Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 7 - Section 7.5 - Conditional Probability and Independence - Exercises - Page 507: 18

Answer

$\dfrac{1}{6}$

Work Step by Step

Here, $A$ denotes the red die is $4$ and $B$ denotes the green die is $4$. Our aim is to calculate the conditional probability $P(A|B)$. This can be found as: $P(A|B)=\dfrac{P(A \cap B)}{P(B)} ~~~~(1)$ We can see that $A \cap B$ shows of all outcomes that red die is $4$ and the green die is $4$. This means that $A \cap B=\{(4,4)\}$ So, $P(A \cap B)=\dfrac{n(A \cap B)}{n(S)}=\dfrac{1}{36}$ and $P(B)=\dfrac{n(B)}{n(S)}=\dfrac{6}{36}$ Thus, the equation (1) becomes: $P(A~|~B)=\dfrac{P(A \cap B)}{P(B)} =\dfrac{1/36}{6/36}=\dfrac{1}{6}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.