Suppose that A1, A2, A3, A4, and A5 are five mutually exclusive and exhaustiveevents in a sample space S, and suppose that P(A1)=0.2, P(A2)=0.1, P(A3) =0.15, P(A4)=0.3, and P(A5)=0.25. Another event E in S is such that P(E|A1) =0.2, P(E|A2)=0.1, P(E|A3)=0.35, P(E|A4)=0.3, and P(E|A5)=0.25. Find theprobabilities P(A1|E), P(A2|E), P(A3|E), P(A4|E), and P(A5|E). Suppose that A₁, A2, A3, A4, and A5 are five mutually exclusive and exhaustiveevents in a sample space S, and suppose that P(A₁) = 0.2, P(A₂) = 0.1, P(A3) =0.15, P(A₁) = 0.3, and P(A5) = 0.25. Another event E in S is such that P(E|A₁) =0.2, P(E|A₂) = 0.1, P(E|A3) = 0.35, P(E|A₁) = 0.3, and P(E|A5) = 0.25. Find theprobabilities P(A₁|E), P(A₂|E), P(A3|E), P(A₁|E), and P(A5|E).:

The given probabilities are Also, for an event in , the conditional probabilities are:

Answered: Suppose that A1, A2, A3, A4, and A5 are…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.8: Probability

Problem 29E

See similar textbooks

Related questions

Question

Suppose that A1, A2, A3, A4, and A5 are five mutually exclusive and exhaustive
events in a sample space S, and suppose that P(A1)=0.2, P(A2)=0.1, P(A3) =
0.15, P(A4)=0.3, and P(A5)=0.25. Another event E in S is such that P(E|A1) =
0.2, P(E|A2)=0.1, P(E|A3)=0.35, P(E|A4)=0.3, and P(E|A5)=0.25. Find the
probabilities P(A1|E), P(A2|E), P(A3|E), P(A4|E), and P(A5|E).

Suppose that A₁, A2, A3, A4, and A5 are five mutually exclusive and exhaustive
events in a sample space S, and suppose that P(A₁) = 0.2, P(A₂) = 0.1, P(A3) =
0.15, P(A₁) = 0.3, and P(A5) = 0.25. Another event E in S is such that P(E|A₁) =
0.2, P(E|A₂) = 0.1, P(E|A3) = 0.35, P(E|A₁) = 0.3, and P(E|A5) = 0.25. Find the
probabilities P(A₁|E), P(A₂|E), P(A3|E), P(A₁|E), and P(A5|E).
:

Expert Solution

Step 1: Given information.

The given probabilities are
$P open parentheses A subscript 1 close parentheses equals 0.2 P open parentheses A subscript 2 close parentheses equals 0.1 P open parentheses A subscript 3 close parentheses equals 0.15 P open parentheses A subscript 4 close parentheses equals 0.3 P open parentheses A subscript 5 close parentheses equals 0.25$

Also, for an event $E$ in $S$ , the conditional probabilities are:
$P open parentheses E vertical line A subscript 1 close parentheses equals 0.2 P open parentheses E vertical line A subscript 2 close parentheses equals 0.1 P open parentheses E vertical line A subscript 3 close parentheses equals 0.35 P open parentheses E vertical line A subscript 4 close parentheses equals 0.3 P open parentheses E vertical line A subscript 5 close parentheses equals 0.25$