## Thursday, 7 August 2014

### chapter-14 Worked Out Examples

 Example: 11
 For two events $A$ and $B$, find the probability that exactly one of the two events occur.
 Solution: 11
This can happen in two ways:
 Way Set of Outcomes $A$ occurs, $B$ doesn’t $A \cap \bar B$ $B$ occurs, $A$ doesn’t $\bar A \cap B$
Note that the two ways are $ME$. Thus, the required probability is $P\left( {A \cap \bar B} \right) + P\left( {\bar A \cap B} \right)$, which from the second relation becomes $P(A) - P\left( {A \cap B} \right) + P\left( B \right) - P\left( {A \cap B} \right)$ $= P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right)$
Justify this last expression using a Venn diagram, by shading the area it represents.
 Example: 12
For two events $A$ and $B$, show that $P\left( B \right) = P\left( A \right) \cdot P\left( {B/A} \right) + P\left( {\bar A} \right) \cdot P\left( {B/\bar A} \right)$
 Solution: 12
This is straightforward, since $P\left( A \right) \cdot P\left( {B/A} \right) = P\left( {A \cap B} \right)$ recall the discussion in the previous section and $P\left( {\bar A} \right) \cdot P\left( {B/\bar A} \right) = P\left( {\bar A \cap B} \right)$
Now, since $A\,\,{\rm{and}}\,\,\bar A$ are mutually exclusive, we must have $P\left( {A \cap B} \right) + P\left( {\bar A \cap B} \right) = P\left( B \right)$ 