saarthak299 wrote:
Bunuel wrote:
There are 12 balls in an urn, out of which 4 balls are picked up at random. Is the probability of all the 4 balls being red greater than 1/33?
(1) If 2 balls are picked up, the probability of both being red is 5/33 -->\(\frac{red}{12}*\frac{(red - 1)}{11} = \frac{5}{33}\) . We can get the number of red balls from that equation and answer the question. Sufficient.
(2) There are 7 blue balls. This implies thatat most there are 5 red balls. So, the probability of all the 4 balls being red isat most
(1) If 2 balls are picked up, the probability of both being red is 5/33 -->\(\frac{red}{12}*\frac{(red - 1)}{11} = \frac{5}{33}\) . We can get the number of red balls from that equation and answer the question. Sufficient.
(2) There are 7 blue balls. This implies thatat most there are 5 red balls. So, the probability of all the 4 balls being red isat most
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