1. An entertainment agency commissions a market survey to evaluate the public's interest indoor vs outdoor concerts. 35% of the respondents say that they have attended an outdoor concert in the last year. 22% of the respondents say that they have attended an indoor concert in the last year. 15% of the respondents have attended both types of events in the same time period. (a) What perce..tage of respondents did not attend any concerts in the last year? (b) What percentage of respondents have attended an indoor concert but not an outdoor one? (c) What is the probability that a respondent have been to an indoor concert, given that they have not been to an outdoor one?
1. An entertainment agency commissions a market survey to evaluate the public's interest indoor vs outdoor concerts. 35% of the respondents say that they have attended an outdoor concert in the last year. 22% of the respondents say that they have attended an indoor concert in the last year. 15% of the respondents have attended both types of events in the same time period. (a) What perce..tage of respondents did not attend any concerts in the last year? (b) What percentage of respondents have attended an indoor concert but not an outdoor one? (c) What is the probability that a respondent have been to an indoor concert, given that they have not been to an outdoor one?
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...
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Please answer all parts if that’s possible for you please. Attached is the formula sheet
![Axioms of Probability
Also Note:
1. P(S)=1
For any two events A and B,
2. For any event E, 0S P(E) S1
P(A) - P(ANB) + P(ANB')
3. For any two mutually exclusive events,
and
P(EUF) = P(E) + P(F)
P(AN B) = P(A|B)P(B).
Events A and B are independent if:
Addition Rule
P(EUF) - P(E) + P(F) - P(ENF)
P(A|B) = P(A)
or
Conditional Probability
P(ANB) = P(A)P(B).
P(B|A) =
PLAOB)
%3D
Bayes' Theorem:
Total Probability Rule
P(A|B)P(B)
P(A|B)P(B) + P(A|B)P(B)
P(A) = P(A|B)P(B) + P(A|B') P(B')
P(B|A)=
Similarly,
Similarly,
P(A) =P(A|E)P(E) + P(A|E)P(E)+
P(B|E,)P(E.)
...+ P(AE)P(E.)
P(BE,)P(E)+ P(B|E,)P(E,) +-… + P(B\E)P(E)
Probability Mass and Density Functions
Cumulative Distribution Function
If X is a discrete r.v:
• F(r) - P(X S =)
P(X - z) - f(z)
• lim,- F(a) -0
Ele) =1 (total probability)
• lim,- F(r) =1
If X is a continuous r.v.
• F(r) = , f (y)dy if X is a contimuous r.v
P(X = z) =0
• F(r) = Es, S(x) if X is a discrete r.v.
f(z)dz = 1 (total probability)
• P(a < X sb) = F(b) – F(a)
Expected Value and Variance
Expected Value of a Function of a RV
• EX) -E, zf(z) if X is a discrete r.v.
• Elh(X)] = E, h(z)/(x) if X is a discrete r.v.
• E[X] = If(z)dz if X is a contimous r.v.
• Eh(X)) = (z)/()dz if X is a contim-
ous r.v.
• Var(X) = E[X) - ELX]?
• ElaX + = aE[X] +6
• Var(aX + b) - aVar(X)
• Var(X) - E[(x - E|X])
Derivatives and Integrals of Common Functions
ar
• Se*dr =
• Sre'dz =ez - Je'dz - ze-e (using integration by parts)
. dale -!
• Sdz - In(r)
Common Discrete Distributions
• X- Bernoulli(p).
if r= 1;
l1-P ifz=0"
f(2) =
E[X] = P. Var(X) = p(1 - p).
• X- Geometrie(p),
f(2) = (1- p)-'p, ze (1,2, .). E[X] =; Var(X) = .
Geometrie Series: E - for 0 <q<1
• X- Binomial(n, p).
S(2) = ()(1- p)"-"p", z € {0,1, n}. E[X] = np, Var(X) = np(1 - p).
• X- Negative Binomial(r, p),
S(z) = ((1 - p)"-, E[X] = 5, z € {r.r+1,.), Var(x) = e.
• X- Hypergeometrie(n, M, N),
f(2) = CE), ELX] = n f, Var(X) = =»(1-4).
%3D
• X- Poisson(At),
f(z) =A Ie (0, 1,.), ELX] = At, Var(X) = At.
Common Contimuous Distributions
• X- Exponential(A),
f(2) = Ae-, z€ (0, oc) EJX] = Var(X) = .
• X- Erlang(r, A),
S(2) = , z € [0, 0), E[X] = {. Var(X) = .
• X- Weibull(a, 3).
