Chapter 15 Worksheet
16 September 2011
MA203J Contemporary Mathematics
(1) We’re going to work with an experiment where we flip a coin 10 times in a row, and the observation
is what side the coin lands on “H” or “T.”
(a) Briefly describe the sample space.
The sample space consists of all strings of 10 H’s and T’s. For example, HHHTHTHHHT is in
the sample space, but HHHTH, and HT4HTT are not.
(b) How many different outcomes are there for this experiment?
For each coin flip, there are two possible outcomes. As there are 10 total coin flips, we use the
multiplication rule, and multiply 2 by itself 10 times:
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 210 = 1024
(c) Let E be the event described by “at least 9 heads are flipped.” Write out this event.
E=
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HHHHHHHHHH,
T HHHHHHHHH,
HT HHHHHHHH,
HHT HHHHHHH,
HHHT HHHHHH,
HHHHT HHHHH,
HHHHHT HHHH,
HHHHHHT HHH,
HHHHHHHT HH,
HHHHHHHHT H,
HHHHHHHHHT
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(d) Let F be the event described by “exactly 9 heads are flipped.” Write out this event as well.
F =
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T HHHHHHHHH,
HT HHHHHHHH,
HHT HHHHHHH,
HHHT HHHHHH,
HHHHT HHHHH,
HHHHHT HHHH,
HHHHHHT HHH,
HHHHHHHT HH,
HHHHHHHHT H,
HHHHHHHHHT
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(e) Describe (in your own words) why flipping a coin is an equiprobable space.
Flipping a coin is an equiprobable space because all possible outcomes are equally likely.
(f) Determine the probability of event F . Recall that (since flipping coins is an equiprobable space)
this is done by taking the number of outcomes in the event F divided by the total number of
possible outcomes.
Here, we need to take the number of outcomes listed in event F and divide that by the total
number of possible outcomes. In part (d), we listed all 10 outcomes in event F , and in part (b)
we computed the total number of possible outcomes (there were 1024), so the probability of event
F is:
10
.
P r(F ) =
1024
Chapter 15 Worksheet
16 September 2011
MA203J Contemporary Mathematics
(g) Determine the odds for event F . (This one is a bit trickier. I’d suggest looking at the definition
on page 573.)
The number of favorable outcomes is 10 and the total number of outcomes is 1024, so the number
of unfavorable outcomes is 1024 − 10 = 1014. Hence, the odds for event F are 10 : 1014.
(h) Also determine the odds against event F . (Again, this is trickier, and the definition is on page
573.)
We’ve already determined the number of favorable and unfavorable outcomes, so the odds against
are 1014 : 10.
(2) Time for a new experiment. I own 3 pairs of jeans, 4 pairs of brown pants, and 2 black skirts. I also
own 18 green shirts, 5 purple shirts, and 1 red shirt.
(a) Under the assumption that none of my clothes clash with each other, how many different outfits
can I wear?
I must choose an outfit by choosing a shirt, and a bottoms. There are 18 + 5 + 1 = 24 total shirts,
and 3 + 4 + 2 = 9 total bottoms. We apply the multiplication rule here. With 24 ways to choose
a shirt, and 9 ways to choose a bottom, there are 24 ∗ 9 = 216 total possible outfits.
(b) I am going on vacation for a week (7 days), and I want to bring only green shirts. How many
different ways can I decide on which 7 green shirts I will bring?
Since here we only care about the shirts that go into the suitcase, and not the order they go into
the suitcase, or the order in which I wear them, we should use combinations. There are 18 shirts
here, and we are choosing 7 of them. Hence, there are 18 C7 ways to choose the shirts that go into
the suitcase. We’ll compute that now:
18 C7
=
18!
(18 − 7)! × 7!
18 × 17 × 16 × 15 × 14 × 13 × 12 × 11!
11! × 7!
18 × 17 × 16 × 15 × 14 × 13 × 12
=
7!
