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Procedure.
Questions.
Answers.
3.
Does a person’s birth year
matter? No.
How many days are in a year?
365. Is a person’s birthday
equally likely to be any day of the year?
Yes.
Does the date of one
person’s birthday affect the date of another person’s birthday?
No. If two people were both born
on May 12, is that a match? Yes.
If nobody else in the class
has the same birthday as me (August 2), does that mean that there are no
matches in the classroom? No.
6.
Does that mean that there
will always be a match in a group of people this large?
No. Is the probability of a match
high with this number of people? We
can’t tell. Does that mean that there
will never be a match in a group of people this large?
No. Is the probability of a match
low with this number of people? We
can’t tell.
9.
Why does the program use 50
trials? So that the
experimental probabilities will more accurately represent the theoretical
probabilities.
12.
Is the simulation’s answer
the best answer to the problem? No.
Can we be sure that this is
the answer? Why or why not?
No, because the simulation can only give experimental
probabilities, not theoretical probabilities.
13.
Can we calculate the
probability of at least one match? Yes,
but it would be difficult. What makes calculating this
difficult? We would have
to calculate the probability of one match, of two matches, of three
matches, etc. Is it easier to calculate a
different probability? Can
you suggest another probability we could calculate? Yes, the probability of no
matches.
What is the relationship
between these two probabilities? They
are complements.
14.
What is the probability that
the first two people do not share a birthday? 364/365.
15.
Assuming that the first two
do not share a birthday, what is the probability that the third person
does not share either of their birthdays?
363/365. What is the probability that
none of the three people share a birthday? (364 * 363)/(3652).
16. Assuming none of the first
three people share a birthday, what is the probability that the fourth
person does not share any of the first three birthdays?
362/365. What is the probability that
none of the four people share a birthday?
(364 * 363 * 362)/(3653).
17.
Can we generate a formula for
the kth person not sharing any of the first k-1
birthdays? [365 * 364 *
363 * … (366-k)]/[365k] or [364 * 363 *
… (366-k)]/[365k-1].
18.
What is the probability that
there is at least one birthday match in a group of k people?
1 – {[365 * 364 * 363 * … (366-k)]/[365k]}
or 1 – {[364 * 363 * … (366-k)]/[365k-1]}.
21A. Should these graphs be
continuous or discrete? Why?
Discrete, because an x-value represents a number of people and
has to be a nonnegative integer. Why is the first graph
increasing? Why is the second
one decreasing? As more
people are added to the group, it becomes more likely that there will be a
birthday match, and less likely that there will not be a birthday match.
21B. What will the resulting graph
look like if we added these two plots together? The horizontal line, y = 1.
21C.
Calculate the point at which
the first graph is increasing the fastest. k = 20. At which point is the second
graph decreasing the fastest? Also
at k = 20. How are the derivatives of
the two graphs related? They
are additive inverses.
22.
If there are two people in
the room, how many possible birthday matches are there?
1. What is the probability that
this match would not occur? 364/365.
23.
How many possible birthday
matches are there for three people? 3.
Are these possible matches
independent events? Yes.
What is the probability that
none of these matches would occur? (364/365)3.
24.
Repeat this for four and five
people. 4 people, 6
matches, (364/365)6. 5
people, 10 matches, (364/365)10.
25.
Can we write this probability
as a general equation? Let k be the number of people and n(k)
be the number of opportunities for birthday matches for k people.
Then n(1) = 0 and n(k+1) = n(k)
= k + 1. Then P(no
birthday matches occur among k people) = (364/365)n(k).
26.
Why are the probabilities
from this method different from those in Method 2?
What assumption is different?
In Method 2, the birthday matches are dependent, while in Method
3, the birthday matches are treated as independent events.
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to The Birthday
Problem Lesson Plan
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