Introduction to Probability - Chapter 10 pptx
Introduction to Probability - Chapter 10 pptx
... -5 0
00000000000 0-1 00
111111249797300
233420000000 0
100 000000000 -5 0
2100 00000000 0
3347111714111 1101 6251300
00000000000 0-1 00
12211 3100 000 -5 0
00000000000 0-1 00
2 3100 0000000 0
3100 00000000 50
100 000000000 -5 0
344 7101 191112141 310 550
13349579886 3100
104 66 9101 30000 ... follows:
365
392 CHAPTER 10. GENERATING FUNCTIONS
Z
1
Z
2
Z
3
Z
4
Z
5
Z...
Introduction to Probability - Chapter 7 pptx
... X
3
.Thus
P (S
3
=3) = P(S
2
=2)P (X
3
=1)
300 CHAPTER 7. SUMS OF RANDOM VARIABLES
-1 5 -1 0
-5 5
10 15
0.025
0.05
0.075
0.1
0.125
0.15
0.175
n = 5
n = 10
n = 15
n = 20
n = 25
Figure 7.7: Convolution ... customer at a bank is exponentially dis-
tributed with mean service time 2 minutes. Let X be the total service time
for 10 customers. Estimate the probability that X>2...
Introduction to Probability - Chapter 1 pps
... PROBABILITIES 5
5
10 15 20 25 30 35 40
-1 0
-8
-6
-4
-2
2
4
6
8
10
Figure 1.1: Peter’s winnings in 40 plays of heads or tails.
One can understand this calculation as follows: The probability that ... more appropriate to assign a distribution
function which assigns probability .513 to the outcome boy and probability .487 to
the outcome girl than to assign pro...
Introduction to Probability - Chapter 2 docx
... with careful planning one would have to be extremely lucky to be able to stop
so cleverly.
The second author likes to trace his interest in probability theory to the Chicago
World’s Fair of 1933 ... referring to a continuous random variable X (say with a uniform density
function), it is customary to say that “X is uniformly distributed on the interval
[a, b].” It is also custo...
Introduction to Probability - Chapter 3 ppt
... =
1234
2143
,
indicating that a
1
went to a
2
, a
2
to a
1
, a
3
to a
4
, and a
4
to a
3
.
Ifwealwayschoosethetoprowtobe1234then, to prescribe the permutation,
we need only give the bottom row, with the understanding ... Birthday problem.
We now turn to the topic of permutations.
Permutations
Definition 3.1 Let A be any finite set. A permutation of A is a one -to- one mapping
of...
Introduction to Probability - Chapter 4 doc
... ✷
146 CHAPTER 4. CONDITIONAL PROBABILITY
Number having The results
Disease this disease ++ +– –+ ––
d
1
3215 2 110 301 704 100
d
2
2125 396 132 1187 410
d
3
4660 510 3568 73 509
Total 100 00
Table ... drawn is red?
4.3. PARADOXES 175
13 Write a program to allow you to compare the strategies play-the-winner and
play-the-best-machine for the two-armed bandit problem of Example...
Introduction to Probability - Chapter 5 docx
... result of 100 Bernoulli trials
with p =1 /100 0. The expected value of S
100
is λ = 100 (1 /100 0) = .1. The exact
probability that S
100
= j is b (100 , 1 /100 0,j), and the Poisson approximation is
e
−.1
(.1)
j
j!
.
In ... distribution to X? If it is appropriate,
what probability should we assign to each outcome?
(a) Take the first 100 students who enter the cafeteria to eat...
Introduction to Probability - Chapter 6 doc
... 61 62 63 64
1 /10 2 /10 4 /10 2 /10 1 /10
.
(a) Find E(F ) and V (F ).
(b) Define T = F − 62. Find E(T ) and V (T ), and compare these answers
with those in part (a).
(c) It is decided to report the ... real-valued random vari-
able. One way to determine the expected value of φ(X) is to first determine the
distribution function of this random variable, and then use the definition of...
Introduction to Probability - Chapter 8 docx
... ≤
.21
n(.1)
2
=
21
n
.
Thus, if n = 100 ,
P (|A
100
− .3|≥.1) ≤ .21 ,
or if n = 100 0,
P (|A
100 0
− .3|≥.1) ≤ .021 .
These can be rewritten as
P (.2 <A
100
<.4) ≥ .79 ,
P (.2 <A
100 0
<.4) ≥ .979 .
These ... x) to compute the exact
probability that you estimated in Exercise 1. Compare the two results.
3 Write a program to toss a coin 10, 000 times. Let S
n
be the n...
Introduction to Probability - Chapter 9 pps
... want to estimate the probability
P (S
1700
> 106 0) = P (S
1700
≥ 106 1)
= P
S
∗
1700
≥
106 0.5 102 0
20
= P (S
∗
1700
≥ 2.025) .
From Table 9.4, if we interpolate, we would estimate this probability ... or greater. Estimate the probability that, for this to happen,
(a) more than 210 tosses are required.
(b) less than 190 tosses are required.
(c) between 180 and 210 toss...
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