Download CLABE Statistics Homework assignment

.
CLABE Statistics
Homework assignment - Problem sheet 3
.
1.
A
(a) A box contains 5 tickets
and 7 tickets
B . What is the probability that a randomly
selected set of 2 tickets (extracted simultaneously) will include 1
A
(b) A box contains 6 tickets
and 4 tickets
A
and 6 tickets
and 1
B?
B . What is the probability that a randomly
selected set of 3 tickets (extracted simultaneously) will include 1
(c) A box contains 10 tickets
A
A
and 2
B 's?
B . What is the probability that a randomly
selected set of 4 tickets (extracted simultaneously) will include 2
A's
and 2
B 's?
SOLUTION
(a) We rst number the tickets as
A1 , A2 , . . . A5
and
B1 , B2 , . . . B7
describe the sample space as a set of equally likely results.
so that it is possible to
In this problem it is more
convenient to consider unordered sequence so that, for instance, we don not distinguish
between
(A1 , A3 )
(A3 , A1 ). In this case the sample space is made up of 12
= 66
2
the event E = the selected sequence include one A and one B can be
and
unordered pairs and
written as
E = {(Ai , Bj ) | i = 1, . . . , 5
and therefore it contains
5×7
and
j = 1, . . . , 7}
equally likely pairs. Hence,
P (E) =
5×7
35
=
= 0.53.
66
66
Note that one can also consider equally likely ordered pairs so that, for instance, both
(A1 , A3 ) and (A3 , A1 ) belong to the sample space. In this case, the
12 × 11 equally likely pairs and the event E can be written as
sample space is made
up of
E = {(Ai , Bj ), (Bj , Ai ) | i = 1, . . . , 5
and therefore it contains
2×5×7
and
equally likely sequences. Hence,
35
2×5×7
=
= 0.53.
12 × 11
66
10
In this case, the sample space contains
= 120 equally likely sequences whereas the
3
4
event E = the selected sequence 1 A and 2 B 's is made up of 6 ×
2 = 36 equally likely
P (E) =
(b)
j = 1, . . . , 7}
sequences so that
P (E) =
36
3
=
= 0.3.
120
10
1
If one prefers to consider ordered sequences, then the sample space contains
equally likely sequences whereas the event
E
is made up of
3×6×4×3
10 × 9 × 8
equally likely
sequences so that
P (E) =
3×6×4×3
3
=
= 0.3.
10 × 9 × 8
10
16
(c) In the case of 4 extractions, the sample space contains
the event
4
equally likely sequences whereas
E = the selected sequence includes 2 A and 2 B 's is made up of
10
2
×
6
2 equally
likely sequences so that
10
2
P (E) =
×
16
6
2
= 0.371.
4
Also in this case it is possible to consider ordered sequences and to compute the probability
of
E
as
4
2
× 10 × 9 × 6 × 5
135
=
= 0.371.
16 × 15 × 14 × 13
364
P (E) =
2.
If a group of 6 people is in a room, what is the probability that at least two of them have the
same birthday?
Ignore leap years and assume that each day in the year is equally likely as a
birthday. Compute the same probability for a group of 50 people.
SOLUTION
E = at least two people have the same birthday ,
the same birthday . Since P (E) = 1 − P (Ē) one can compute
Let
E
be the event
P (Ē) =
so that
P (E) = 1 − 0.96 = 0.04.
and for
n = 50
it holds that
Ē = no two people have
365 × 364 × · · · × 360
= 0.96
3656
Similarly, for
P (Ē) =
then
n
people in a room one has
365 × 364 × · · · × (365 − n + 1)
365n
P (E) = 1 − 0.03 = 0.97.
3.
A box of 50 printer cartridges is used to stock the shelves. 10 of the cartridges are defective.
(a) A customer buys 5 cartridges from the 50.
What is the probability that all the ones he
bought are good?
(b) What is the probability that at least one of the cartridges bought by the rst customer is
defective?
(c) A second customer buys 5 more cartridges from the same shelf. What is the probability
that he got all good ones too, assuming the rst customer got all good ones?
2
SOLUTION
(a) Let
E1
E1 = the 5 cartridges of the rst customer are all good . There are 50
5
40
combinations of 5 cartridges and
5 combinations of 5 good cartridges. Hence
be the event
dierent
40
5
50
5
P (E1 ) =
=
40 × 39 × · · · × 36
= 0.31.
