A jar contains 16 red balls and 8 white balls. If 3 balls ar

Max@Math Revolution wrote:[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

P(1st marble is white) = 8/24. (Of the 24 marbles, 8 are white.)
P(2nd marble is white) = 7/23. (Of the 23 remaining marbles, 7 are white.)
P(3rd marble is white) = 6/22. (Of the 22 remaining marbles, 6 are white.)
To combine these probabilities, we multiply:
8/24 * 7/23 * 6/22 = 1/3 * 7/23 * 3/11 = 7/(23*11) = 7/253 ≈ 7/250 = 28/1000 = 0.28 ≈ 0.03.

The correct answer is B.

Max@Math Revolution wrote:[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.

Brent@GMATPrepNow wrote: ASIDE: rather than perform long division, we can take 7/253 and create an equivalent fraction by multiplying top and bottom by 4 to get...
7/253 = (7)(4)/(253)(4)
≈ 28/1000
≈ 0.028
≈0.03

Long division is something that must be avoided. That´s why GMATH´s approach deals with what we call "breaking numbers".

It is not comfortable to do (say) 253/7 by long division - and we did not - this can be done WITHOUT APPROXIMATIONS easily, as we did and the way we repeat below:

253 = 210 + 42 + 1 , this choice is "smart" because 210 and 42 are not only divisible by 7 (obviously), but also because their quotients are trivial (30 and 6)...

Therefore 253/7 = 210/7 + 42/7 + 1/7 = 36 + 1/7 , without FULL PRECISION!

One last detail: when we approximate (say) 7/253 by 7/250 , we must be cautious because approximations in denominators may have "small errors" that propagate according to the "magnitude" (absolute value) of the numerator... that´s why we believe the approach we have taken is safer, although in this case the approximation 7/253 by 7/250 was sufficiently good for the alternative choices.

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

=>

There are 24C3 ways of choosing 3 balls from the 24 in the jar, and 8C3 ways of choosing 3 balls from the 8 white balls. Therefore, the probability of choosing 3 white balls from the jar is:
8C3 / 24C3 = { (8*7*6) / (1*2*3) } / { (24*23*22) / (1*2*3) } = (8*7*6) / (24*23*22) = 7 / ( 23* 11 ) ≈ 1/ (3*11) ≈ 0.03

Therefore, the answer is B.
Answer: B