A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?
$$(1)The\ probability\ that\ one\ ball\ selected\ at\ random\ is\ either\ red\ or\ white\ is\ \frac{2}{3}$$
$$(2)There\ are\ nineballs\ in\ the\ barrel.$$
OA C
Source: Princeton Review
A barrel contains only red balls, white balls, and brown
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Say there r numbers of red, w numbers of white, and b numbers of brown balls.BTGmoderatorDC wrote:A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?
$$(1)The\ probability\ that\ one\ ball\ selected\ at\ random\ is\ either\ red\ or\ white\ is\ \frac{2}{3}$$
$$(2)There\ are\ nineballs\ in\ the\ barrel.$$
OA C
Source: Princeton Review
Thus, the total number of balls = r + w + b
We have to find out:
Probability that two brown balls will be selected = bC2 / (r + w + b)C2 = [b*(b - 1) / (r + w + b)*(r + w + b - 1)]*(1*2/1*2)
= [b*(b - 1) / (r + w + b)*(r + w + b - 1)]
Let's take each statement one by one.
(1)The probability that one ball selected at random is either red or white is 2/3.
=> The probability that one ball selected at random is brown = 1 - 2/3 = 1/3.
=> bC1 / (r + w + b)C1 = 1/3
b / (r + w + b) = 1/3
=> r + w + b = 3b
Let's find out the value of [b*(b - 1) / (r + w + b)*(r + w + b - 1)] .
[b*(b - 1) / (r + w + b)*(r + w + b - 1)] = [b*(b - 1) / (3b)*(3b - 1)] = (b - 1) / (3b - 1)
We do not have the value of b. Insufficient.
(2)There are nine balls in the barrel.
=> r + w + b = 9.
Can't get the value of [b*(b - 1) / (r + w + b)*(r + w + b - 1)]. Insufficient.
(1) and (2) together
We have r + w + b = 9 = 3b => b = 3
Probability that two brown balls will be selected = (b - 1) / (3b - 1) = (3 - 1) / (3*3 - 1) = 2/8 = 1/4. Sufficient.
The correct answer: [spoiler][/spoiler]
Hope this helps!
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Let there be x red balls, y, white balls and z brown balls.
From the statement 1, we have that the probability of selection either red or white is 2/3
or (x+y)/(x+y+z)=2/3 . . . We see that x+y=2z or total number of balls = 3z.
So, the probability of drawing 2 brown balls will be zc2/3zc2 or z(z-1)/(3z-1)(3z-2) . . . Thus till we know z we can find the probability.
The statement 2 says that x+y+z=9 so, not sufficient.
Combining we see that 3z=9 and z=3 so probability will be 3/9 or 1/3.
Hence, C is the correct answer. Regards!
From the statement 1, we have that the probability of selection either red or white is 2/3
or (x+y)/(x+y+z)=2/3 . . . We see that x+y=2z or total number of balls = 3z.
So, the probability of drawing 2 brown balls will be zc2/3zc2 or z(z-1)/(3z-1)(3z-2) . . . Thus till we know z we can find the probability.
The statement 2 says that x+y+z=9 so, not sufficient.
Combining we see that 3z=9 and z=3 so probability will be 3/9 or 1/3.
Hence, C is the correct answer. Regards!
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\[?\,\,\,\, = \,\,\,P\left( {2\,\,br\,\,{\text{out}}\,\,{\text{of}}\,\,2\,\,{\text{sequential}}\,\,{\text{extractions}}\,\,{\text{no}}}\,\,{\text{replacements}} \right)\,\,\,\,\,\,\,\]BTGmoderatorDC wrote:A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?
(1)The probability that one ball selected at random is either red or white is 2/3
(2)There are nine balls in the barrel.
Source: Princeton Review
\[\left( 1 \right)\,\,P\left( {re\,\,{\text{or}}\,\,wh\,\,{\text{out}}\,\,{\text{of}}\,\,1\,\,{\text{extraction}}} \right) = \frac{2}{3}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,P\left( {br\,\,{\text{out}}\,\,{\text{of}}\,\,1\,\,{\text{extraction}}} \right) = \frac{1}{3}\]
\[\left\{ \begin{gathered}
\,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {1,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0 \hfill \\
\,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {2,2,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? \ne 0 \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,\,r + w + b\,\, = \,\,9\]
\[\left\{ \begin{gathered}
\,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {7,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 0 \hfill \\
\,{\text{Take}}\,\,\,\left( {r,w,b} \right) = \left( {6,1,2} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,? \ne 0 \hfill \\
\end{gathered} \right.\]
\[\left( {1 + 2} \right)\,\,\,br = \frac{1}{3}\left( 9 \right) = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\, = \,\,{\text{unique}}\,\,\,\,\left( {\frac{3}{9} \cdot \frac{2}{8}} \right)\]
The above follows the notations and rationale taught in the GMATH method.
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BTGmoderatorDC wrote:A barrel contains only red balls, white balls, and brown balls. If two balls are selected at random without replacing each ball, what is the probability that two brown balls will be selected?
$$(1)The\ probability\ that\ one\ ball\ selected\ at\ random\ is\ either\ red\ or\ white\ is\ \frac{2}{3}$$
$$(2)There\ are\ nineballs\ in\ the\ barrel.$$
Statement One Alone:
The probability that one ball selected at random is either red of white is 2/3.
That means the probability that one ball selected at random is brown is 1/3. However, since we don't know how many brown balls there are, we can't determine the probability that two brown balls will be selected without replacement. For example, if there are 2 brown balls in the barrel (before any selection), the probability that two brown balls will be selected without replacement is 2/6 x 1/5 = 1/3 x 1/5 = 1/15. However, if there are 3 brown balls in the barrel (before any selection), the probability that two brown balls will be selected without replacement is 3/9 x 2/8 = 1/3 x 1/4 = 1/12.
Statement one alone is not sufficient.
Statement Two Alone:
There are nine balls in the barrel.
Without knowing the number of balls of each color (especially the color brown) or the probability of selecting a certain color (especially the color brown), we can't determine the probability that two brown balls will be selected without replacement.
Statement two alone is not sufficient.
Statements One and Two Together:
With the two statements, we can determine that there are 3 brown balls in the barrel (before any selection) since 1/3 x 9 = 3. Knowing that there are 3 brown balls in the barrel (before any selection), we can also determine the probability that two brown balls will be selected without replacement. That probability is is 3/9 x 2/8 = 1/3 x 1/4 = 1/12.
Answer: C
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