why the part I circled is k-1 rather just k? I made my assumption using red words, please tell me whether my assumption is true. The k-1 problem really bothers m
why the part I circled is k-1 rather just k? I made my assumption using red words, please tell me whether my assumption is true. The k-1 problem really bothers m
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
Related questions
Question
Could u please tell me why the part I circled is k-1 rather just k? I made my assumption using red words, please tell me whether my assumption is true. The k-1 problem really bothers me, thanks a lot! :)
![12:01 Sun Apr 30
<
Probability Notes
Date
C
A child buys coupons to try to complete a coupon book. There are N different types of
coupons. They are sold in sealed envelopes; each envelope contains a single coupon, which is
equally likely to be of any of the N types. Let X be the number of coupons bought until the
collection is completed. We will compute the expectation of X.
E[X] = E[Y₁+ Y₂+...+YN]
lectures_week3
We define auxiliary random variables Y₁,..., YN. We let Y₁ 1 (that is, Y₁ is actually de-
terministic, always equal to 1). For k = {2,..., N}, we define Y as the number of coupons
bought from (and not including) the moment when her collection reached k − 1 different types
to (and including) the moment when her collection reached k different types. To illustrate this,
assume that the coupon types are A, B, C, and the sequence of coupons bought by the child
new
are:
new
A, A, A, B, B, A, A, B, C.
In this case, we have that Y₁ = 1, Y₂ = 3 and Y3 5. We now make two observations.
=
First, we have
=
= E[Y₁] + E[Y₂] + ... + E[YN]
N
N
N
N
+
N - 1 N - 2
N
k=1
X = Y₁+
+ YN.
• Second, for k ≥ 2, the distribution of Y is Geometric (N=k+1) To see this, assume that
the child just bought the coupon that made her collection grow to k - 1 different types.
•E(X) of geometric distribution Geo (p) is 1/p:
why is k-1 ?
+ +
Probability Notes
N
1
=
T
N
k=1
exercises_week3
1
k
N-(k-1)
لر
number of type
The number is very close to f₁1dx = log(N). Hence, E[X] is very close to N log(N).
è a
Calculus Notes
we don't have.
total number of type
Ys,
If Ys there are 10 types in total.
We obviously still have 10-5=5 types
to be collected why
10- (5-1) = 6 ?
72%
need](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb41de797-8c36-43f3-a49e-0d77bbbd163e%2Fa0f531c6-fc34-4fb4-9f5a-ff703b660aa4%2F4udzgw_processed.png&w=3840&q=75)
Transcribed Image Text:12:01 Sun Apr 30
<
Probability Notes
Date
C
A child buys coupons to try to complete a coupon book. There are N different types of
coupons. They are sold in sealed envelopes; each envelope contains a single coupon, which is
equally likely to be of any of the N types. Let X be the number of coupons bought until the
collection is completed. We will compute the expectation of X.
E[X] = E[Y₁+ Y₂+...+YN]
lectures_week3
We define auxiliary random variables Y₁,..., YN. We let Y₁ 1 (that is, Y₁ is actually de-
terministic, always equal to 1). For k = {2,..., N}, we define Y as the number of coupons
bought from (and not including) the moment when her collection reached k − 1 different types
to (and including) the moment when her collection reached k different types. To illustrate this,
assume that the coupon types are A, B, C, and the sequence of coupons bought by the child
new
are:
new
A, A, A, B, B, A, A, B, C.
In this case, we have that Y₁ = 1, Y₂ = 3 and Y3 5. We now make two observations.
=
First, we have
=
= E[Y₁] + E[Y₂] + ... + E[YN]
N
N
N
N
+
N - 1 N - 2
N
k=1
X = Y₁+
+ YN.
• Second, for k ≥ 2, the distribution of Y is Geometric (N=k+1) To see this, assume that
the child just bought the coupon that made her collection grow to k - 1 different types.
•E(X) of geometric distribution Geo (p) is 1/p:
why is k-1 ?
+ +
Probability Notes
N
1
=
T
N
k=1
exercises_week3
1
k
N-(k-1)
لر
number of type
The number is very close to f₁1dx = log(N). Hence, E[X] is very close to N log(N).
è a
Calculus Notes
we don't have.
total number of type
Ys,
If Ys there are 10 types in total.
We obviously still have 10-5=5 types
to be collected why
10- (5-1) = 6 ?
72%
need
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps
Recommended textbooks for you
A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON
A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON