# M31 - Probabilities

```Math 408, Actuarial Statistics I
A.J. Hildebrand
Combinatorial Probabilities
Key concepts
• Permutation: arrangement in some order.
• Ordered versus unordered samples: In ordered samples, the order of the elements
in the sample matters; e.g., digits in a phone number, or the letters in a word. In
unordered samples the order of the elements is irrelevant; e.g., elements in a subset, or
lottery numbers.
• Samples with replacement versus samples without replacement: In the first
case, repetition of the same element is allowed (e.g., numbers in a license plate); in the
second, repetition not allowed (as in a lottery drawing—once a number has been drawn,
it cannot be drawn again).
Formulas
• Number of permutations of n objects: n!
• Number of ordered samples of size r, with replacement, from n objects: nr
• Number of ordered samples of size r, without replacement, from n objects:
n(n − 1) &middot; &middot; &middot; (n − r + 1) =
n!
= n Pr .
(n − r)!
• Number of unordered samples of size r, without replacement, from a set of n objects
(= number of subsets of size r from a set of n elements) (combinations):
n
n!
n(n − 1) . . . (n − r + 1)
n Pr
=
=
=
.
r
r!
r!(n − r)!
r!
(See back of page for properties of these binomial coefficients.)
• Number of subsets of a set of n elements: 2n
1
Math 408, Actuarial Statistics I
A.J. Hildebrand
Binomial coefficients
n
n!
.
• Definition: For n = 1, 2, . . . and k = 0, 1, . . . , n,
=
k!(n − k)!
k
(Note that, by definition, 0! = 1.)
• Alternate notations: n Ck or C(n, k)
n
n(n − 1) . . . (n − k + 1)
• Alternate definition:
=
.
k
k!
(This version is convenient for hand-calculating binomial coefficients.)
n
n
• Symmetry property:
=
k
n−k
n
n
n
n
• Special cases:
=
= 1,
=
=n
0
n
1
n−1
n X
n k n−k
n
• Binomial Theorem: (x + y) =
x y
k
k=0
• Binomial Theorem, special case:
n X
n
k=0
k
pk (1 − p)n−k = 1
n
• Combinatorial Interpretations:
represents
k
1. the number of ways to select k objects out of n given objects (in the sense of
unordered samples without replacement);
2. the number of k-element subsets of an n-element set;
3. the number of n-letter HT sequences with exactly k H’s and n − k T’s.
• Binomial distribution: Given a positive integer n and a number p with 0 &lt;p &lt; 1, the
n k
binomial distribution b(n, p) is the distribution with density (p.m.f.) f (k) =
p (1 −
k
p)n−k , for k = 0, 1, . . . , n.
2
Probability Examples
●
●
A jar contains 30 red marbles, 12 yellow
marbles, 8 green marbles and 5 blue marbles
What is the probability that you draw and
replace marbles 3 times and you get NO red
marbles?
●
There are 55 marbles, 25 of which are not red
●
P(getting a color other than red) = P(25/55) ≈ .455
●
Probability of this happening 3 times in a row is
found by .455*.455*.455 ≈ .094
Example 2: At least 1 Red
●
●
A jar contains 30 red marbles, 12 yellow
marbles, 8 green marbles and 5 blue marbles
What is the probability that you draw and
replace marbles 3 times and you get at least 1
Red?
●
●
It's easier to calculate the probability of getting NO
red marbles, and subtract that from 1 (we use the
complement rule : P(AC) = 1 – P(C)
From previous example, it is 1 - .094 = .906
Example 3: The First Red
●
●
A jar contains 30 red marbles, 12 yellow
marbles, 8 green marbles and 5 blue marbles
You draw and replace marbles 3 times. What is
the probability the third marble is the first red
marble?
●
●
This means the first two are not red. We calculated
P(drawing a non-red) = .455. Therefore,
P(red)=.545
P(non-red &amp; non-red &amp; Red) = P(non-red) * P(nonred) * P(red) = .455 * .455 * .545 = .113
Example 4: Red, Yellow and Blue
●
●
A jar contains 30 red marbles, 12 yellow
marbles, 8 green marbles and 5 blue marbles
You draw and replace marbles 3 times. What is
the probability you draw 1 Red, 1 Yellow, and 1
Blue?
●
●
●
This is harder, because we are drawing marbles in
an order, but we don't care about which order we
get Red, Yellow and Blue, just that there is 1 of
each.
But we can do it!
Example 4: Continued
●
●
●
●
Let RBY = “Draw a Red, then Blue, then Yellow”
So all disjoint events we want to consider are: RBY,
RYB, YRB, YBR, BYR, BRY – there are 6 of them.
P(RBY) = P(R)*P(B)*P(Y) = (30/55)*(5/55)*(12/55)
= .0108
But we have 6 disjoint cases. Because each one is
calculated as a product of the three, and each
disjoint case has the same probability (each order is
equally likely), our answer is 6*.0108 = .0649
```