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:
WaySet of Outcomes
A occurs, B doesn’tA \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)

No comments: