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Math Revolution GMAT math practice question]

A jar contains 16 red balls and 8 white balls. If 3 balls are selected at random from the jar, what is the approximate probability that all balls selected are white?

A. 0.02

B. 0.03

C. 0.04

D. 0.05

E. 0.06

\({\text{Jar}}\,\,\left\{ \begin{gathered}

16\,r \hfill \\

8\,w \hfill \\

\end{gathered} \right.\,\,\,\,\, \Rightarrow \,\,\,3\,\,{\text{simultaneous}}\,\,{\text{extractions}}\)

\({\text{? = P}}\left( {{\text{all}}\,\,w} \right)\)

\({\text{total}} = C\left( {16 + 8,3} \right) = C\left( {24,3} \right)\,\,\,\,{\text{equiprobable}}\)

\({\text{favorable}} = C\left( {8,3} \right)\)

\(? = \frac{{C\left( {8,3} \right)}}{{C\left( {24,3} \right)}} = \cdots = \frac{7}{{253}}\)

\({\left( ? \right)^{ - 1}} = \frac{{210 + 42 + 1}}{7} = \boxed{36\frac{1}{7}}\)

\({\left( A \right)^{ - 1}} = {\left( {\frac{2}{{100}}} \right)^{ - 1}} = 50\)

\({\left( B \right)^{ - 1}} = {\left( {\frac{3}{{100}}} \right)^{ - 1}} = \frac{{99 + 1}}{3} = \boxed{33\frac{1}{3}}\)

From the fact that 33 1/3 is less than 36 1/7 , we are sure the other alternative choices (all of them are in increasing order) are not closer to our FOCUS.

(In other words, the reciprocals of (C), (D) and (E) are certainly less than 33 1/3 , hence the closest to 36 1/7 is 33 1/3.)

The above follows the notations and rationale taught in the GMATH method.

_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)

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