4 Chapter Homework key
[Over 72 points]
2011/09/20
403. [Chapter 4; 16 p.] (Definitions)
a) [2 p.] What is a statistical experiment?
A natural or artificial process which produces, each time it is invoked, one
outcome out of a small set of possible outcomes, with a specific probability
distribution.
b) [2 p.] What is the sample space for a statistical experiment?
The sample space is the set of possible outcomes of the experiment.
c) [2 p.] What do we mean by independent statistical experiments?
If we do the same experiment twice in a row, the probabilities of the
different outcomes on the second try are not influenced by whatever
happened on the first try. If we carry out two different experiments, the
outcomes of one are not affected by whatever outcome resulted from the
other.
d) [2 p.] Given an example of independent statistical experiments.
Rolling two dice, or rolling one die twice. Drawing two cards from a deck, or
two balls from an urn, with replacement (you replace the card or ball you
drew on the first try, and shuffle, before your second try.)
e) [2 p.] Given an example of non-independent statistical experiments.
The usual state lottery setup: you have a bunch of numbered balls in an urn,
you draw a sequence of balls without replacing the balls drawn. As you draw
each ball, the probabilities for the following draws change.
f) [2 p.] For two sets A and B, define the Union
AB
The union of two sets is defined as the (larger) set of items that belong to
either A or B (or both).
g) [2 p.] For two sets A and B, define the Intersection
AB
The intersection of two sets is defined as the (smaller) set of items that
belong to both A and B.
1
h) [2 p.] For set S and a subset A, define the Complement
Aor Ac
If A is a subset of S, then the complement of A with respect to S is defined
as the set of items that belong to S that are not in A.
414. [Ch. 4, 14 points] (Probability)
a) [2 p.] You roll a die. What is the probability of rolling a 3 ?
1/6
b) [2 p.] You roll a die. What is the probability of rolling a 2 or a 3 ?
1/6 + 1/6 = 2/6 = 1/3
c) [2 p.] You roll two dice (or the same die twice in a row). What is the
probability of rolling a 2 and a 3 ?
Ambiguous: either like (d) or like (f)
d) [2 p.] You roll two dice. What is the probability of rolling a 2 then a 3 ?
With two rolls of the die we have 6x6=36 possible sequences.
The sequence (2,3) is one of those, so its probability is 1/36
e) [3 p.] You roll two dice. What is the probability of rolling at least one
two or one three?
At least one two: (2,x) and (x,2) add up to 12
At least one three: (3,x) and (x,3) add up to 12
Then we deduct the ones that have been double-counted: (2,2),
(2,3), (3,2), (3,3) = 4
Answer: (12 + 12 4) / 36 = 20 / 36 = 5/9
This is an example of the addition law:
P( A B) P( A) P( B) P( A B)
[To satisfy yourself, write down all 36 sequences and count how
many satisfy our requirement.]
2
f) [3 p.] You roll two dice. What is the probability of rolling exactly one 2
and one 3 ?
This can be done two ways: (2,3) or (3,2).
Probability 2/36 = 1/18.
502. [Chapter 5; 12 p.] (Discrete Uniform probability)
We have an 8-sided fair die with sides labeled A through H.
a) [2 p.] What is the probability of rolling a G?
For a fair die, the probability of one of 8 outcomes in 1/8 = 0.125.
b) [2 p.] What is the probability of rolling a C?
Same answer = 1/8.
c) [2 p.] What is the probability of rolling any letter other than A?
Use the complement rule: if the probability of rolling an A is 1/8, then
the probability of rolling anything other than A must be 1-1/8 = 7/8 =
0.875.
d) [3 p.] You roll the die 5 times. What is the probability of rolling the
following specific sequence: A-B-C-D-E ?
Since the partial experiments (each roll of the die) are independent, we use
the multiplication rule: the probability of a sequence like this is equal to
the product of the individual probabilities. So:
Pr(ABCDE) = 1/8 x 1/8 x 1/8 x 1/8 x 1/8 = (1/8)5 = 1/32,768 = 0.00003
e) [3 p.] In calculating the answer to (d), what property of the experiment
rolling a die did you rely on?
The independence property.
3
4002. [Ch. 4, 14 points] (Combinations and permutations)
a) [2 points] You have 5 favorite CDs. How many different ways can you
play those CDs in sequence?
5! = 120
b) [2 points] You draw 5 cards from a (normal) deck of cards. What is the
probability that you will draw 5 hearts? (Any 5 hearts)
We have selection without replacement. For this problem, you
can consider each of the 5 draws in a sequence:
13/52 * 12/51 * 11/50 * 10/49 * 9/48 = could 0.0004952
We also considered this problem as a hypergeometric one
(see chapter 5): let the Hearts be the good light bulbs and the
rest be the bad ones; we have 13 good ones and 39 bad ones,
and we want to know the probability of selecting 5 good ones:
13
Pr(5,0) C 5
C 0
C
39
52
5
13 12 11 10 9
0.0004952
52 51 50 49 48
c) [2 points] You draw 5 cards from a normal deck. What is the
probability that you will draw 5 specific hearts (e.g. 3,4,5,6,7) in any
order?
Consider each draw in sequence:
5/52 * 4/51 * 3/50 * 2/49 * 1/48 = 1/2,598,960
(On the first draw, we have 52 cards in the deck and either of
the 5 hearts is acceptable. If we draw one of them, then on the
next draw we have 51 cards in the deck, and only 4 acceptable
hearts. And so on.)
Or, using the formula for combinations: the number of 5-card
combinations (groups) that can be selected from a deck of 52
cards is
4
C
52
5
52!
2,598,960
5!47!
and the probability of hitting one of them is then
P = 1 / 2,598,960.
d) [2 points] You draw 5 cards from a normal deck. What is the
probability that you will draw the 5 specific hearts in a specific order?
The number of sequences we can generate by drawing 5 cards
without replacement from a deck of 52 cards is:
P
52
5
311,875,200
5 specific cards in a specific order is one of those possible
sequences. (The fact that theyre hearts is irrelevant right here.)
So the answer is:
P = 1/311,875,200
e) [2 points] A new state lottery requires you to pick 7 letters or digits
correctly to win (order does not matter). How many possible
combinations are there? What is the probability of picking a winner?
Assume we are doing selection without replacement from a bin
that contains 36 different items (10 digits and 26 letters.) Since
order does not matter, the number of possible combinations
(groups) is:
C
36
7
8,347,680
The probability of picking the correct one is then 1/8,347,680.
f) [2 points] A car race has 7 cars competing. You need to pick which
contestants will win, place, and show (#1, #2, #3). How many tickets
would you need to purchase to win?
This is selection without replacement, and since order matters
we are asking about the number of sequences (permutations).
We have:
5
P
7
3
210.
And that is the number of different tickets youd have to
purchase to be sure that you had the right one.
g) [2 points] As in question (f) but you only have to pick which 3 cars
come in first.
Order does not matter, so we are asking about combinations
(groups):
C
7
3
35
4003. [Ch. 4, 16 points] (Combinations and permutations)
Setting: given a car license plate that has room for 7 symbols. How many
different tags can we make if:
a) [2 p.] Each position may contain a digit {0..9} only, and a digit may
appear more than once.
Unless the problem says otherwise, assume we are asking about
the number of permutations (sequences), since on a license
plate the order matters.
A digit may appear more than once means we have selection
with replacement. Then the correct formula is:
#permutations = 107 = 10,000,000
b) [2 p.] As in (a), except either of the first 6 positions may also contain
a space.
In this case, consider each independent position individually:
11 x 11 x 11 x 11 x 11 x 11 x 10 = 116 x 10 = 17,715,610
c) [2 p.] Each position may contain a digit only, and no digit may appear
more than once.
No digit may appear more than once translates into selection
without replacement. We get:
6
P
10
7
604,800
d) [2 p.] Each position may contain a letter {A..Z} or a digit {0..9}, and
any letter or digit may appear more than once.
Now the set of possible symbols expands to 10+26=36. We have
selection with replacement:
367 = 78,364,164,096
e) [2 p.] Each position may contain a letter or a digit, but no letter or
digit may appear more than once.
We have selection without replacement:
P
36
7
42,072,307,200
f) [2 p.] The pattern on a tag is DDLLDDD (e.g. 43GT505), and any letter
or digit may appear more than once.
Again, consider each position individually:
10 x 10 x 26 x 26 x 10 x 10 x 10 = 105 x 262 = 67,600,000
g) [2 p.] As above, but no letter or digit may appear more than once.
Now we have selection without replacement. We need to
consider the letters and the digits separately and calculate the
respective number of sequences; since these are to occur
simultaneously, we multiply the two answers together:
P
10
5
P 2 30,240 650 19,656,000
26
h) [2 p.] Given a specific 7-place tag, how many different tags can we
generate that use the same 7 different symbols?
7! = 5,040
7