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Notes for MTH 451, No. 2
Repeated independent events
Imagine a coin which has been weighted so that the probability of getting a head
is p, where 0 < p < 1, and thus the probability of a tail is q = 1 − p. Suppose
the coin is tossed repeatedly and let H1 : heads of toss no. 1, H2 : heads on toss
no. 2, etc. We assume that the successive coin tosses are independent. This
implies that, for example, that the probability of getting 4 heads in a row must
be
P (H1 H2 H3 H4 ) = P (H1 )P (H2 )P (H3 )P (H4 ) = p4 .
(Remember that H1 H2 H3 H4 = H1 ∩ H2 ∩ H3 ∩ H4 .)
Likewise, the probability of getting exactly one tail in 4 tosses would be 4p3 q
since the event A : exactly one tail in 4 tosses is the union of 4 mutually exclusive
events,
A = H1c H2 H3 H4 ∪ H1 H2c H3 H4 ∪ H1 H2 H3c H4 ∪ H1 H2 H3 H4c .
and each of these events has probability p3 q.
Suppose we toss the coin until a tail appears. First we want to consider the
probability that a tail eventually does appear. Let Ai be the event that a tail
first appears on toss number i. The event that a tail eventually appears in then
the event
E = A1 ∪ A2 ∪ A3 · · ·
where, of course, the Ai ’s are mutually exclusive. Recall
Axiom 4. For a sequence of pairwise mutually exclusive events,
P (A1 ∪ A2 ∪ . . .) = P (A1 ) + P (A2 ) + . . . .
If there are infinitely many events the right side of this equation is an infinite
series, and we can write the equation more concisely as
P (∪∞
i=1 Ai ) =
∞
X
P (Ai ).
i=1
P∞
P∞
Now P (Ai ) = pi−1 q, so P (E) = i=1 P (Ai ) = i=1 pi−1 q. You should recognize this as a geometric series and verify that its sum is indeed 1.
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Consider the event B: the first tail appears on an even numbered toss. Check
that P (B) = pq + p3 q + p5 q + · · · and find the value of P (B) by summing a
geometric series.
Fundamental Counting Principle
Suppose procedure has k steps. If the first step can be done in n1 ways, and
after that, the second can be done in n2 ways, and after that, the third can
be done in n3 ways, etc., then the entire procedure can be done in n1 n2 · · · nk
ways.
Permutations
Consider a set with n elements. An ordered choice of k elements from this set
is called a permutation of length k. The number of such permutations is given
by n(n − 1)(n − 2) · · · (n − k + 1) = n!/(n − k)!.
In particular, the number of permutations of length n, i.e. of all the elements
is n!
Permutations with repetitions
Consider a set with n = n1 + n2 + · · · nk elements where n1 elements are of the
same kind and indistinguishable, n2 are of another kind and indistinguishable,
etc. The number of distinguishable ways to arrange these n elements is
n!
.
n1 !n2 ! · · · nk !
(Note: In many applications some of the ni ’s are equal to 1.)
Example: How many distinct rearrangements of the word MISSISSIPPI are
11!
there? Answer: 1!4!4!2!
since there is a single M, 4 I’s, 4 S’s, and 2 P’s.
To see how the formula works in this case, think of the letters as being distinguishable by subscripts:
M I1 S1 S2 I2 S3 S4 I3 P1 P2 I4
There are 11! arrangements of these 11 distinguishable “letters,” but, for example, the P ’s could occur in the order P1 , P2 or P2 , P1 but these would be indistinguishable without the subscripts. Likewise, the I’s can occur as I1 I2 I3 I4 ,
I4 I3 I2 I1 . . . – a total of 4! ways indistinguishable when we remove the subscripts;
the same applies to the S’s. Therefore to get the number of indistinguishable
arrangements we divide 11! by 2!4!4!.
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Combinations
Consider a set with n elements. An unordered choice of k elements from this
set is called a combination of size k. This is exactly the same as a subset with
k elements. The number of such combinations is given by
n
n!
=
k
(n − k)!k!
.
In other words, nk is the number of k-elements subsets of a set with n elements.
The numbers nk are called binomial coefficients because of their appearence in:
The Binomial Theorem: (x + y)n =
Pn
k=1
n
k
xk y n−k .
The total number of subsets of a set with n elements, of any size including 0
and n, is 2n . To see this, imagine the n elements arranged in a row. Picking
a subset amounts to deciding, one element at a time, whether to include that
element. Since we have 2 choices for each element, the Fundamental Counting
Formula says that there are 2n possible ways of choosing a subset. (Note that
the empty set is the result of choosing not to include each element.) Since each
subset has either 0, 1, 2 . . . or n elements, we see that
n
n
n
n
+
+
+ ··· +
= 2n .
0
1
2
n
We can also verify this formula but setting x = y = 1 in the Binomial Theorem.
An important identity about the binomial coefficients is
n+1
n
n
=
+
.
k
k
k−1
Here’s a simple way to see this: Given a set, A, with n + 1 elements, remove
one element and call it x. The number of size k subsets of A that don’t contain
element x is nk since we would choose the size k subset from the remaining n
elements of
A. The number of size k subsets of A that do contain element x is
n
then k−1
since we would form such a subset of A by choosing k − 1 elements
from the remaining ones and then throwing element x into the subset.
Another simple, but very useful fact is
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n
n
=
.
k
n−k
This is easy to check from the formula for nk but is also obvious because the
number of k-subsets of an n-set must be the same as the number of (n − k)subsets since the number of subsets should equal the number of the complements
of these subsets.
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