Thursday, 7 August 2014

chapter-11 -worked-out-examples-4

    Example: 7      

A pair of dice is rolled until a sum of either 5 or 7 is obtained. Find the probability that 5 comes before 7.
Solution: 7

A sum of 5 can be obtained in 4 ways, namely
\{ (1,\;4)\;(2,\;3)\;(3,\;2)\;(4,\;1)\}
from a total number of 36 ways of throwing a pair of dice. If we let E denote the event 5 of obtaining a sum of 5, we have
P(E) = \dfrac{4}{{36}} = \dfrac{1}{9}
Similarly, let the event F be that of obtaining a sum of 7; this can happen in 6 ways, namely
\left\{ {(1,\;6)\;(2,\;5)\;(3,\;4)\;(4,\;3)\;(5,\;2)\;(6,\;1)} \right\}
so that
P(F) = \dfrac{6}{{36}} = \dfrac{1}{6}
Finally, if we let G be the event of obtaining neither a 5 or a 7, we have
P(G) = 1 - P(E) - P(F)
 = 1 - \dfrac{1}{9} - \dfrac{1}{6}
 = \dfrac{{13}}{{18}}
Now, we want a sum of 5 to come before a sum of 7. Think about how this can happen. Every time you roll the pair of dice, you should either get a sum of 5  or you should get neither a sum of 5 nor 7.
Therefore, the following (mutually exclusive) sequences of throws leads us to a sum of 5before a sum of 7.
{\rm{Sequence}}  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P({\rm{Sequence}})
E \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{1}{9}
G, \,{\rm{then}}\,E\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \dfrac{{13}}{{18}} \times \dfrac{1}{9}({\rm{Successive \,Rolls\, are\, independent\, events!}})
G \,{\rm{then}} \,G,\, {\rm{then}} \, E \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{13}}{{18}} \times \dfrac{{13}}{{18}} \times \dfrac{1}{9}
G\,{\rm{then}}\, G,\, {\rm{then}}\, G, \,{\rm{then}} \, E \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \dfrac{{13}}{{18}} \times \dfrac{{13}}{{18}} \times \dfrac{{13}}{{18}} \times \dfrac{1}{9}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\vdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\vdots
The required probability is obtained by adding the terms in the right column.
P(5\,{\rm{before}}\,7)=\dfrac{1}{9} + \dfrac{{13}}{{18}}\; \cdot \;\dfrac{1}{9} + {\left( {\dfrac{{13}}{{18}}} \right)^2}\; \cdot \;\dfrac{1}{9} + {\left( {\dfrac{{13}}{{18}}} \right)^3}\; \cdot \;\dfrac{1}{9} + \;\ldots \infty
 = \dfrac{1}{9}\left( {1 + \dfrac{{13}}{{18}}\; + \;{{\left( {\dfrac{{13}}{{18}}} \right)}^2}\; + {{\left( {\dfrac{{13}}{{18}}} \right)}^3} + \;\ldots \infty } \right)
 = \dfrac{1}{9}\; \cdot \;\dfrac{1}{{1 - \dfrac{{13}}{{18}}}}
 = \dfrac{2}{5}
In passing, note that the probability of obtaining a 7 before 5 is simply
{\rm{P (7 \,{\rm{before}}\, 5)  =  1  -  P(5\,{\rm{ before}}\, 7)}}
 = \dfrac{3}{5}
Can you appreciate why obtaining a 7 before 5 is more likely than obtaining a 5 before 7?
     Example: 8      

Two unit squares are chosen at random from a standard chessboard. What is the probability that the two squares have exactly one corner in common?
Solution: 8

The total number of ways of selecting two squares from a standard chessboard is simply {}^{64}{C_2} = 2016.
Now, let us find the number of ways of selecting a pair with exactly one corner in common. For that, consider the following figure.
It is immediately evident that the number of favorable ways is 49 + 49 = 98.
The required probability is \dfrac{{98}}{{2016}} = \dfrac{7}{{144}}

No comments: