11Daddition


 * The Addition Law of Probability **


 * If we have two events, A and B then
 * PR(A È B) = Pr(A) + Pr(B) - Pr(A Ç B)


 * the probability of A __**or**__ B is equal to
 * the probability of A plus the probability of B
 * but this counts the intersection between A and B twice, so
 * minus the probability of A __**and**__ B


 * ** Demonstration **
 * If we have two sets as shown in the Venn Diagram, such that
 * E = {1, 2, 3, 4, 5, 6, 7, 8}
 * A = {1, 3, 5, 7}
 * B = {1, 2, 3, 4}


 * We can see that

math . \qquad \qquad \qquad Pr(A) = \dfrac{4}{8} = \dfrac{1}{2} math . math . \qquad \qquad \qquad Pr(B) = \dfrac{4}{8} = \dfrac{1}{2} math


 * so

math . \qquad \qquad \qquad Pr(A) + Pr(B) = \dfrac{1}{2} + \dfrac{1}{2} = 1 math


 * but, from the Venn Diagram, we can see that:

math . \qquad \qquad \qquad Pr(A \, \cup \, B) = \dfrac{6}{8} = \dfrac{3}{4} math


 * this is because A Ç B = {1, 3} has been counted twice
 * so we have to subtract Pr(A Ç B)

math . \qquad \qquad \qquad Pr(A \, \cap \, B) = \dfrac{2}{8} math . math . \qquad \qquad \qquad Pr(A \, \cup \, B) = Pr(A) + Pr(B) - Pr(A \, \cap \, B) \\.\\ . \qquad \qquad \qquad Pr(A \, \cup \, B) = \dfrac{4}{8} + \dfrac{4}{8} - \dfrac{2}{8} \\.\\ . \qquad \qquad \qquad Pr(A \, \cup \, B) = \dfrac{6}{8} = \dfrac{3}{4} math


 * ** Example **


 * A single dice is rolled
 * Let A = the event that an odd number occurs
 * Let B = the event that a number less than 5 occurs
 * Find the probability that either A __**or**__ B occurs.


 * Solution:
 * A = {1,3, 5}
 * B = {1, 2, 3, 4}
 * A Ç B = {1, 3}

math . \qquad \qquad \qquad Pr(A \, \cup \, B) = Pr(A) + Pr(B) - Pr(A \, \cap \, B) \\. \\ . \qquad \qquad \qquad Pr(A \, \cup \, B) = \dfrac{3}{6} + \dfrac{4}{6} - \dfrac{2}{6} \\.\\ . \qquad \qquad \qquad Pr(A \, \cup \, B) = \dfrac{5}{6} math .