Download Lecture 5: Symmetric Probability Spaces 1. Symmetric Probability

Lecture 5: Symmetric Probability Spaces
1. Symmetric Probability Spaces: Suppose that S = {1, · · · , N } is finite and let F = 2S . If
the probability of every element is equal, then (S, F, P) is said to be a symmetric probability
space. In this case, we have
1 = P(S) =
N
X
P({i}) = N · P(i)
i=1
and so P({i}) = 1/N for each i ∈ S. Furthermore, if E ⊂ S, then
P(E) =
|E|
1
|E| =
.
N
|S|
2. Examples
Example 5a: (i) If two dice are rolled, what is the probability that the sum of the numbers is
equal to 7?
There are 6 possible outcomes that satisfy the condition that the sum be 7, out of 36 total
possible outcomes, so the probability is 6/36 = 1/6.
Example 5a (ii): If two dice are rolled, what is the probability that the first number is less
than the second number?
Since the first number is either less than, equal to, or greater than the second number and the
probability that the first number is less than the second number is equal to the probability that
it is greater than it, the probability in question is
1
6
5
1−
= .
2
36
12
Exercise 49: If a group of 6 men and 6 women is randomly divided into two groups of size 6
each, what is the probability that both groups have the same number of men?
There are 12
6 ways of choosing the two groups. Since both groups can contain the same number of men only if each contains three men, in which case each also contains three women, the
2
2
number of such choices is equal to 63 , giving 63 / 12
6 = 400/924 for the probability.
Example: If a fair coin is flipped n times, what is the probability of getting heads an even
number of times?
Using the fact that
Pn
n
k≡0 mod 2 k
=
Pn
n
k≡1 mod 2 k
1
, this probability is equal to 1/2.
Example 5b (i): If 3 balls are sampled without replacement from an urn containing 5 black
balls and 6 white balls, what is the probability that the sample contains one white ball and two
black balls?
Notice that this problem can be solved in two ways. If we take the order of sampling into
account, then |S| = 11 ∗ 10 ∗ 9 = 990 and |E| = 6 ∗ 5 ∗ 4 + 5 ∗ 6 ∗ 4 + 5 ∗ 4 ∗ 6 = 360, so that the
probability is 360/990 = 4/11. Alternatively, if we neglect the order, then S = 11
3 = 165 and
E = 61 · 52 = 60, so the probability is 60/165 = 4/11.
Example 5b (ii): Suppose that 5 people are chosen at random from a group of 20 individuals
consisting of 10 couples. What is the probability that the sample contains no couple?
10
First observe that there are 20
5 possible samples and that there are 5 ways of choosing 5
distinct couples and 25 = 32 ways of choosing one member of each of these couples. Consequently,
the probability is
10 5
5 2
20 .
5
Alternatively, if we consider ordered samples, then there are 20 ∗ 19 ∗ 18 ∗ 17 ∗ 16 possibilities, of
which 20 ∗ 18 ∗ 16 ∗ 14 ∗ 12 contain no couples. These two approaches give the same answer.
Example 5d: An urn contains n − 1 red balls and 1 blue ball. If k of these balls are sampled
without replacement, what is the probability that the sample contains the blue ball?
1
n
n−1
In this case |S| = nk and |E| = n−1
·
,
so
the
probability
is
k−1
1
k−1 / k = k/n. Alternatively,
let Ai be the event that the i’th ball chosen is blue and notice that the Ai are mutually exclusive
and that P(Ai ) = 1/n. Since E = A1 ∪ · · · ∪ Ak , it follows that P(A) = k/n.
Example 5g: A 5-card poker hand is a full house if it contains 3 cards of one denomination
and 2 cards of a second denomination? What is the probability of being dealt a full house?
Note that there are 52
hands neglecting order. Also, there are 13 ∗ 12 (ordered) pairs of
5
distinct denominations and, for each such pair, 43 42 ways of being dealt three cards from the
first denomination and two cards from the second. Thus the probability is
13 · · · 12 · 43 · 42
≈ 0.0014.
52
5
Example 5i: If n people are present in a room, what is the probability that no two of them
celebrate their birthday on the same day of the year?
The probability is
when n = 23.
365·364·363···(365−k+1)
365k
=
Qk−1
i=0 (1 − i/365).
This probability is first less than 1/2
Example 5j: Suppose that deck of 52 cards is shuffled and that the cards are turned up one
at a time until the first ace appears. Is the next card more likely to be an ace of spades or the
2
two of clubs?
In both cases, the probability is
51!
52!
=
1
52 .
Example 5m: Suppose that a deck of N cards is randomly shuffled. What is the probability
that no card is left in place?
Let Ei be the event that the i’th card is left in place. Using the inclusion-exclusion formula, the
probability that at least one card is left in place is
!
N
N
X
[
X
X
P(∩k Eik )
P
=
Ei
P(Ei ) −
P(Ei1 ∩ Ei2 ) + · · · + (−1)n+1
i=1
i=1
i1 <i2
N +1
+ · · · + (−1)
i1 <···<in
P(E1 ∩ · · · ∩ EN ).
Observe that
(N − n)!
N!
since any of the (N − n)! orders
of the elements in {1, · · · , N } − {i1 , · · · , in } are possible. Also,
P
N
as there are n terms in i1 <···<in P(Ei1 ∩ · · · ∩ Ein ), it follows that
P(Ei1 ∩ · · · ∩ Ein ) =
X
i1 <···<in
Thus
P
1
N (N − n)!
P(Ei1 ∩ · · · ∩ Ein ) =
= .
N!
n!
n
N
[
!
Ei
=1−
i=1
1
1
1
+ − · · · + (−1)N +1 ,
2! 3!
N!
and so the probability that no card is left in place is
1−1+
1
1
1
− − · · · + (−1)N
.
2! 3!
N!
Notice that this quantity is approximately e−1 when N is large.
Example 5n: What is the probability that if 10 married couples are seated at random in a
circle, then no couple are seated next to each other?
If Ei is the event that the i’th couple are seated next to each other, then the desired probability
is 1 − P(∪i Ei ). As in the previous example, we will use the inclusion-exclusion formula. For this
we need to calculate the probability P(Ei1 ∩ · · · ∩ Ein ). To do so, first notice that there are 19!
possible arrangements of the individuals around the table (seats are unlabeled). Next, notice
that there are 2n · (19 − n)! arrangements that result in a specified group of n couples sitting
together at the table: there are 2n orders of the individuals in the specified couples and (19 − n)!
arrangements of the n couples and the remaining 20 − 2n individuals. Consequently,
P(Ei1 ∩ · · · ∩ Ein ) =
3
2n (19 − n)!
19!
and so
10 X
10
2n (19 − n)!
.
P(∪i Ei ) =
(−1)n+1
19!
n
n=1
Exercise 35: Seven balls are sampled without replacement from an urn containing 12 red, 16
blue, and 18 green balls. Find the probability that:
(a) 3 red, 2 blue, and 2 green balls are sampled;
(b) at least 2 red balls are sampled;
(c) all sampled balls are the same color;
(d) either exactly three red or exactly three blue balls are sampled.
4