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Thursday, 7 August 2014

chapter 8 - Worked Out Examples 1

     Example: 1  

(a) A bag contains 5 red and 10  white balls, all identical in shape and size. A person puts his hand in and without looking, pulls out a ball. What is the probability of it being red?
(b) In the same bag, he puts his hand in and draws out a pair of balls. What is the probability of both of them being red?
Solution: 1-(a)

It should be obvious that drawing a white ball is more likely than drawing a red one, since there are more white balls. Precisely,
P({\rm{red}}\;{\rm{ball}}) = \dfrac{5}{{15}}\,\,\begin{array}{*{20}{c}}  { \to \,\,{\rm{red}}\;{\rm{balls}}}\\  {\, \to \,\,{\rm{total}}\;{\rm{balls}}}  \end{array}\;
 = \dfrac{1}{3}
and P({\rm{white}}\;{\rm{ball}}) = \dfrac{{10}}{{15}}\begin{array}{*{20}{c}}  { \to {\rm{white balls}}}\\  { \to {\rm{total}}\;{\rm{balls}}}  \end{array}\;
 = \dfrac{2}{3}
so that,
P({\rm{red\,ball}}) + P({\rm{white\,ball}}) = 1 as expected
Solution: 1-(b)

There are 5 red balls, so there are {}^5{C_2} pairs of red balls possible. (Some readers might raise an objection here: if the balls are identical, how can there be {}^5{C_2} ways of forming pairs? There should be just one way, as discussed in P&C. These readers should try to appreciate the difference between number of pairs and number of ways of forming pairs. The former is {}^5{C_2}, the latter is 1) .
The total number of pairs possible is {}^{15}{C_2}. Thus
P({\rm{red\,pair}}) = \dfrac{{^5{C_2}}}{{^{15}{C_2}}} = \dfrac{2}{{21}}
Similarly, we will have
P ({\rm{white\,pair}}) = \dfrac{{^{10}{C_2}}}{{^{15}{C_2}}} = \dfrac{3}{7}
and  P ({\rm{one \,red,\, one \, white\, pair}})= \dfrac{{5 \times 10}}{{^{15}{C_2}}} = \dfrac{{10}}{{21}}
Note that
P ({\rm{red \, pair}}) + P ({\rm{white \, pair}}) + P ({\rm{one\, red,\, one\, white\, pair}})
 = \dfrac{2}{{21}}\;\; + \;\;\dfrac{3}{7}\;\; + \;\;\dfrac{{10}}{{21}}
=1
     Example: 2    

There are two bags, one containing 4 white and 5 black balls, and the other containing 3 white and 2 black balls. A person draws one ball at random from each bag. Find the probability that
(a) both balls are white
(b) one ball is white and one is black
Solution: 2-(a)

Let the two bags be labelled A and B respectively. Note that drawing a ball from A is independent of drawing a ball from B. Thus,
P\,({\rm{both}}\;{\rm{white}}) = P\left\{ \begin{array}{l}  {\rm{ball}}\;{\rm{from}}\;{\rm{bag}}\\  A\;{\rm{is}}\;{\rm{white}}  \end{array} \right\} \times P\left\{ \begin{array}{l}  {\rm{ball}}\;{\rm{from}}\;{\rm{bag}}\\  \;B\;{\rm{is \,white }}  \end{array} \right\}
 = \,\,\,\,\dfrac{4}{9} \times \dfrac{3}{5}
 = \dfrac{{12}}{{45}}
Solution: 2-(b)

We note that the case {one white and one black} is possible in two ways:
X : Ball from bag A white, ball from bag B black
Y : Ball from bag A black, ball from bag B white
Note that X and Y are mutually exclusive events.
Thus,
P ({\rm{one\, white,\, one \,black}}) = P(X)\;\; + \;\;P(Y)
 = \dfrac{4}{9}\; \times \;\dfrac{2}{5}\; + \;\dfrac{5}{9}\; \times \;\dfrac{3}{5}
 = \dfrac{{23}}{{45}}
We could also have calculated this probability by noting that
P ({\rm{both \, white}}) + P ({\rm{one \, white, \, one \, black}}) + P ({\rm{both \, black}}) = 1
and since P({\rm{both\, black}}) = \dfrac{5}{9}\; \times \;\dfrac{2}{5}\; = \;\dfrac{{10}}{{45}}, we have
P({\rm{one\, white,\, one\, black}}) =1 - P ({\rm{both\, white}}) - P({\rm{both\, black}})
 = 1 - \dfrac{{12}}{{45}}\; - \;\dfrac{{10}}{{45}}
 = \dfrac{{23}}{{45}}
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