50 × 49 × · · · × 46
D is the event at least one of the cartridges of the rst
D = Ē1 so that P (D) = P (Ē1 ) = 1 − P (E1 ) = 1 − 0.31 = 0.69.
(b) It is sucient to notice that if
customer is defective then
(c) Assume that the rst customer bought 5 good cartridges and let
5 cartridges of the second customer are all good .
35
5
45
5
P (E2 ) =
=
E2
be the event
E2 = the
Then,
35 × 34 × · · · × 31
= 0.266.
45 × 44 × · · · × 41
4.
There are 40 dierent versions of an online homework.
students.
These are assigned randomly to the
If a group of 3 classmates decide to work on the homework together, what is the
probability that at least two of them receive the same version?
SOLUTION
From the probabilistic point of view, this is the same as the
compute the probability of the complementary event. If
same version birthday problem
and it is easier to
E = at least 2 of the 3 students have the
then
P (E) = 1 − P (Ē)
40 × 39 × 38
= 1−
403
= 0.074.
5.
Suppose you are dealt a hand of 4 cards at random from a deck of 32 cards consisting of 4 aces,
4 2s, 4 3s, and so on up to 4 8s.
(a) What is the probability that your hand will have at least 3 hearts?
(b) What is the probability that your hand will have at least 3 cards in the same suit?
SOLUTION
3
(a) The sample space consists of choosing 4 cards from 32, that is, there are
32
4
= 35960
dierent (equally likely) hands. Now, we have to count the number of ways of choosing at
least 3 heart cards. This happens in exactly two ways. We may have 3 hearts and one non
heart, or all four hearts. Since there are 8 heart cards, we can choose three in
forth card is one of the 24 non hearts, and therefore there are
8
4
hearts. Furthermore, there are
3
P (H) =
S
denote the event × 24
× 24 +
32
4
8
3
ways. The
hands with exactly 3
H
is the event
then
8
=
4
(b) Let
3
ways of choosing all four hearts. Hence if
your hand will have at least 3 hearts (♥) 8
8
56 × 24 + 70
= 0.0393.
35960
your hand will have at least 3 spades (♠) whereas let
D
and
C
denote the same event for the remaining suits, diamonds (♦) and clubs (♣), respectively.
H ∪ S ∪ D ∪ C . Since H S D and C are
P (H) = P (S) = P (D) = P (C) it follows that
We want to compute the probability of the event
mutually exclusive events and, furthermore,
P (H ∪ S ∪ D ∪ C) = P (H) + P (S) + P (D) + P (C) = 4 × 0.0393 = 0.157.
(1)
6.
Computer processors are shipped in lots of 70 from a factory.
Before being shipped, 25 are
randomly tested from each lot. If any of these 25 fail, the entire lot is not shipped. What is the
probability that a lot containing exactly 4 bad processors gets shipped?
SOLUTION
Let
E
be the event a lot containing exactly 4 bad processors gets shipped , then
66
25
70
25
P (E) =
66! 25! 45!
25! 41! 70!
45 × 44 × 43 × 42
=
70 × 69 × 68 × 67
= 0.162
=
7.
The diagram below shows a model railway track.
is
At each of the junctions
P, Q
and
R
the
2
3 and the probability of it branching to the right
probability of a train going straight ahead is
1
3 . The direction taken at every junction is independent of the direction taken at the previous
junctions. A train starts at point
A
b
A.
Pb
Q
b
b
R
b
b
B
b
C
b
b
4
D
(a) What is the probability that it reaches point
C?
(b) What is the probability that it reaches point
D?
SOLUTION
We denote by P s the event the train go straight ahead at junction P and by P r the event the
train branch to the right at junction P . The events Qs, Qr, Rs and Rr are dened similarly.
(a)
P (the
train reaches point
C) = P (P r ∩ Qs ∩ Rs)
= P (P r) × P (Qs) × P (P s)
1 2 2
=
× ×
3 3 3
4
=
27
(b)
P (the
train reaches point
D) = P [(P r ∩ Qr) ∪ (P r ∩ Qs ∩ Rr)]
= P (P r ∩ Qr) + P (P r ∩ Qs ∩ Rr)
1
2
=
+
9 27
5
=
27
8.
