Download presentation

Presentation is loading. Please wait.

Published bySamuel Thompson Modified over 4 years ago

1
MATHPOWER TM 12, WESTERN EDITION 8.4.1

2
8.4.2 Conditional Probability If A and B are events from an experiment, the conditional probability of B given A (P(A|B)), is the probability that Event B will occur given that Event A has already occurred. The conditional probability is equal to the probability that B and A will occur divided by the probability that B will occur. This is given in Bayes Formula:

3
8.4.3 Conditional Probability Determine the conditional probability for each of the following: a) Given P(B and A) = 0.725 and P(B) = 0.78, find P(A|B). P(A|B) = 0.9295 a)Given P(blonde and tall) = 0.5 and P(blonde) = 0.73, find P(A|B). P(A|B) = 0.6849

4
8.4.4 Finding Conditional Probability It is known that 10% of the population has a certain disease. For a patient without the disease, a blood test for the disease Shows not positive 95% of the time. For a patient with the Disease, the blood test shows positive 99% of the time. What is the probability that a person whose blood test is positive for the disease actually has the disease? 0.10 0.90 sick not sick test positive test negative P(sick and positive) P(sick and negative) P(not sick and positive) P(not sick and negative) 0.99 0.01 0.95 0.05 = 0.10 x 0.99 = 0.099 = 0.10 x 0.01 = 0.001 = 0.90 x 0.05 = 0.045 = 0.90 x 0.95 = 0.855

5
8.4.5 Finding Conditional Probability [contd] P(B and A) = P(sick and positive) = 0.099 P(B) = P(positive) P(positive) = P(sick and positive) or P(not sick and positive) = 0.099 + 0.045 = 0.144 Therefore, the probability of the person testing positive and actually having the disease is 0.6875.

6
A new medical test for cancer is 95% accurate. If 0.8% of the population suffer from cancer, what is the probability that a person selected at random will test negative and actually have cancer? 8.4.6 Finding Conditional Probability 0.008 0.992 cancer not cancer test positive test negative P(sick and positive) P(sick and negative) P(not sick and positive) P(not sick and negative) 0.95 0.05 0.95 0.05 = 0.008 x 0.95 = 0.0076 = 0.008 x 0.05 = 0.0004 = 0.992 x 0.05 = 0.0496 = 0.992 x 0.95 = 0.9424

7
8.4.7 Finding Conditional Probability [contd] P(B and A) = P(cancer and negative) = 0.0004 P(B) = P(negative) P(negative) = P(cancer and negative) or P(not cancer and negative) = 0.0004 + 0.9424 = 0.9428 Therefore, the probability of the person testing negative and actually having the disease is 0.0004.

Similar presentations

OK

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.1 Chapter Six Probability.

Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 6.1 Chapter Six Probability.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on bank lending regulations Ppt on social contract theory ethics Ppt on hplc method development and validation Ppt on rc coupled amplifier experiment Ppt on electronic payment systems Ppt on human chromosomes have hundreds Ppt on charge-coupled device illustration Ppt on ac series motor Ppt on child labour in english Ppt on is matter around us pure