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16 Sep 2014, 00:39
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C
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A bowl contains only green and blue chips. If two chips are drawn from the bowl at random and without replacement, what is the probability that both chips will be blue?
(1) The ratio of blue chips to green chips in the bowl is 3:4.
(2) There are 5 more green chips in the bowl than blue chips.
(1) The ratio of blue chips to green chips in the bowl is 3:4.
(2) There are 5 more green chips in the bowl than blue chips.
16 Sep 2014, 00:39
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Official Solution:
A bowl contains only green and blue chips. If two chips are drawn from the bowl at random and without replacement, what is the probability that both chips will be blue?
(1) The ratio of blue chips to green chips in the bowl is 3:4.
If \(B=3\) and \(G=4\), then \(P=\frac{3}{7}*\frac{2}{6}\), but if \(B=6\) and \(G=8\), then \(P=\frac{6}{14}*\frac{5}{13}\). We get two different answers, so this information is not sufficient.
(2) There are 5 more green chips in the bowl than blue chips.
So, \(G=B+5\). This information is clearly insufficient by itself.
(1)+(2) Combining the information, we have \(B:G=3:4\) and \(G=B+5\). Solving this system of equations gives us: \(B=15\) and \(G=20\). With this information, we can find the probability: \(P=\frac{15}{35}*\frac{14}{34}\). Sufficient.
Answer: C
A bowl contains only green and blue chips. If two chips are drawn from the bowl at random and without replacement, what is the probability that both chips will be blue?
(1) The ratio of blue chips to green chips in the bowl is 3:4.
If \(B=3\) and \(G=4\), then \(P=\frac{3}{7}*\frac{2}{6}\), but if \(B=6\) and \(G=8\), then \(P=\frac{6}{14}*\frac{5}{13}\). We get two different answers, so this information is not sufficient.
(2) There are 5 more green chips in the bowl than blue chips.
So, \(G=B+5\). This information is clearly insufficient by itself.
(1)+(2) Combining the information, we have \(B:G=3:4\) and \(G=B+5\). Solving this system of equations gives us: \(B=15\) and \(G=20\). With this information, we can find the probability: \(P=\frac{15}{35}*\frac{14}{34}\). Sufficient.
Answer: C
10 Oct 2017, 23:04
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This is how I approached this.
Given: Bowl contains only Blue and Green Chips. Replacement is not allowed. This is Value question and not yes/no
S1: Blue Chips : Green Chip = 3x : 4x
=> Probability of picking two blue =\(\frac{3x}{7x} * \frac{(3x-1)}{(7x-1)}\). This cannot be solved without knowing the value of x. Insufficient.
S2: Clearly insufficient. Blue and Green could be any number. The probability will be different for different cases.
S1+S2 : We have \(4x - 3x = 5 => x = 5\). Knowing x and using S1, we can now find the probablity. Hence sufficient.
Final answer: C
Given: Bowl contains only Blue and Green Chips. Replacement is not allowed. This is Value question and not yes/no
S1: Blue Chips : Green Chip = 3x : 4x
=> Probability of picking two blue =\(\frac{3x}{7x} * \frac{(3x-1)}{(7x-1)}\). This cannot be solved without knowing the value of x. Insufficient.
S2: Clearly insufficient. Blue and Green could be any number. The probability will be different for different cases.
S1+S2 : We have \(4x - 3x = 5 => x = 5\). Knowing x and using S1, we can now find the probablity. Hence sufficient.
Final answer: C