An insurance company estimated that 30% of all automobile accidents were partly caused by
weather conditions and that 20% of all automobile accidents involved bodily injury. Further, of
those accidents that involved bodily injury, 40% were partly caused by weather conditions.
(a) What is the probability that a randomly chosen accident both was partly caused by weather
conditions and involved bodily injury?
(b) Are the events
W = Partly caused by weather conditions and
B = Involved bodily injury independent?
(c) A randomly chosen accident was partly caused by weather conditions, what is the probability
that it involved bodily injury?
(d) What is the probability that a randomly chosen accident both was not partly caused by
weather conditions and did not involve bodily injury?
SOLUTION
The information provided in the text of the exercise can be formalized as follows
P (W ) = 0.3 P (B) = 0.20
(a)
P (W ∩ B) = P (W |B)P (B) = 0.08;
5
and
P (W |B) = 0.4
(b) the events
B
and
W
are not independent because
P (W ) 6= P (W |B);
(c)
P (B|W ) =
P (W |B)P (B)
= 0.267.
P (W )
(d) We want to compute the probability of the event
W̄ ∩ B̄
that is the complement of
W ∪ B.
Hence
P (W̄ ∩ B̄) = 1 − P (W ∪ B)
= 1 − [P (W ) + P (B) − P (W ∩ B)]
= 0.58
9.
It is known that a student who does his online homework on a regular basis has a chance of 83
percent to get a good grade (A or B); but the chance drops to 58 percent if he doesn't do the
homework regularly. John has been very busy with other courses and an evening job and gures
that he has only a 69 percent chance of doing the homework regularly. What is his chance of not
getting a good grade in the course?
SOLUTION
If
G
denotes the event homework regularly John will get a good grade then we know that
H denotes the event John will do his
P (G|H) = 0.83, P (G|H̄) = 0.58 and P (H) = 0.69.
and
Hence
P (G) = P (G|H)P (H) + P (G|H̄)P (H̄)
= 0.7525
and
P (Ḡ) = 1 − 0.7525 = 0.2475.
10.
1
3
10 if he goes out with his friends and 5 if he
3
does not go out with his friends. The probability that Marco goes out with his friends is . What
4
The probability that Marco does his homework is
is the probability that Marco does his homework?
SOLUTION
If
H
denotes the event with his friends Marco does his homework then we know that
and F denotes the event Marco goes out
P (H|F ) = 1/10, P (H|F̄ ) = 3/5 and P (F ) = 3/4. Hence
P (H) = P (H|F )P (F ) + P (H|F̄ )P (F̄ )
9
=
40
= 0.225.
6
11.
Imagine that, while in Mexico, you also took a side trip to Las Vegas, to pay homage to the TV
show CSI. Late one night in a bar you meet a guy who claims to know that in the casino at
the Tropicana there are two sorts of slot machines: one that pays out 10% of the time, and one
that pays out 20% of the time (note these numbers may not be very realistic). The two types
of machines are colored red and blue. The only problem is, the guy is so drunk he can't quite
remember which color corresponds to which kind of machine. Unfortunately, that night the guy
becomes the victim in the next CSI episode, so you are unable to ask him again when he is sober.
Next day you go to the Tropicana to nd out more.
You nd a red and a blue machine side
by side. You toss a coin to decide which machine to try rst; based on this you then put the
coin into the red machine. It does not pay out. How should you update your estimate of the
probability that this is the machine you are interested in? What if it had paid out - what would
be your new estimate then?
SOLUTION
Consider the following events:
R= the red machine is the good one B = the blue machine is the good one M = the red machine pays out then we know that
P (M |R) = 0.2, P (M |B) = 0.1
and, before we play, P(R)=0.5. The latter
probability is updated as follow:
P (M̄ |R)P (R)
P (M̄ )
0.8 × 0.5
=
0.85
= 0.47
P (R|M̄ ) =
where
P (M̄ ) = P (M̄ |R)P (R) + P (M̄ |B)P (B)
= 0.8 × 0.5 + 0.9 × 0.5
= 0.85.
If the red machine had paid out, then
P (R|M ) = 0.67.
7