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Procedure.
Questions.
Answers.
2.
If I reach in to the bag
without looking and choose a jellybean at random, what is the probability
that it is a grape one? ½,
because two of the four available jellybeans are grape.
What is the probability that
I chose a licorice jellybean? ½,
because two of the four available jellybeans are licorice.
3.
What is the probability that
I pull out the other grape one? 1/3,
because there are only three jellybeans left and only one is grape.
What is the probability that
I take a licorice one? 2/3,
because two of the three jellybeans left are licorice.
Why do the probabilities
change from what they were before? Taking
the first jellybean from the bag has changed the problem situation.
Instead of four jellybeans (two of each flavor), there are now only
three jellybeans (two licorice and one grape).
The probabilities of the flavor of the second jellybean are
dependent on the flavor of the first jellybean.
4.
5.
What does the first node of
the diagram represent? The
first jellybean I take from the bag.
How many branches does it
have? 2, one for each
flavor.
How do we label them?
Let G1 = the event that the first jellybean taken is grape and
L1 = the event that the first jellybean taken is licorice.
Label the branches G1 and L1.
What probabilities do we
assign to each branch? P(G1)
= ½ and P(L1) = ½. Note
that these are the only possible outcomes and that the sum of their
probabilities equals 1.
6.
What does the node at the end
of this branch represent? The
second jellybean I take (assuming the first one is grape).
How many branches does it
have? 2, one for each
flavor.
How do we label them?
Let G2 = the event that the second jellybean taken is grape and
L2 = the event that the second jellybean taken is licorice.
Label the branches G2 and L2.
What probabilities do we
assign to each branch? P(G2)
= 1/3 and P(L2) = 2/3. Note that these are the only possible outcomes and
that the sum of their probabilities equals 1.
7.
Repeat these steps for the
other branch of the first node (the one that assumes that the first
jellybean I take is licorice). The
answers are the same as the ones in Procedure 6 except that P(G2) = 2/3
and P(L2) = 1/3.
8.
Label them.
Let G1L2 = the first jellybean I take is grape and the second is
licorice. Therefore, we can
label the final outcomes G1G2, G1L2, L1G2, and L1L2.
How can you find the
probability associated with each of these branches? The probabilities for the possible outcomes of the second
jellybean (G2 and L2) were affected by the outcome of the first jellybean
(either G1 or L1). Since the
flavor of the second jellybean is dependent on the flavor of the first, we
need the following formula: P(AB) = P(A|B) * P(B).
Example:
P(G1G2) = P(G2|G1) * P(G1) = 1/3 * ½ = 1/6.
P(G1L2)
= P(L2|G1) * P(G1) = 2/3 * ½ = 1/3.
P(L1G2)
= P(G2|L1) * P(L1) = 2/3 * ½ = 1/3.
P(L1L2)
= P(L2|L1) * P(L1) = 1/3 * ½ = 1/6.
Note
that the sum of the probabilities of these final outcomes equals 1.
9.
What is the probability that
the jellybean in my pocket is also grape?
1/3. There are
three jellybeans whose specific flavors are unknown, but we know that only
one of them is grape.
What is the probability that
it is a licorice jellybean? 2/3.
Of the three jellybeans whose specific flavors are unknown, two are
licorice.
10.
Why isn’t it equally likely
that the jellybean in my pocket is either flavor when there was two of
each flavor in the bag when I took it?
A common misconception among students is that the probabilities
in this situation are still ½ for each flavor of jellybean because there
were equal numbers of each flavor in bag when the jellybean was taken. They also assume that there is no way that knowing the flavor
of the second jellybean can influence the probability of the first
jellybean’s flavor. They
think an event cannot retroactively affect another event that has already
happened. Explain to any of
your students who think this that the event of choosing the second
jellybean does not actually affect the selection of the first jellybean,
but that our knowledge of the second jellybean’s flavor affects the
probabilities of which jellybean we selected first.
Learning the flavor of the second jellybean gives us additional
information that helps us determine what the flavor of the first jellybean
could be. Help your students
see that this situation is really the same as the one above where we knew
the flavor of the first jellybean, but not the flavor of the second.
In each case, we know that one of the jellybeans is grape and that
there are three other jellybeans whose flavors we do not know.
