) =
3
1
20