160, 392, 960
=
5, 040
= 31, 824
=
(c) On my vacation, I’ve determined that I need 4 pairs of pants1 , and a skirt. How many ways can
I choose my pants? What about my skirt?
Here, there are 7 total pairs of pants, and we are again choosing 4 pairs, meaning that order does
not matter. Thus, there are
7 C4
=
7!
7×6×5×4×3×2×1
5, 040
=
=
= 35
(4 − 3)! × 4!
(3 × 2 × 1) × (4 × 3 × 2 × 1)
144
ways to choose the pants. For the skirts, there are only 2 choices, and we’re choosing 1, so we
can consider this as 2 choices (skirt a or skirt b), or we can say that there are 2 C1 = 2 choices.
(d) For my weeklong vacation2 , how many total ways are there for me to choose what to pack?
I know that there are
1
2
18 C7
ways for me to choose my shirt, 7 C4 ways to choose pants, and 2 ways
Here, I meant for pants to mean both the jeans and the brown pants, but forgot to say that
This is still the same vacation
Chapter 15 Worksheet
16 September 2011
MA203J Contemporary Mathematics
to choose the skirt. We now apply the multiplication rule, to get that there are
2, 227, 680 ways I can choose clothes to take on my trip.
18 C7
×7 C4 × 2 =
(e) On my vacation, I’ve now learned that I will need to wear my red shirt on Wednesday3 , and I
want to plan all of my outfits. How many ways can I determine which clothes I will wear each
day (not just which ones to pack)?
Let’s assume that my vacation starts on Monday and goes through Sunday (it will not actually
matter).
• On Monday, there are 18 shirt options and 9 bottoms options, giving 18 × 9 = 162 options
for Monday’s outfit.
• On Tuesday there are only 17 shirt options (I’ve already used one shirt on Monday), and 8
remaining bottoms options, for 17 × 8 = 136 outfit options for Tuesday.
• On Wednesday I must wear my red shirt, but I can wear any of my remaining 7 bottoms, so
there are 7 outfit options left for Wednesday.
• On Thursday I can wear any of my remaining 16 green shirts, and any of my remaining 6
bottoms, so I have 16 × 6 = 96 outfit options for Thursday.
• On Friday I can wear any of the remaining 15 green shirts, and any of the remaining 5
bottoms, so I have 15 × 5 = 75 outfit options for Friday.
• On Saturday, I can wear any of the remaining 14 green shirts and 4 bottoms, so I have
14 × 4 = 56 outfit options for Saturday.
• On Sunday, I can wear any of the remaining 13 green shirts and 3 bottoms, so I have 13×3 = 39
outfit options for Sunday.
Now, we can use the multiplication rule, to obtain that there are
162 × 136 × 7 × 96 × 75 × 56 × 39 = 2, 425, 141, 555, 200
total choices for outfits for the week.4
(f) My apartment this morning had no power, so I got dressed in the dark, but know that I choose
pants and a shirt to wear. Assuming I cannot tell any of my clothes apart in the dark, what is the
probability that I chose a pair of jeans, and a green shirt? (Hint: Are these events independent?)
There are 24 total shirts, and so under the assumption that I can’t tell my clothes apart in the
dark, all of the shirts are equally likely. Also, there are 7 total pairs of pants, all of which are
18
equally likely. Since there are 18 green shirts, the probability of choosing a green shirt is 24
and
3
since there are 3 pairs of jeans, the probability of choosing a pair of jeans is 7 . These two events
are independent, and so we can say that
P r(choosing jeans & a green shirt) = P r(choosing jeans) × P r(choosing a green shirt).
Thus, the probability of choosing jeans and a green shirt is
3 18
3 × 18
54
×
=
=
.
7 24
7 × 24
168
3
I have not yet actually packed for my trip, but will still be bringing as many green shirts as possible, and I will be wearing
different bottoms every day now.
4
My calculator gave me a different looking answer. I used a calculator on a computer since the numbers here were so big.
My TI−83 had 2.425141555E12.