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.

Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Chapter11: Data Analysis And Probability
Section11.8: Probabilities Of Disjoint And Overlapping Events
Problem 2C
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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)
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