PROBABILITY AND DECISION MAKING
45730
QUIZ 1 (OFFLINE) SOLUTIONS
Problem 1
(
Sample Spaces
)
.
Four friends toss an unbiased coin each to determine a winner. The per
son whose coin shows up different from all the other three wins. The outcome of any toss is independent
of the others.
(1.1) What is the sample space of this random experiment?
Solution.
There are 2
4
= 16 possible outcomes in tossing four coins.
/
(1.2) What is the probability that the four coins show up all heads or all tails?
Solution.
The probability that the four coins show up all heads is
1
16
, and there is the same
probability that they show up all tails.
Thus the probability that the four coins show up all
heads or all tails is
1
16
+
1
16
=
1
8
.
/
(1.3) What is the probability that a coin shows up different from all the other three?
Solution.
The only outcomes where we have one coin different from all other three are: HTTT,
THTT, TTHT, TTTH,THHH, HTHH, HHTH, HHHT. So the probability is
8
16
=
1
2
.
/
Problem 2
(
Sample Spaces and Set Theory
)
.
Two games are supposed to take place between
teams A and B. The probability that team A wins the first game is 0.3, and the probability that it wins
the second is 0.5. Suppose the probability that team B wins BOTH games is 0.5.
(2.1) What is the appropriate sample space to analyze this problem?
Solution.
An appropriate sample space would be to consider the outcomes of both games. So we
have
AA, AB, BA, BB
, where
AB
represents the event that team A wins the first game and B
the second. For other outcomes is similar. The events
AA, AB, BA, BB
are mutually exclusive
and collectively exhaustive.
/
(2.2) What is the probability of team A winning both games?
Solution.
From the data we have
P
(
BB
) = 0
.
5,
P
(
AA
∪
BA
) = 0
.
5 and
P
(
AA
∪
AB
) = 0
.
3.
Note that
P
(
BB
)+
P
(
AA
∪
BA
) = 1
.
0, and this implies that we must have
P
(
AB
) = 0
.
0. Since
the considered events are mutually exclusive. This implies that
P
(
AA
) = 0
.
3.
/
(2.3) Are the events that Team A wins the first game, and Team A wins the second game independent?
Solution.
Let
A
# be the event that Team A wins the first game and #
A
the event that Team
A wins the second game.
A
# and #
A
are not independent since the condition
P
(
AA
) =
P
(
A
#
∩
#
A
) =
P
(
A
#)
·
P
(#
A
) does not hold.
/
(2.4) What is the probability of B winning the first game?
Solution.
From 2.2 we have that
P
(
AA
) = 0
.
3. So we get that
P
(
BA
) = 0
.
5

0
.
3 = 0
.
2. The
probability of B winning the first game is
P
(
BA
) +
P
(
BB
) = 0
.
2 + 0
.
5 = 0
.
7.
/
(2.5) What is the probability of B winning the second game?
Solution.
We have that
P
(
AA
) = 0
.
3,
P
(
BA
) = 0
.
2,
P
(
AB
) = 0
.
0 and
P
(
BB
) = 0
.
5. So the
probability that B is winning the second game is 0
.
5.
/
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QUIZ 1 (OFFLINE) SOLUTIONS
Problem 3
(
Counting I
)
.
We draw at random two cards out of a standard deck of 52 cards without
replacement.
(3.1) What is the probability that the first card is red (the second can be red or black)?
Solution.
Since the order matters, the size of the sample space is the number of possibilities for
the first card (52) times the number of possibilities for the second (51, since the first card is
not replaced). Of this the number where the first card is red (and the second card is any of the
remaining) is 26 times 51. The required probability is
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 Spring '08
 ravi
 Conditional Probability, Probability, Decision Making, Probability theory

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