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

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A jar contains 16 red balls and 8 white balls. If 3 balls are selected  [#permalink]

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New post 29 Aug 2018, 23:53
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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

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A jar contains 16 red balls and 8 white balls. If 3 balls are selected  [#permalink]

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New post 30 Aug 2018, 00:12
MathRevolution 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


3 balls selected without replacement.
P(WWW) = 8/24 * 7/23 *6/22 = 7/253 = 0.027
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Re: A jar contains 16 red balls and 8 white balls. If 3 balls are selected  [#permalink]

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New post 30 Aug 2018, 02:21
MathRevolution 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) = \(\frac{8}{24}\) (Of the 24 marbles, 8 are white.)
P(2nd marble is white) = \(\frac{7}{23}\) (Of the 23 remaining marbles, 7 are white.)
P(3rd marble is white) = \(\frac{6}{22}\) (Of the 22 remaining marbles, 6 are white.)
To combine these probabilities, we multiply:
\(\frac{8}{24} * \frac{7}{23} * \frac{6}{22} = \frac{1}{3} * \frac{7}{23} * \frac{3}{11} = \frac{7}{23*11} = \frac{7}{253} ≈ \frac{7}{250} = \frac{28}{1000} = 0.028 ≈ 0.03\)


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A jar contains 16 red balls and 8 white balls. If 3 balls are selected  [#permalink]

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New post 30 Aug 2018, 03:25
MathRevolution 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.
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A jar contains 16 red balls and 8 white balls. If 3 balls are selected  [#permalink]

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New post Updated on: 30 Aug 2018, 09:35
Top Contributor
MathRevolution 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


There are 24 balls in the jar (before any are removed)

P(all 3 balls are white) = P(1st ball is white AND 2nd ball is white AND 3rd ball is white)
= P(1st ball is white) x P(2nd ball is white) x P(3rd ball is white)
= 8/24 x 7/23 x 6/22
= 1/3 x 7/23 x 3/11
= 7/253

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

Aside: I'm not suggesting that it would be wrong to use long division to convert 7/253 to a decimal. That's a perfectly valid approach. However, if you're looking for an alternative approach, then we can also apply the above approach.

Answer: B

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Originally posted by GMATPrepNow on 30 Aug 2018, 06:47.
Last edited by GMATPrepNow on 30 Aug 2018, 09:35, edited 1 time in total.
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Re: A jar contains 16 red balls and 8 white balls. If 3 balls are selected  [#permalink]

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New post 30 Aug 2018, 07:31
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 purposes.


The above follows the notations and rationale taught in the GMATH method.
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Re: A jar contains 16 red balls and 8 white balls. If 3 balls are selected  [#permalink]

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New post 02 Sep 2018, 17:50
=>

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 = \({ (\frac{8*7*6)}{(1*2*3)} } / {\frac{(24*23*22)}{(1*2*3)} } = \frac{(8*7*6)}{(24*23*22)} = \frac{7}{( 23* 11 )} ≒ \frac{1}{(3*11)} ≒ 0.03\)

Therefore, the answer is B.
Answer: B
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A jar contains 16 red balls and 8 white balls. If 3 balls are selected  [#permalink]

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New post 02 Sep 2018, 19:08
MathRevolution 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



Now picking up three balls in 24 balls is a very basic question in GMAT and is equal to \(\frac{8C3}{24C3}\)..
This as shown above equals \(\frac{8*7*6}{24*23*22}\)=\(\frac{7}{253}\)..
Of course the biggest question is how do you simplify 7/253..
All methods shown above are correct..

1) 7/253 can be easily written as 7*4/250*4=28/1000=0.028 As also shown by GMATPrepNow
A decrease of 3 in 253 is perfectly fine . Of course don't decrease by 3 if the denominator is small, say make 13 as 10 in denominator as it will have a huge effect.

2) you can get the denominator to slightly friendly terms when compared to numerator 7.
253 lies between 210 and 280, so answer should be between 7/210 and 7/280.
7/210=1/30=0.033 and 7/280=1/40=0.025
So all values between 210 and 280 will be approximated to 0.03
Thus our answer is 0.03

You have to look at the way that suits you and can vary from person to person
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A jar contains 16 red balls and 8 white balls. If 3 balls are selected  [#permalink]

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New post 04 Sep 2018, 03:49
MathRevolution 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


The probability of selecting 3 white balls is:

8/24 x 7/23 x 6/22 = 1/3 x 7/23 x 3/11 = 7/23 x 1/11 = 7/253 = 0.0277 ≈ 0.03

Alternate Solution:

The total number of ways that we can draw 3 balls from 24 balls is 24C3 = 24!/(3! x 21!) = 24 x 23 x 22 / 3 x 2 x 1 = 8 x 23 x 11 = 2024.

The number of ways we can pick 3 white balls out of 8 white balls is 8C3 = 8!/(3! x 5!) = 8 x 7 x 6 / 3 x 2 x 1 = 4 x 7 x 2 = 56. The total number of ways we can pick 0 red balls out of 16 red balls is 1.

Thus, the number of ways to pick 3 white balls out of 3 picks, when there are 8 white balls and 16 red balls is by getting 3 white balls and no red balls, out of a total of 24 balls to choose from: (8C3 x 16C0) / 24C3 = 56 x 1 / 2024 = 0.0277 ≈ 0.03

Answer: B
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Re: A jar contains 16 red balls and 8 white balls. If 3 balls are selected &nbs [#permalink] 04 Sep 2018, 03:49
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