1. Units: Counting
Geometric Sequences
Logs & Exponential
Probability
Birthday Girl! *^(@@)^*
1

2. Zoe wakes up. She sleepily steps out of bed. As part
of her daily morning routin. She begins to put water
in a pot to boil for her tea. Then she goes to brush her
teeth. Brushing her teeth takes her two minutes.
If the starting temperature of the water is 15 0C and
the temperature of the water increases by 10% every
10 seconds, what is the temperature when she
finishes brushing her teeth? How long will take the
water to reach the boiling point (100 0C)?
2

3. 1st term: 15 0C starting temperature
2nd term: 16.5 0C after 10 seconds, increased by 10%
3rd term: 18.15 0C after 20 seconds, increased by 10%
. . . of 16.5 0C
If the temperature is increased by 10%, the new temperature
will be 110% of the original temperature. The common ratio
will be 1.10.
3

4. 47.08 0C
What is the temperature when she finishes brushing
her teeth?
We know the 1st term is 15 and the common ratio is
1.10. We are trying to find the temperature when she
finishes brushing her teeth which is the 13th term.
( 2 minutes = 120 seconds, 120sec / 10sec = 12 terms,
plus the 1st term is 13 terms)
4

5. 15, 16.5, 18.15 . . .
The formula for this is : tn = t1*r(n­1)
t1 = 15 t13 = 15*1.10(13­1)
r = 1.10 = 15*1.1012
= 47.08
The termperature when she finishes brushing
her teeth is about 47.08 0C.
5

6. approximately 3 minutes and 20 seconds
How long will take the water to reach the boiling
point (100 0C)?
The formula for this is : tn = t1*r(n­1)
100 = 15*1.10(n­1)
tn = 100 100 = 1.10(n­1)
t1 = 15 15
r = 1.10 log(1.10)100 = n­1
15 change of base law:
log 100 b logcb
15 loga =
= n­1 logca
log 1.10
~
19.9047 n­1
~
~
20.9047 n
~
n ~ 21
~
6

7. From this, we can conclude that the water will be boiling
just before the 21st term which is 20*10sec = 200 sec
( the 1st term is not counted because it starts at 0
seconds.)
Therefore water will be boiling just before 3 minutes and
20 seconds.
7

8. While she is sipping on her tea, she starts thinking
about her 18th birthday which is coming up soon. She is
planning to go out for dinner on her birthday. She calls
the restaurant and the waitress tells her that the table
can only fit in 15 people. She decides to go with her
parents and 12 of her friends.
8

9. 50388 ways
How many ways can she choose 12 people
from her 20 closest friends.
(a) Amy doesn't want to go,
When Amy doesn't want to go, the number of
people that can choose from is 19.
You have to choose 12 people from 19 people.
19!
19C12 = 7! 12! = 50388
There are 50388 ways to choose 12 people
from her 20 closest friends if Amy doesn't
go.
9

10. 43758 ways
How many ways can she choose 12 people
from her 20 closest friends.
(b) Bill and Kat insist on going,
Bill and Kat insist on going, so you only
have to choose 10 people from 18 friends.
18 !
18C10 = = 43758
8! 10!
There are 43758 ways to choose 10 other people to go.
10

11. 19488 ways
How many ways can she choose 12 people
from her 20 closest friends.
(c) Amy doesn't go and Bill and Kat go.
If Amy doesn't want to go, that means can only
choose from 19 people.
Bill and Kat want to go, that means you can only
choose 10 out of 17. ( Amy and Bill and Kat are
already taken 20 ­ 3 = 17 )
17!
17C10 = = 19448
7! 10!
There are 19448 ways to choose other 10 people
if Amy doesn't want to go and Bill and Kat insist
on going.
11

12. On the day of her birthday Zoe and her parents
and 12 friends celebrate at the restaurant. She
notices that there was another person there
celebrating their birthday.
Then she wonders what are the chances that
another person shares the same birthday as her.
(a) What is the probability that another person in
the restaurant has the same birthday as her.
(same birthday as Zoe)
(b) What is the probability that two people in the
restaurant have the same birthday.
(any two people share the same birthday)
( There are 35 people in the restaurant.)
12

13. 7
73
(a) What is the probability that another person
in the restaurant has the same birthday as her.
(not counting leap years)
The probability that another person has the same
birthday of 1/365 as Zoe, because there is only one
birthday in a year.
This probability is the same for every person in the
restaurant since you are looking for one specific
birth date.
There are 35 people in the restaurant, so ...
1 1 1 . . . 1 35 7
365 365 365 365 365 73
7
The probability is
73 .
13