S(2) = -49", z E (0, 0), E[X] = 3r(1 + 4). Var(X) = °{r(1 + 2) – r(1 + 29°}
Here I'(-) is the Gamma function, which satisfies:
r(r) = y-e-dy for z>0.
r(r + 1) = (r)r(r) for r>0
I'(n) = (n- 1)! if n is a positive iuteger.
3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F377b8d6e-0b43-439e-adad-4590c50894c7%2Fa62be4a8-1ef6-46ea-b773-6fe52a4ad8b2%2Flqg87ar_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Axioms of Probability
Also Note:
1. P(S)=1
For any two events A and B,
2. For any event E, 0S P(E) S1
P(A) - P(ANB) + P(ANB')
3. For any two mutually exclusive events,
and
P(EUF) = P(E) + P(F)
P(AN B) = P(A|B)P(B).
Events A and B are independent if:
Addition Rule
P(EUF) - P(E) + P(F) - P(ENF)
P(A|B) = P(A)
or
Conditional Probability
P(ANB) = P(A)P(B).
P(B|A) =
PLAOB)
%3D
Bayes' Theorem:
Total Probability Rule
P(A|B)P(B)
P(A|B)P(B) + P(A|B)P(B)
P(A) = P(A|B)P(B) + P(A|B') P(B')
P(B|A)=
Similarly,
Similarly,
P(A) =P(A|E)P(E) + P(A|E)P(E)+
P(B|E,)P(E.)
...+ P(AE)P(E.)
P(BE,)P(E)+ P(B|E,)P(E,) +-… + P(B\E)P(E)
Probability Mass and Density Functions
Cumulative Distribution Function
If X is a discrete r.v:
• F(r) - P(X S =)
P(X - z) - f(z)
• lim,- F(a) -0
Ele) =1 (total probability)
• lim,- F(r) =1
If X is a continuous r.v.
• F(r) = , f (y)dy if X is a contimuous r.v
P(X = z) =0
• F(r) = Es, S(x) if X is a discrete r.v.
f(z)dz = 1 (total probability)
• P(a < X sb) = F(b) – F(a)
Expected Value and Variance
Expected Value of a Function of a RV
• EX) -E, zf(z) if X is a discrete r.v.
• Elh(X)] = E, h(z)/(x) if X is a discrete r.v.
• E[X] = If(z)dz if X is a contimous r.v.
• Eh(X)) = (z)/()dz if X is a contim-
ous r.v.
• Var(X) = E[X) - ELX]?
• ElaX + = aE[X] +6
• Var(aX + b) - aVar(X)
• Var(X) - E[(x - E|X])
Derivatives and Integrals of Common Functions
ar
• Se*dr =
• Sre'dz =ez - Je'dz - ze-e (using integration by parts)
. dale -!
• Sdz - In(r)
Common Discrete Distributions
• X- Bernoulli(p).
if r= 1;
l1-P ifz=0"
f(2) =
E[X] = P. Var(X) = p(1 - p).
• X- Geometrie(p),
f(2) = (1- p)-'p, ze (1,2, .). E[X] =; Var(X) = .
Geometrie Series: E - for 0 <q<1
• X- Binomial(n, p).
S(2) = ()(1- p)"-"p", z € {0,1, n}. E[X] = np, Var(X) = np(1 - p).
• X- Negative Binomial(r, p),
S(z) = ((1 - p)"-, E[X] = 5, z € {r.r+1,.), Var(x) = e.
• X- Hypergeometrie(n, M, N),
f(2) = CE), ELX] = n f, Var(X) = =»(1-4).
%3D
• X- Poisson(At),
f(z) =A Ie (0, 1,.), ELX] = At, Var(X) = At.
Common Contimuous Distributions
• X- Exponential(A),
f(2) = Ae-, z€ (0, oc) EJX] = Var(X) = .
• X- Erlang(r, A),
S(2) = , z € [0, 0), E[X] = {. Var(X) = .
• X- Weibull(a, 3).
S(2) = -49", z E (0, 0), E[X] = 3r(1 + 4). Var(X) = °{r(1 + 2) – r(1 + 29°}
Here I'(-) is the Gamma function, which satisfies:
r(r) = y-e-dy for z>0.
r(r + 1) = (r)r(r) for r>0
I'(n) = (n- 1)! if n is a positive iuteger.
3D
Transcribed Image Text:1. An entertainment agency commissions a market survey to evaluate the public's interest indoor vs
outdoor concerts. 35% of the respondents say that they have attended an outdoor concert in the last
year. 22% of the respondents say that they have attended an indoor concert in the last year. 15% of
the respondents have attended both types of events in the same time period.
(a) What perce..tage of respondents did not attend any concerts in the last year?
(b) What percentage of respondents have attended an indoor concert but not an outdoor one?
(c) What is the probability that a respondent have been to an indoor concert, given that they have
not been to an outdoor one?
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