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Theory of Probability : Recitation 2(Feb13)
1. Solutions for examples
How many 5-digit numbers can be formed from the integers 1, 2,...., 9 if no digit can appear
more than twice? (For instance, 41434 is not allowed)
1.
Solution. There are only three types of such numbers.
1. All digits are dierent
Obviously, we have 9 × 8 × 7 × 6 × 5 numbers in this case.
2. Only one pair of digits are same (For example, aabcd)
We choose a number a(9 ways) and nd two digits(places) for this number a ( 52 ways).
Then, choose 3 numbers other than a ( 83 ways) and arrange them in remaining 3 places
(3! ways). Therefore, we have 9 × 52 × 83 × 3! numbers in this case.
3. Two pairs of digits are same (For example, aabbc)
We choose two numbers
a, b ( 92 ways). Find two places to put a ( 52 ways) and then
two places to put b ( 32 ways). We select a number other than
a, b (7 ways) and put this
9
number into the last remaining seat. Therefore, we have 2 × 52 × 32 × 7 numbers in
this case.
Therefore, we have 9 × 8 × 7 × 6 × 5 + 9 ×
5
2
×
8
3
× 3! +
9
2
×
5
2
×
3
2
× 7 = 52920 numbers.
A group of individuals containing b boys and g girls is lined up in random order; that is, each
of the (b + g)! permutations is assumed to be equally likely. What is the probability that the
person in ith position, 1 ≤ i ≤ b + g , is a girl?
2.
Solution. We have |S| = (b + g)! and therefore enough calculate |E| where
E = {ith position is a girl}.
We have g ways to nd a girl at the ith position. Then, we can line up remaining (b + g − 1)
persons in (b + g − 1)! ways. Therefore,
|E| = g × (b + g − 1)!
and the probability is
|E|
g × (b + g − 1)!
g
=
=
.
|S|
(b + g)!
b+g
Note here that this probability is just the proportion of females in the group.
An instructor gives her class a set of 10 problems with the information that the nal exam
will consist of random selection of 5 of them. If a student has gured out how to do 7 of the
problems, what is the probability that he or she will answer correctly (a) all 5 problems? (b)
at least 4 of the problems? (c) at least 3 of the problems?
4.
1
Solution. Obviously, |S| = 105 . Let Ek be an event that k out of 7 studied problems are
chosen. Then,
we can select an outcome of Ek by choosing k problems fromthe 7 studied prob
3
7
ways). Therefore,
lems ( k ways) and 5 − k problems from the 3 unstudied problems ( 5−k
7
3
|Ek | = k 5−k . It is also obvious that Ek 's are pairwise disjoint.
(a)
In this case, E = E5 and therefore, |E| =
(b) E = E4 ∪ E5
and hence |E| =
(c) E = E3 ∪ E4 ∪ E5
7
5
3
0
and hence |E| =
+
7
5
7 3
21
5 0 = 21. Thus, the answer is (10)
5
7 3
126
1
=
126
.
The
answer
is
=
.
4 1
2
(10
5)
7 3
7 3
3
0 + 4 1 + 3 2 = 231. The answer
=
1
12 .
is
11
12 .
2. Solutions for the selected problems from Assignment 2
We can count the number of two pairs(patterns of the denomination is aabbc) in the following
steps :
3.
1. Select two denominations a and b :
ways
13
2
2. Select c from the remaining 11 denominations :
3. Choose two cards of denomination a :
4
2
4. Choose two cards of denomination b :
4
2
5. Choose a card of denomination c :
4
1
ways
ways
ways
11
1
ways
11 4 4 4
Therefore, the number of two pairs is given by 13
2
1
2 2 1 whereas the sample space
52
consists of 5 possible outcomes. Hence, the desired probability is
13
2
11
1
4 4 4
2 2
52
5
1
∼ 0.0475 = 4.75%.
Remark. You can easily compute the probability for the one pair(0.422=42.2%), full house(0.00144=0.144%)
and four of a kind(0.00024=0.02%) by the same argument.
In this case, |S| = 12n . Let E be an event that two persons have the same birth-month.
Instead of computing |E| directly, we will compute |E c |, the complement of E . Note that E c
indicates the event that n persons have dierent birth-months. To count this event, we rst
choose n months out of 12 months ( 12
n ways) and distribute these selected n months to n
5.
persons (n! ways). Thus, |E c | =
12
n
n! and P (E c ) =
n!
(12
n)
12n
. By the property of the probability,
12
n
P (E) = 1 − P (E c ) = 1 −
n!
12n
.
Note that our answer is 0 if n > 12 and which is consistent with the simple observation that E
is empty when n > 12. (since 13 persons cannot have dierent birth-months)
2
Remark. If we assume that all individual birthdays are equally likely, then the birth-months are
not equally distributed (and the answer should be changed). For example, the February has the
28
31
probability 365
whereas July has 365
(for the common years). In addition, there is an interesting
result about the distribution of the birthday. A statistical test shows that the birthdays are not
equally distributed but show the seasonal variation. Interestingly enough, the September shows
the highest positive deviation. (Why?)
7.
Let's simply denote pj = (1 − r)j r. Then, for any subset E ⊂ Ω, we dene P (E) by
P (E) =
X
pj
j∈E
and we will prove that P (·) satises the axiom of probability.
•
Axiom 2 : P (Ω) = 1
This directly follows from
P (Ω) =
X
pj =
j∈Ω
•
∞
X
pj =
j=0
∞
X
(1 − r)j r =
j=0
1
r=1
1 − (1 − r)
Axiom 1 : 0 ≤ P (E) ≤ 1 for each E
P
Note that since pj > 0 for all j , we have p(E) = j∈E pj ≥ 0. (Equality holds when
E = φ) The reverse inequality follows from
X
P (E) =
pj ≤
j∈E
X
pj = 1
j∈Ω
by Axiom 2.
•
Axiom 3 : Countable additivity
Let E1 , E2 , · · · is a sequence of mutually exclusive events. Then,
∞
X
k=0
P (Ek ) =
∞ X
X
k=0 j∈Ek
X
pj =
X
pj =
pj = P
S
j∈ ∞
k=0 Ek
j∈Ek for some k
∞
[
!
Ek
.
k=0
Note that the second equality is valid since Ei 's are pairwise disjoint.
10. We choose n balls from 2n dierent balls consisting
of n red balls and n blue balls. We
denote such an event by E then obviously |E| = 2n
.
Now
let's count |E| in a dierent manner.
n
Let's denote the event that we choose k red balls and n − k blue balls by Ek . Then it is obvious
that
n
E = E0 ∪ E1 ∪ · · · ∪ En =
[
k=0
and Ei 's are pairwise disjoint. Thus, we have
|E| =
n
X
k=0
3
|Ek |.
Ek
2
n
= nk and therefore we have |E| =
Note nally that |Ek | = nk n−k
countings on E complete the proof.
Remark. Such a technique is so-called double counting.
4
n 2
k=0 k .
Pn
These two