THE LAW OF TOTAL PROBABILITY

Let E1, E2, E3, …………. En mutually exclusive and exhaustive events and A is dependent event occurs after E1, E2, E3, …………. En.

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Probability of event A

P\left( A \right) = \,P\left( {A \cap {E_1}} \right) + P\left( {A \cap {E_2}} \right) + P\left( {A \cap {E_3}} \right) + ............ + P\left( {A \cap {E_n}} \right)

By multiplication theorem

P\left( A \right) = P\left( {{E_1}} \right).P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right).P\left( {A/{E_2}} \right) + .......\, + P\left( {{E_n}} \right).P\left( {A/{E_n}} \right) .

Where P\left( {A/{E_i}} \right) stands for probability of event A, if it is given that event {E_i} already occurs.

 

 

Example 1:  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.
Solution:

Let

E1 : Selecting bag 1

E2 : Selecting bag 2

A : Drawing red ball

P\left( {{E_1}} \right) = \frac{1}{2}

P\left( {{E_2}} \right) = \frac{1}{2}

P\left( {A/{E_1}} \right) = \,{\rm{Probability of}}\,{\rm{drawing}}\,{\rm{a}}\,{\rm{red}}\,{\rm{ball,}}\,{\rm{if}}\,{\rm{first}}\,{\rm{bag}}\,{\rm{is}}\,{\rm{selected}}

                          \,\, = \,\frac{4}{7}

P\left( {A/{E_2}} \right)\, = \,{\rm{Probability of}}\,{\rm{drawing}}\,{\rm{a}}\,{\rm{red}}\,{\rm{ball,}}\,{\rm{if}}\,{\rm{second}}\,{\rm{bag}}\,{\rm{is}}\,{\rm{selected}}

                          \,\, = \,\frac{2}{6}

\therefore \,\,\,P\left( A \right) = P\left( {{E_1}} \right).P\left( {A/{E_1}} \right) + P\left( {{E_2}} \right).P\left( {A/{E_2}} \right)

                      = \,\frac{1}{2} \times \frac{4}{7} + \frac{1}{2} \times \frac{2}{6}

                    \,\, = \,\frac{4}{{14}} + \frac{2}{{12}}

                      = \,\frac{2}{7} + \frac{1}{6}

                      = \,\frac{{19}}{{42}}

 

 

Post Author: E-Maths