# BU7527 Example sheet — 1 Mike Peardon — School of Mathematics, TCD

```BU7527
Example sheet — 1
Mike Peardon — mjp@maths.tcd.ie
School of Mathematics, TCD
Monday, 28th September
Try to answer these questions before tomorrow’s lectures. We will go through solutions
in class.
1. Two dice are rolled. What is the probability the biggest number showing on either of
the dice was more than four given the smallest number was more than two?
Deﬁne two events, A ≡ “biggest number &gt; 4” and B ≡ “smallest number &gt; 2”. Now
the theorem of conditional probability tells us
P (A|B) =
12/36
3
P (A ∩ B)
=
=
P (B)
16/36
4
2. A group of friends, three men and six women are going out and call a taxi. A random
selection of four friends get into the taxi. Find the probability the taxi contains
a. Two men and two women.
b. Four women.
c. At least two men.
The possible combinations are (M, W ) ∈ {(3, 1), (2, 2), (1, 3), (0, 4)} and their probabilities are
1
= 21
(3, 1): P (3, 1) = 1 C4 &times; 39 &middot; 82 &middot; 17
(2, 2):
P (2, 2) = 2 C4 &times; ( 93 &middot; 82 ) &times; ( 67 &times; 56 )
=
(1, 3):
P (1, 3) = 1 C4 &times; ( 93 ) &times; ( 67 &times; 56 &times; 54 )
=
(0, 4):
P (0, 4) = 0 C4 &times; ( 96 &times; 85 &times; 47 &times; 36 )
=
5
14
10
21
5
42
So the three probabilities are:
a. Two men and two women has probability 5/6
b. Four women has probability 5/42
c. At least two men has probability 1/21 + 5/14 = 17/42
1
ı9em In a game, a player takes turns to roll a die and keeps a count of N , the number
of turns taken. If the number showing is less than or equal to N , the game stops. Find
P (N = n) at the end of the game for all possible values of n.
n P (N = n)
1
2
3
4
5
6
1
6
5
6
5
6
5
6
5
6
5
6
=
&middot;
2
6
4
6
4
6
4
6
4
6
&middot; &middot;
=
3
6
3
6
3
6
3
6
&middot; &middot; &middot;
=
4
6
2
6
2
6
=
&middot; &middot; &middot; &middot;
&middot; &middot; &middot; &middot;
5
6
1
6
=
=
1
6
5
18
5
18
5
27
50
648
10
648
= 16.7%
= 27.8%
= 27.8%
= 18.5%
= 7.7%
= 1.5%
3. Two fair dice are rolled. If we deﬁne two events A and B such that A is said to have
occurred if the ﬁrst roll is a six and B is said to have occured if the sum of the two
rolls is seven then are A and B independent? What if the sum of the rolls is ten?
Deﬁne events A ≡“ﬁrst roll is 6” and B ≡“sum of rolls is 7” then we see P (A) =
1/6, P (B) = 1/6 and P (A ∩ B) = 1/36 so since P (A) &times; P (B) = 1/36 = P (A ∩ B)
the events are independent.
4. Bayes theorem: The probability your friend ﬁnds a mathematics lecture interesting
is 95% and mathematics lectures make up 10% of their modules this semester. The
probability they ﬁnd a lecture for any other course interesting is 5%. You overhead
your friend saying they have just enjoyed an interesting lecture. What is the probability
they have just attended a mathematics lecture?
Bayes theorem tells us
P (A)
P (B|A).
P (B)
Deﬁne A as the event “friend goes to maths lecture” and B is the event “friend goes
to an interesting lecture”. So we have P (A) = 0.1. Then we also have P (B|A) = 0.95
and P (B|Ac ) = 0.05. We can compute the probability of attending an interesting
lecture as
P (A|B) =
P (B) = P (B|A)P (A) + P (B|Ac )P (Ac )
= 0.95 &times; 0.1 + 0.05 &times; 0.9
= 0.14
Finally, using Bayes theorem gives
P (A|B) =
P (A)
0.1
P (B|A) =
&times; 0.95 = 67.9%...
P (B)
0.14
2
5. Binomial experiment: To improve proﬁts, the airline BinomialAir sells more tickets
for each ﬂight than there are seats on its planes. Each plane has 20 seats and the
airline always sells 24 tickets. On average, 19 passengers actually check in. What
is the probability that on a particular ﬂight, more passengers check in than can be
accommodated on the plane?
Need to consider the probability either 24,23,22 or 21 passengers turn up. This is a
. So
binomial experiment, with p = 19
24
24!
24! 0!
24!
P (23) =
23! 1!
24!
P (22) =
22! 2!
24!
P (21) =
21! 3!
P (24) =
p24 q 0 = 0.37%
p23 q 1 = 2.32%
p22 q 2 = 7.02%
p21 q 3 = 13.5%
The sum P (24) + P (23) + P (22) + P (21) is then 23.3%.
3
```