# BD1, the binomial distribution robot

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BD1, the binomial distribution robot

The robot BD1 was activated on the corner of a neighborhood
with streets that looked like a grid of squares. The robot moved
without turning, but by using two set of wheels—one set for
North/South and one for East/West. Unfortunately, both sets
were damaged, so it could only go North and East. If the robot’s
batteries only had enough energy to move one block, it had two
ﬁnal destinations, (1, 0) and (0, 1). Each destination has one
path: to get to (1, 0), move East 1 block. To get to (0, 1), move
North 1 block.
1.    Suppose BD1 had enough energy to move two blocks.
(a) What ﬁnal destinations could it have?
(b) For each destination, describe all the possible paths it
could take to arrive there. For example, one path it              You could use a shorthand to
could take is North 1 block, then East 1 block.                   write the paths. For example,
North 1 block then East 1 block
could be NE. East two blocks
2.    What if BD1 had enough energy to move three blocks?                  could be EE.
Find all the paths for that case.
3.    One more time: Find all the paths BD1 could take if it
had enough energy to move exactly four blocks.
4.    Look just at the paths the robot could take if it could move
four blocks.
(a) How many of the paths allowed BD1 to move North
exactly 0 times? Where did it arrive each time?
(b) How many allowed BD1 to move North exactly 1 time?
Where did it arrive each time?
(c) Complete columns two and three in the following table.

Number of           Number of paths Destinations
times North
0
1
2
3
4

(d)    Suppose that, before traveling a block, BD1 chose which
direction to move completely at random. That is, it

Problems with a Point: February 4, 2002                        c EDC 2002
BD1, the binomial distribution robot: Problem                    2

was just as likely to go North as it was to go East. Find
the probability that BD1 arrived at each destination,
and add a “Probability” column to the table.
(e) Make a probability distribution graph showing the prob-
ability that BD1 moved a certain number of time North.

5.    Suppose you ﬂip a fair coin four times and count the num-      A “fair” coin is the ideal coin—a
ber of heads. Explain why this would have the same prob-       ﬂip has exactly the same chance
of being heads as being tails.
ability distribution graph as how many times BD1 travels
North on a four block trip.
6.    Make a probability distribution graph for ﬂipping a coin
ﬁve times and counting the number of heads.

7.    Compare the graphs you made, and list at least two fea-        You may want to make more
tures that these graphs have in common (but other graphs       graphs, using your answers to the
ﬁrst few problems or imagining
might not have). Explain why the situations make the           ﬂipping a coin more than ﬁve
graphs have that feature.                                      times.

Whenever you have two choices to make multiple times, the
probability distribution is called a binomial distribution. So far
you’ve only looked at cases where the two choices were equally
likely. If the two choices are not equally likely, you still have
a binomial distribution—but how will the distribution graph
change?

Problems with a Point: February 4, 2002                  c EDC 2002
BD1, the binomial distribution robot: Problem                             3

2
Suppose that BD1 was programmed to prefer North travel, so                3
of the time it chose North, and 1 of the time it chose East.
3
8.    First think about a three-block trip. One of the possible
destinations for the trip is (1, 2).
(a) List again the possible paths to (1, 2).
(b) What is the probability that BD1 takes the ﬁrst of the
paths you listed? What are the probabilities for each
of the other paths?
(c) What is the probability that BD1 gets to (1, 2)?
(d) Complete the following table to show the number of
times moving North and associated probabilities.
Number of           Number of paths Probability
times North
0
1
2
3

(e)    Make a probability distribution graph for this case.

9.    Now do the same for a four-block trip.
(a) First complete the table.
Number of           Number of paths Probability
times North
0
1
2
3
4

Problems with a Point: February 4, 2002                           c EDC 2002
BD1, the binomial distribution robot: Problem                       4

(b)    Next, make a probability distribution graph.

10.    Compare these binomial distributions to the binomial dis-
tributions when the two choices were equally likely. How
are they diﬀerent? How are they the same?
11.    Challenge: In problems 8e and 9b, you created graphs for
the distribution of times North, for three- and four-block
trips. Suppose you made similar graphs for the number of
times East.
(a) Compare the East graphs to the North graphs. (How
would they be the same? How would they be diﬀerent?
(b) Why does this happen?
(c) Relate these graphs to the corresponding graphs when
the choices are equally likely. Do you observe the ex-
act same thing, or does something diﬀerent occur? Ex-
plain.

Problems with a Point: February 4, 2002                     c EDC 2002
BD1, the binomial distribution robot: Hints                   1

Hints
Hint to problems 1–3. You can use the paths for some num-
ber of blocks to ﬁnd ﬁnd the paths for one more (for example,
use the paths for two blocks to ﬁnd the paths for three). For
each of the ﬁnal destinations, what if BD1 could go just one
more block?
Hint to problem 4d. Each path (but not destination!) is
equally likely for the robot to take.
Hint to problem 5. Suppose you were making these walks in-
stead of the robot. How could you choose completely at random
whether to go North or East?
Hint to problem 6. Make a chart of the paths and probabili-
ties, ﬁrst.
Hint to problem 7. Focus on the shape of the graph. Con-
sider: How is getting 3 heads and 2 tails similar to getting 2
heads and 3 tails? If you toss a coin 100 times, about how
many times would you expect to get heads? How can a path
change to go from going North 0 times to going North 1 time?
Hint to problem 8b. Here’s the beginnings of an example:
For the path NEN, the robot ﬁrst must choose North—which
happens 2 of the time. Then, the robot must choose East, which
3
happens 1 of those 2 of all times. So the robot will choose NE
3          3
for the ﬁrst two directions 2 of the time. . . .
9

Hint to problem 11. Create the East graphs if necessary. Is
there a relationship between the number of times East and the
number of times North in any given path?

