2) A randomly chosen moviegoer purchases a soda but not po p) A randomly chosen moviegoer purchases a soda or popcorm c) A randomly chosen moviegoer does not purchase anything.

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...
icon
Related questions
Question
Please answer correctly and show all your work. Attached is the formula sheet you can use.
For each of the events described below, provide an expression using events A and B; and compute their
respective probabilities.
(a) A randomly chosen moviegoer purchases a soda but not popcorn.
(b) A randomly chosen moviegoer purchases a soda or popcorn.
(c) A randomly chosen moviegoer does not purchase anything.
Transcribed Image Text:For each of the events described below, provide an expression using events A and B; and compute their respective probabilities. (a) A randomly chosen moviegoer purchases a soda but not popcorn. (b) A randomly chosen moviegoer purchases a soda or popcorn. (c) A randomly chosen moviegoer does not purchase anything.
Axioms of Probability
Also Note:
1. P(S)=1
For any two events A and B,
2. For any event E, 0< P(E) < 1
P(A) = P(AN B) + P(AN B')
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(AN B) = P(A)P(B).
Р(BJA) —
P(ANB)
P(A)
Bayes' Theorem:
Total Probability Rule
Р(A В)P(В)
Р(А|B)Р(В) + Р(A|B')P(В')
P(A) = P(A|B)P(B)+P(A|B')P(B')
P(B|A)
Similarly,
Similarly,
P(A) =P(A|E1)P(E1) + P(A|E2)P(E2)+
...+ P(A|Ek)P(Ek)
P(B|E1)P(E1)
P(B|E1)P(E1) + P(B|E2)P(E2) + · .+ P(B|ER)P(Ex)
P(E1|B)
Transcribed Image Text:Axioms of Probability Also Note: 1. P(S)=1 For any two events A and B, 2. For any event E, 0< P(E) < 1 P(A) = P(AN B) + P(AN B') 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(AN B) = P(A)P(B). Р(BJA) — P(ANB) P(A) Bayes' Theorem: Total Probability Rule Р(A В)P(В) Р(А|B)Р(В) + Р(A|B')P(В') P(A) = P(A|B)P(B)+P(A|B')P(B') P(B|A) Similarly, Similarly, P(A) =P(A|E1)P(E1) + P(A|E2)P(E2)+ ...+ P(A|Ek)P(Ek) P(B|E1)P(E1) P(B|E1)P(E1) + P(B|E2)P(E2) + · .+ P(B|ER)P(Ex) P(E1|B)
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Recommended textbooks for you
A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON
A First Course in Probability
A First Course in Probability
Probability
ISBN:
9780321794772
Author:
Sheldon Ross
Publisher:
PEARSON