14. 81.4%
(b) What is the probability that two people in the restaurant
have the same birthday. (not counting leap years)
The probability for 2 people to not have the same birthday is
364/365.
The probability for the 3rd person to not have the same birthday
as the first 2 people is 363/365.
So the probability for none of the 3 people to have the same
birthday is :
(364/365)*(363/365) = (364 * 363)/(365^2)
The probability for the 4th person to not have the same birthday is
362/365.
So the probability for none of the 4 people to have the same
birthday is :
(364/365)*(363/365)*(362/365) = (364*363*362)/(365^3)
14

15. So the general formula for the probability that n people will
not share the same birthday is :
(364/365)*(363/365)* . . . * ((366 ­ n)/365)
= (364*363* . . . * (366 ­ n)) / (365^n)
Since there is 35 people in the restaurant, the probability is :
(364*363* . . . *(366 ­ 35)) / (365^35)
= 0.1856
Therefore the probability 2 people will have the same birthday
is:
P(two people share birthday) = 1 ­ P(no two people share birthday).
1 ­ 0.1856 = 81.4%
15

16. # of favorable (365P35) total # of ways for 35 people
outcomes to have different birthdays
P = ( ie. no paired birthdays)
the total # of
possible outcomes
365^35 total # of ways for 35
2nd way:
people to have birthday
(365P35)
36535 (365P35) the probability for 35 people to
365! 365^35 not have the same birthday
=
(365 ­ 35)! 365^35
= 0.1856
P(two people share birthday) = 1 ­ P(no two people share birthday).
1 ­ 0.1856 = 81.4%
16

17. From the gifts she received for her birthday. Zoe finds
that she has gotten $500 in gift money. She wants to
deposit this money into the bank where it accumulates
interest at a certain rate.
There are three different plans for accumulating
interest.
a) plan A at the rate of 3.96% annually
b) plan B at the rate of 3.75% monthly
c) plan C at the rate of 3.74% continuously
Which plan is the best for her if she doesn't plan to
spend any of the money for three years.
17

18. r t*n
A = P ( 1 + n )
A the final amount
P principle / original amount
r the rate of which it will accumulate
n number of times per year
t amount of time that has been accumulating
18

19. $ 557.42
a) plan A at rate of 3.96% annually
A unknown (the final amount)
P $ 500
r 3.96% = 0.0396
n 1 r t*n
A = P ( 1 + n )
t 3 years
3 *1
A = 500 ( 1 + 0.0369 )
1
= 500 (1.0369) 3
= 557.42
19

20. $ 559.44
b) plan B at the rate of 3.75% monthly
A unknown (the final amount)
P $ 500
r 3.75% = 0.0375 r t*n
n 12 A = P ( 1 + n )
t 3 years
A = 500 (1 + 0.0375 3*12
)
12
36
= 500 (1.003125)
= 559.44
20

21. $ 559.37
c) plan C at the rate of 3.74% continuously
Continuously : A = Pe rt
A unknown (the final amount)
P $ 500 (0.0374*3)
r 3.74% = 0.0374 A = 500 e
t 3 years
0.1122
= 500 e
= 559.37
21

22. a) plan A at the rate of 3.96% annually
$557.42
b) plan B at the rate of 3.75% monthly
$559.44
c) plan C at the rate of 3.74% continuously
$559.37
The best plan for Zoe is plan B with the
rate of 3.75% monthly.
22

23. After deciding how she is going to deal with her money,
she finds that it's already late. She is happy with her day
and glad she is 18 now. Being tired, she gets ready for bed
and has a nice long sleep filled with sweet dreams.
The End . . .
23

24. Reflection
Why did you choose the concepts you did to create your problem set?
I never really understood probability that well so I decided to use that concept as part
of my problem set. I was looking forward to explore more on this unit to gain a better
understanding of probability. Hopefully I know more about these concepts more than
before.
How do these problems provide an overview of your best mathematical
understanding of what you have learned so far?
Solving these problems have given me more confidence in approaching problems of
these types. I have always had difficulty correctly answering these questions, but now
I feel that I can do them better.
Did you learn anything from this assignment? Was it educationally valuable to
you?
Yes, I believe that by doing this assignment I am able to think more thoroughly about
a problem than before. I think that I will think more logically when dealing with
problems and be able to fully understand and answer them correctly.
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