Problems with a Point: February 4, 2002               c EDC 2002
BD1, the binomial distribution robot: Answers                     1

1. (a) Possible ﬁnal destinations are (0, 2), (1, 1), and (2, 0).
(b) (0, 2): NN                                                     Paths are all described using the
(1, 1): NE and EN                                              shorthand mentioned in the
margin of the problem. For
(2, 0): EE                                                     example, EN means East one
block then North one block. NN
2.    (0, 3):    NNN                                                  means North two blocks.
(1, 2):    ENN, NEN, NNE
(2, 1):    EEN, ENE, NEE
(3, 0):    EEE
3.    (0, 4):    NNNN
(1, 3):    ENNN, NENN, NNEN, NNNE
(2, 2):    EENN, ENEN, ENNE, NEEN, NENE, NNEE
(3, 1):    EEEN, EENE, ENEE, NEEE
(4, 0):    EEEE
4. (a) Only one path, ending at (4, 0).
(b) Four path, all ending at (3, 1).
(c) Here is the completed table, including the last column
(part d).
Number of Number of paths Destinations Probability
times North
1
0                1              (4, 0)         16
4
1                4              (3, 1)         16
6
2                6              (2, 2)         16
4
3                4              (1, 3)         16
1
4                1              (0, 4)         16
(d) See the table above. Equivalent fractions or decimals
can also be used.
(e)

5.    There are two outcomes, heads or tails; BD1 had two di-
rections, North or East. Before deciding whether to go
North or East, it’s as if BD1 ﬂipped a coin: heads means

Problems with a Point: February 4, 2002                   c EDC 2002
BD1, the binomial distribution robot: Answers                          2

go North, tails means go East. To go North twice, BD1
would have to ﬂip heads twice. The results are identical.
6.

7.    Lists may vary. Here are some important features:                    Teacher’s Note: Consider a
student successful if he or she can
present one of these, and
Symmetry. Since each situation has two equally likely                adequately explain it. You may
choices, getting one choice a certain number of times will           want to have a class discussion
around these features. However,
give the same probability as getting the other choice that           unless you’re prepared to let
name number of times. So, for example, getting 2 heads on            students explore the binomial
theorem or Pascal’s triangle, you
n coin tosses has the same probability as getting 2 tails—           may want to keep the pattern
but that means you got n − 2 heads. So x and n − x will              discussion lighter than what’s
given here.
always have the same probability.

Increasing then decreasing. Since each choice is equally
likely, for n trials (coin tosses, for example), it’s most likely
that half will be one choice and half will be the other. So
the middle—for example, 3 of 6, 3 and 4 of 7, 4 of 8, and
so on, will have the highest probability. The further you
get from the middle, the lower the probability.

Non-linear change. The pattern—even just looking at
the increasing side—is not linear. The pattern is not easy
to see in the situation. However, consider BP1 moving
can only have 1 possible path. To go to exactly 1 time
North, BP1 has n places where it could do that. So there             You might think there are
are now n paths with 1 time North. To go to 2 times                  n(n − 1) paths, but each of these
is duplicated—if you ﬁrst change
North, each of those n paths have n − 1 places that can be           the second direction from East to
changed from East to North.                                          North, then change the third one,
you get the same thing as if you
8. (a) ENN, NEN, NNE                                                       changed the third direction and
4                          then the second one. See more
(b) The probability for each path is        27
.                        about the binomial theorem and
(c) 12
27
Pascal’s triangle.

Problems with a Point: February 4, 2002                        c EDC 2002
BD1, the binomial distribution robot: Answers                      3

(d)    Here is the table for the three-block trip:
Number of Number of paths Probability
times North
1
0                  1        27
6
1                  3        27
12
2                  3        27
8
3                  1        27
(e)

9. (a)      Here is the table for the four-block trip:
Number of Number of paths Probability
times North
1
0                  1        81
8
1                  4        81
24
2                  6        81
32
3                  4        81
16
4                  1        81
(b)

10.    The symmetry is gone—when the choices are not equally
likely, the distribution is skewed (in this case, toward more
times North). The graphs still increase then decrease, and

Problems with a Point: February 4, 2002                    c EDC 2002
BD1, the binomial distribution robot: Answers                       4

the pattern is still non-linear. (In fact, the pattern is even
more complicated than before.)
11. (a)  The graphs have the same shape, but reversed—a re-
ﬂection about the exact center (the center bar if there’s
an odd number of bars, or between the center two if
there’s an even number.)
(b) For a trip of n blocks, moving x times North means
moving n − x times East. For example, on a four-
block trip, if BD1 moves North 3 times, it must move
East once. So the probability of moving 0 times North
is the same as the probability of moving n times East;
1 time North is the same as n−1 times East, and so on.
So the bars will have the same heights, but in reversed
order.
(c) For equally likely choices, the graphs will be identical.
This may seem diﬀerent, but it’s actually the same—
because for equally likely choices, the graphs are sym-
metric. (See the answer to problem 7.) The graph
actually does get reversed, but you can’t tell because
the left and right halves are reﬂections of each other.

Problems with a Point: February 4, 2002                     c EDC 2002

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