We do know that only one can be grape and the other two are
licorice, therefore it is more likely that the other jellybean is
licorice.
11.
What probabilities do we want
to know? P(G1|G2) or
P(L1|G2).
What else do we need to use
the formula? P(G1G2),
P(L1G2), and P(G2).
Which of these do we already
know? P(G1G2) = 1/6 and
P(L1G2) = 1/3.
How can we find P(G2) from
the tree diagram? P(G2) =
P(G1G2) + P(L1G2) = 1/6 + 1/3 = ½.
Now, use the formula.
P(G1|G2)
= P(G1G2) / P(G2) = (1/6) / (1/2) = 1/3;
P(L1|G2)
= P(L1G2) / P(G2) = (1/3) / (1/2) = 2/3.
12.
Are there any branches that
are not necessary in this situation (Situation 2)?
Yes, G1L2 and L1L2.
Why?
These branches involve a licorice jellybean being chosen second,
but we know that the second jellybean has to be grape.
What branches are still
possible? G1G2 and L1G2.
These are the branches that involve a grape jellybean being chosen
second.
Is there a special
relationship between these two events?
They’re complementary. The
first jellybean must be either grape (G1G2) or licorice (L1G2), but it
cannot be both.
What is the relationship
between the probabilities of these events?
P(L1G2) = 2 * P(G1G2). L1G2
is twice as likely to occur as G1G2.
What probabilities can we
assign to two complementary events where one event is twice as likely to
occur as the other event? 2/3
and 1/3, in this case, P(L1|G2) = 2/3 and P(G1|G2) = 1/3.
14.
What probabilities are we
looking for? P(G1|G2) or P(L1|G2).
What probabilities do we know
from the Tree Diagram? P(L1),
P(G1), P(G2|L1), P(G2|G1).
Use Bayes Theorem to find the
two probabilities we’re looking for.
P(G1|G2) = P(G1G2) / P(G2) = [P(G2|G1) * P(G1)] / [P(G2|G1) * P(G1) +
P(G2|L1) * P(L1)] =
[1/3 * ½] / [1/3 * ½ + 2/3 * ½] = [1/6] / [½] =
1/3
P(L1|G2) = P(L1G2) / P(G2) = [P(G2|L1) * P(L1)] / [P(G2|G1) * P(G1) +
P(G2|L1) * P(L1)] =
[2/3 * ½] / [1/3 * ½ + 2/3 * ½] = [1/3] / [½] =
2/3
Are the answers the same as
we found from the other methods? Yes.
17.
What should appear first in
the tree diagram: whether or not a person has the disease or whether or
not a person tests positive for the disease?
Why? Whether or not
a person has the disease. A
person’s test result is dependent on whether or not they have the
disease, not the other way around.
How do we label the two
branches of the first node? Let
A = the person has the disease and ~A = the person does not have the
disease. Label the branches A
and ~A.
What probabilities go with
each branch? P(A) = .01
and P(~A) = .99.
How do we know the
probability that somebody in Atlantis does not have the Atlantis Death
Flu? It is the complement
of having the disease, so P(~A) = 1 - P(A).
18.
How do we label the two
branches of this node? Let
+ = the person has a positive test result and - = the person has a
negative test result. Label
the branches + and -.
What probabilities go with
each branch? P(+) = .95
and P(-) = .05.
How do we know the
probability that somebody with the Atlantis Death Flu tests negative?
It is the complement of that same person testing positive.
Testing positive and testing negative are complementary events.
Therefore, P(-) = 1 - P(+).
19.
Repeat these steps with the
node at the end of the branch that assumes that a person does not have the
Atlantis Death Flu. The
answers are the same as the ones in Procedure 18 except that P(+) = .06
and P(-)
= .94.
20.
What probability are we
looking for? P(A|+).
What probabilities do we know
from the Tree Diagram? P(A),
P(~A), P(+|A), P(+|~A). Use Bayes Theorem to answer
the question. P(A|+) = P(A+)
/ P(+) = [P(+|A) * P(A)] / [P(+|A) * P(A) + P(+|~A) * P(~A)] = [.95 * .01]
/ [.95 * .01 + .06 * .99] = .0095 / .0689 = .13788.
Back
to the Time-Axis Fallacy & Bayes Theorem
Lesson Plan
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