For mutually exclusive events R₁, R₂, and R3, we have P(R₁) = 0.05, P(R₂) = 0.3, and P(R3) = 0.65. Also,P(Q|R₁) = 0.6, P (Q|R₂) = 0.5, and P (Q | R3) = 0.4. Find P (R3 | Q).P(R31Q) =(Type an integer or a simplified fraction.)

For mutually exclusive events R₁, R₂, and R3, we have P(R₁) = 0.05, P(R₂) = 0.3, and P(R3) = 0.65. Also, P(Q|R₁) = 0.6, P (Q | R₂) = 0.5, and P P (Q | R₂) = = 0.4. Find P (R3 | Q). P(R31Q) = (Type an integer or a simplified fraction.)

For mutually exclusive events R₁, R₂, and R3, we have P(R₁) = 0.05, P(R₂) = 0.3, and P(R3) = 0.65. Also, P(Q|R₁) = 0.6, P (Q | R₂) = 0.5, and P P (Q | R₂) = = 0.4. Find P (R3 | Q). P(R31Q) = (Type an integer or a simplified fraction.)

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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Question

For mutually exclusive events R₁, R₂, and R3, we have P(R₁) = 0.05, P(R₂) = 0.3, and P(R3) = 0.65. Also,
P(Q|R₁) = 0.6, P (Q|R₂) = 0.5, and P (Q | R3) = 0.4. Find P (R3 | Q).
P(R31Q) =
(Type an integer or a simplified fraction.)

Expert Solution

Step 1

Given that,

The events $R_{1}, R_{2}, R_{3}$ are mutually exclusive

$\begin{array}{rcl} P (R_{1}) & = & 0.05 \\ P (R_{2}) & = & 0.3 \\ P (R_{3}) & = & 0.65 \\ P (Q | R_{1}) & = & 0.6 \\ P (Q | R_{2}) & = & 0.5 \\ P (Q | R_{3}) & = & 0.4 \end{array}$

If two events $A$ and $B$ are mutually exclusive then

$P (A \cap B) = 0$

The Bayes formula is

$P (A | B) = \frac{P (B | A) P (A)}{P (B)}$

Step by step

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