{"id":6261,"date":"2021-10-11T02:17:51","date_gmt":"2021-10-10T20:47:51","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6261"},"modified":"2022-01-16T17:05:06","modified_gmt":"2022-01-16T11:35:06","slug":"total-probability-theorem","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/total-probability-theorem\/","title":{"rendered":"Total Probability Theorem – Definition and Examples"},"content":{"rendered":"
Here you will learn law of total probability theorem with examples.<\/p>\n
Let’s begin –<\/p>\n
Let an event A of an experiment occurs with its n mutually exclusive and exhaustive events \\(B_1,B_2,B_3….B_n\\) then total probability of occurrence of event A is<\/p>\n
\nP(A) = \\(P(AB_1)\\) + \\(P(AB_2)\\) +……..+ \\(P(AB_n)\\)<\/p>\n
P(A) = P(\\(B_1\\))P(\\(A\/B_1\\)) + P(\\(B_2\\))P(\\(A\/B_2\\)) +……..+ P(\\(B_n\\))P(\\(A\/B_n\\))<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.<\/p>\n
Solution<\/span><\/strong> : A red ball can be drawn in two mutually exclusive ways.<\/p>\n
(I) Selecting bag I and then drawing a red ball from it.<\/p>\n
(II) Selecting bag II and then drawing a red ball from it.<\/p>\n
Let \\(E_1\\), \\(E_2\\) and A denote the events defined as follows :<\/p>\n
\\(E_1\\) = Selecting bag I,<\/p>\n
\\(E_2\\) = Selecting bag II<\/p>\n
A = Drawing a red ball<\/p>\n
Since one of the two bags is selected randomly.<\/p>\n
\\(\\therefore\\) P\\(E_1\\) = \\(1\\over 2\\) and P\\(E_2\\) = \\(1\\over 2\\)<\/p>\n
Now, P(A\/\\(E_1\\)) = Probability of drawing a red ball when the first bag has been chosen.<\/p>\n
= \\(4\\over 7\\) [ because First bag contains 4 red and 3 black balls ]<\/p>\n
and, P(A\/\\(E_2\\)) = Probability of drawing a red ball when the second bag has been chosen.<\/p>\n
= \\(2\\over 6\\) [ because Second bag contains 2 red and 4 black balls ]<\/p>\n
Using the law of total probability, we have<\/p>\n
Required Probabilty = P(A) = P(\\(E_1\\)) P(A\/\\(E_1\\)) + P(\\(E_2\\)) P(A\/\\(E_2\\))<\/p>\n
= \\(1\\over 2\\) \\(\\times\\) \\(4\\over 7\\) + \\(1\\over 2\\) \\(\\times\\) \\(2\\over 6\\) = \\(19\\over 42\\)<\/p>\n
Example<\/strong><\/span> : A purse contains 4 copper and 3 silver coins and another purse contains 6 copper and 2 silver coins. One coin is drawn from any one of the these two purses. The probability that it is a copper coin is-<\/p>\n
Solution<\/span><\/strong> : Let A = event of selecting first purse<\/p>\n
B = event of selecting second purse<\/p>\n
C = event of selecting copper coin<\/p>\n
Then given event has two disjoint cases : AC and BC<\/p>\n
P(C) = P(AC + BC) = P(AC) + P(BC) = P(A)P(C\/A) + P(B)P(C\/B)<\/p>\n
= \\(1\\over 2\\).\\(4\\over 7\\) + \\(1\\over 2\\).\\(6\\over 8\\) = \\(37\\over 56\\)<\/p>\n\n\n