1MA01: Mathematical Methods Tutorial Sheet
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1. Determine if the following events are (a) independent and (b) disjoint.
(a) A coin is flipped twice. Let A be the event “the first flip is heads”
and B is the event “the second flip is tails”.
(a) independent and (b) not disjoint.
(b) A die is rolled. Let A be the event “the die is a four” and B is
the event “the die is a five”.
(a) one event so independence not defined and (b) disjoint.
2. Two tickets are drawn from the box containing 5 tickets labelled:
1, 2, 3, 4 and 5. The first ticket is not returned to the box before the
second is drawn. What is the probability of the events
(a) the first ticket is a 1?
(b) the first ticket is not a 2?
(c) the first ticket is a 2 or 3?
(a) P (1) =
1
5
= 0.2
(b) P (not 2) = 1 − P (2) = 1 −
(c) P (2 or 3) = P (2) + P (3) =
1
5
1
5
= 0.8
+
1
5
= 0.4
3. Consider rolling a (fair) die four times.
(a) what is the probability that all rolls are fives?
(b) what is the probability that the rolls are all the same?
(c) what is the probability that rolls are all different?
(d) what is the probability that a face (one side) appears more than
once?
(a) The probability that all rolls are fives is P (5 and 5 and 5 and 5) =
P (5)P (5)P (5)P (5) since the probability to roll five is independent
of each previous roll.
Then, P (rolling 5)P (rolling 5)P (rolling 5)P (rolling 5) = 61 16 16 16 =
1
1296 .
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Sinéad Ryan, ryan@maths.tcd.ie, see also http://www.maths.tcd.ie/˜ryan/123.html
(b) there are 6 ways (ie 6 different numbers) that can all be the same
so
6
P (all the same) =
1296
(c) P (all rolls are different) = 1 65 46 36 =
5
18
(d) P (face appears more than once) = P (not all different) = 1−P (all different) =
5
1 − 18
= 13
18
4. Each student in a large class is given an identical box containing three
tickets marked 1, 2 and 3. Students are asked to draw two tickets from
the box without replacing the first.
(a) what proportion of students draw a 3 first?
(b) of the students that drew 3 first, what proportion draw a 1 second?
(c) what proportion of students in the class draw 3 first followed by
1?
(a) P (3) = 1/3
(b) from the cohort that drew a 3, the probability of drawing 1 (without replacing the first card) P (1) = 1/2.
(c) P (3 and 1) = P (3)P (1|3) = 1/3(1/2) = 1/6
5. Consider a five-day school week. The probability that it is Friday and
that a student is absent is 0.03. What is the probability that a student
is absent given that today is Friday?
Considering a 5-day week, the probability today is Friday is 1/5 = 0.2.
The probability that it is Friday and a student is absent is 0.03.
Then P (Friday and Absent) = P (Friday)P (Absent|Friday) so
P (Absent|Friday) = P (Friday and Absent)/P (Friday) = 0.03/0.2 = 0.72 = 72.