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A certain jar contains 5 marbles, r of which are red. If two marbles a
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08 Mar 2021, 07:41
08 Mar 2021, 07:41
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A certain jar contains 5 marbles, r of which are red. If two marbles are to be selected at random, and the probability that both marbles will be red is \(\frac{1}{10}\) , what is the value of r?
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Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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5
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Re: A certain jar contains 5 marbles, r of which are red. If two marbles a
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08 Mar 2021, 08:19
08 Mar 2021, 08:19
1
Carcass wrote:
A certain jar contains 5 marbles, r of which are red. If two marbles are to be selected at random, and the probability that both marbles will be red is \(\frac{1}{10}\) , what is the value of r?
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Approach 1
Number of ways of picking \(2\) marbles from \(r\) is \(^rC_2\)
Total number of ways of picking \(2\) marbles from \(5\) is \(^5C_2\)
So, \(\frac{^rC_2}{^5C_2} = \frac{1}{10}\)
Upon solving further;
\(r(r-1) = 2\)
\(r^2 - r - 2 = 0\)
\((r+1)(r-2) = 0\)
i.e. \(r = -1 or 2\)
\(r\) can only be \(2\)
Approach 2
Plug in the values
A. \(\frac{^1C_2}{^5C_2}\), doesn't make sense! How can we pick 2 marbles when there is only 1?
B. \(\frac{^2C_2}{^5C_2} = \frac{1}{10}\), we have our answer!
Don't need to check C, D, and E
Hence, option B
General Discussion
A certain jar contains 5 marbles, r of which are red. If two marbles a
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29 Apr 2021, 15:10
29 Apr 2021, 15:10
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I solved this by setting up a probability equation:
Since it was 'and' you multiply the two events, which equal \(\frac{1}{10}\)
\((\frac{x}{5})*(\frac{x-1}{4}) = \frac{1}{10}\)
Cross multiply and simplify to same equation as above: x\(^2\)-x-2=0
You can't have just one negative red ball in this scenario, so the other solution to this quadratic is 2. Boom.
Since it was 'and' you multiply the two events, which equal \(\frac{1}{10}\)
\((\frac{x}{5})*(\frac{x-1}{4}) = \frac{1}{10}\)
Cross multiply and simplify to same equation as above: x\(^2\)-x-2=0
You can't have just one negative red ball in this scenario, so the other solution to this quadratic is 2. Boom.
