|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
26 May 2017, 04:41
Kudos
Bookmarks
There are \(10\) red and blue marbles in a box. If two marbles are randomly taken out from the box, is the probability of picking two red marbles greater than \(1/3\)?
(1) The probability of getting both blue marbles is less than \(2/3\)
(2) There are more red marbles than blue marbles in the box.
(1) The probability of getting both blue marbles is less than \(2/3\)
(2) There are more red marbles than blue marbles in the box.
26 May 2017, 04:42
Kudos
Bookmarks
Official Solution:
When you take out the marbles, you use \(_{n}C_{r}\), the combination. You can use the 1st step of the variable approach and modify the original condition and the question. If you assume the number of red marbles as \(r\), and the number of blue marbles as \(b\), you get \(\frac{rC2}{10C2} > \frac{1}{3}\)?, \(\frac{\frac{r(r-1)}{2!}}{\frac{(10)(9)}{2!}} > \frac{1}{3}\)?, or \(\frac{r(r-1)}{(10)(9)} > \frac{1}{3}\)?. If you multiply both sides diagonally, you get \(r(r - 1) > 30\)?, then \(r=7, 8, 9, 10\)?.
In the original condition, there are 2 variables \((r, b)\) and 1 equation \((r+b=10)\). In order to match the number of variables and the number of equations, there must be 1 more equation. Therefore, D is most likely to be the answer. In the case of con 1), you get \(\frac{bC2}{10C2} < \frac{2}{3}\) , \(\frac{\frac{b(b-1)}{2!}}{\frac{(10)(9)}{2!}} < \frac{2}{3}\), or \(\frac{b(b-1)}{(10)(9)} < \frac{2}{3}\). If you multiply both sides diagonally, you get \(b(b - 1) < 60\), and \(b = 1, 2, 3, \ldots .,7,8\), so \(r = 9, 8,\ldots .,3,2\). When \(r=6\) is no but \(r=7\) is yes, it is not sufficient.
In the case of con 2), from \(r > b\), you get \(r=6, 7, 8, 9, 10\). When \(r=6\) is no but \(r=7\) is yes, it is not sufficient. Even by solving con 1) & con 2), \(r = 6\) is no but \(r = 7\) is yes, hence it is not sufficient. The answer is E.
Answer: E
When you take out the marbles, you use \(_{n}C_{r}\), the combination. You can use the 1st step of the variable approach and modify the original condition and the question. If you assume the number of red marbles as \(r\), and the number of blue marbles as \(b\), you get \(\frac{rC2}{10C2} > \frac{1}{3}\)?, \(\frac{\frac{r(r-1)}{2!}}{\frac{(10)(9)}{2!}} > \frac{1}{3}\)?, or \(\frac{r(r-1)}{(10)(9)} > \frac{1}{3}\)?. If you multiply both sides diagonally, you get \(r(r - 1) > 30\)?, then \(r=7, 8, 9, 10\)?.
In the original condition, there are 2 variables \((r, b)\) and 1 equation \((r+b=10)\). In order to match the number of variables and the number of equations, there must be 1 more equation. Therefore, D is most likely to be the answer. In the case of con 1), you get \(\frac{bC2}{10C2} < \frac{2}{3}\) , \(\frac{\frac{b(b-1)}{2!}}{\frac{(10)(9)}{2!}} < \frac{2}{3}\), or \(\frac{b(b-1)}{(10)(9)} < \frac{2}{3}\). If you multiply both sides diagonally, you get \(b(b - 1) < 60\), and \(b = 1, 2, 3, \ldots .,7,8\), so \(r = 9, 8,\ldots .,3,2\). When \(r=6\) is no but \(r=7\) is yes, it is not sufficient.
In the case of con 2), from \(r > b\), you get \(r=6, 7, 8, 9, 10\). When \(r=6\) is no but \(r=7\) is yes, it is not sufficient. Even by solving con 1) & con 2), \(r = 6\) is no but \(r = 7\) is yes, hence it is not sufficient. The answer is E.
Answer: E
Moderator:
