Download Chapter 8 Homework Solutions

Chapter 8 Homework Solutions
4.
0 – a If the probability is zero, there are no possible ways for it to occur
0.01 – c This probability isn’t actually zero, but it’s very close to zero, so that means the event is
very unlikely
0.3 – (blank)
0.6 – d Since this probability is greater than 0.5, this event will occur more than half the time
0.99 – (blank)
1 – b A probability of exactly 1 means the event must occur all the time
5.
The directions above the problem say that we have to indicate what would be a reasonable sample
space for each of the phenomenon described below.
a) S = {0, 1, 2, …, 10} counting the # of heads observed
b) S = {0%, 10%, 20%, 30%, …, 100%} or S = {0, .1, .2, .3, …, .9, 1} calculating percentage of 10
flips which are heads
c) S = {yes, no} recording whether or not at least 5 heads occurred
17.
I’m gonna set up a table similar to the one we’ve used in a couple examples. Note: the dice are
weird, so I have to use the face numbers they tell me. The table will show the sums of the two dice.
1
2
2
3
3
4
1
2
3
3
4
4
5
3
4
5
5
6
6
7
4
5
6
6
7
7
8
5
6
7
7
8
8
9
6
7
8
8
9
9
10
8
9
10 10 11
11
12
Next, I’m going to set up probabilities for all of the sums by counting the number of times they
occur in the table (note: |S| = 36 as with regular dice).
sum
probability
2
3
4
5
6
7
8
9
10
11
12
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
Actually (if you look at the table in your notes for standard dice), this is exactly the same probability
model resulting from standard dice. Cute.
18. Similar to the previous problem, I’m gonna set up a table. We have to be careful we understand
what we’re adding up here – there are two dice, each numbered 1 through 4, but when you roll the
dice, there isn’t a number facing up (as there is on a standard die), so if we “get a 3,” that means the
3 is facing down and we need to add up the two numbers on the bottom, then add 1 for some
reason. A little weird, I know, but it is Dungeons & Dragons. Here’s the table I get:
1
2
3
4
1
3
4
5
6
2
4
5
6
7
3
5
6
7
8
4
6
7
8
9
And here are the probabilities:
sum
probability
3
4
5
6
7
8
9
1/16 2/16 3/16 4/16 3/16 2/16 1/16
Based on this table, it’s easy to calculate P(7 or higher) = 3/16 + 2/16 + 1/16 = 6/16.
20. Al, Barb, Carl, Dan, and Earl, I mean, Abby, Boaz, Carmen, Dani, and Eduardo.
a) S = {AB, AC, AD, AE, BC, BD, BE, CD, CE, DE} |S| = 10
b) P(any one of those) = 1/10
c) P(Abby) = 4/10 just counted the number of pairs that include Abby
d) P(not (B or E)) = 3/10 just counted the number without B or E
or
P(not (B or E)) = 1 – P(B or E) = 1 – 7/10 counting the number with B or E, subtract from 1
21. a) There are 2 possibilities for the first flip, 2 for the second flip, 2 for the third flip, etc. So we’ll
have ten 2’s multiplied by each other, so |S| = 2^10
b) P(all H or all T) = P(all H) + P(all T) = 1/(2^10) + 1/(2^10) = 2/(2^10).
25. a) S = {aps, asp, psa, pas, spa, sap} |S| = 6
b) Only spa and sap look like words to me, but the computer says an asp is a small snake and a pas
is a move in ballet, so if you count those, there are 4 of the 6 that are actual English words.
c) So P(actual English word) = 4/6
32. Expected value (mean as the book calls it) is found by multiplying each outcome by its probability.
We can find the expected number of cars, then, like this:
E(X) = 0(.3) + 1(.4) + 2(.2) + 3(.1) = .4 + .4 + .3 = 1.1
Since Linda earns $400 per car she sells, she can expect to make 1.1 * 400 = $440.
34. Another expected value, this time based off of the 4-sided dice from number 18. I can still see the
table with the probabilities, that’s convenient.
E(X) = 3(1/16) + 4(2/16) + 5(3/16) + 6(4/16) +7(3/16) + 8(2/16) + 9(1/16) = 96/16 = 6.
35. This time, we’re looking at the probability tables from number 14. There are two separate tables
here – one for owned houses and one for rented – so we’ll have two separate expected values.
Owned
E(X) = 1(0.000) + 2(0.001) + 3(0.014) + 4(0.099) + 5(0.238) + 6(0.266) + 7(0.178)
+ 8(0.107) + 9(0.050) + 10(0.047)
= 6.248
That was fun, let’s do that again…
Rented
E(X) = 1(0.011) + 2(0.027) + 3(0.229) + … + 10(0.005) = 4.807
This shows that, on average, homes that are rented have fewer rooms than homes which are owned.