tgt

## 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 $5$before 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 $\dfrac{{98}}{{2016}} = \dfrac{7}{{144}}$