ONE MORE STEP TOWARD BETTER TOMORROW : chapter 8

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

	$P({\rm{one\, white,\, one\, black}}) =1 - P ({\rm{both\, white}}) - P({\rm{both\, black}})$
	$= 1 - \dfrac{{12}}{{45}}\; - \;\dfrac{{10}}{{45}}$
	$= \dfrac{{23}}{{45}}$

Thursday, 7 August 2014

chapter 8 - Worked Out Examples 1

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