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A bag contains 6 red chips numbered 1 through 6, respectivel
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05 Aug 2018, 14:28
05 Aug 2018, 14:28
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Question Stats:
75% (01:11) correct
25% (01:14) wrong based on 36 sessions
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A bag contains 6 red chips numbered 1 through 6, respectively, and 6 blue chips numbered 1 through 6, respectively. If 2 chips are to be picked sequentially from the bag of 12 chips, without replacement, what is the probability of picking a red chip and then a blue chip with the same number?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: OA
\(\frac{1}{22}\)
Re: A bag contains 6 red chips numbered 1 through 6, respectivel
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07 Aug 2018, 16:46
07 Aug 2018, 16:46
2
Expert Reply
Explanation
The trap answer in this problem is \(\frac{1}{6}\times \frac{1}{6}\).
This is not the answer to the question being asked—rather, this is the answer to the question, “What is the probability of picking a red chip and then a blue chip that both have #3?” (or any other specific number). This is a more specific question than the one actually asked. In the question, asked, there are six possible ways to fulfill the requirements of the problem, not one, because the problem does not specify whether the number should be 1, 2, 3, 4, 5, or 6.
Thus, any of the 6 red chips is acceptable for the first pick. However, on the second pick, only the blue chip with the same number as the red one that was just picked is acceptable (the chip must “match” the first one picked). Thus:
\(\frac{6}{12}\times \frac{1}{11}=\frac{1}{22}\).
The trap answer in this problem is \(\frac{1}{6}\times \frac{1}{6}\).
This is not the answer to the question being asked—rather, this is the answer to the question, “What is the probability of picking a red chip and then a blue chip that both have #3?” (or any other specific number). This is a more specific question than the one actually asked. In the question, asked, there are six possible ways to fulfill the requirements of the problem, not one, because the problem does not specify whether the number should be 1, 2, 3, 4, 5, or 6.
Thus, any of the 6 red chips is acceptable for the first pick. However, on the second pick, only the blue chip with the same number as the red one that was just picked is acceptable (the chip must “match” the first one picked). Thus:
\(\frac{6}{12}\times \frac{1}{11}=\frac{1}{22}\).
Re: A bag contains 6 red chips numbered 1 through 6, respectivel
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14 Jul 2019, 21:44
14 Jul 2019, 21:44
2
I did it this way:
P( Red)*P(any number from 1 to 6)*P( Blue chip, given that one is already drawn and not replaced)*p( finding the number which appeared on the red chip).
(1/2)*(1)(6/11)(1/6)=1/22
P( Red)*P(any number from 1 to 6)*P( Blue chip, given that one is already drawn and not replaced)*p( finding the number which appeared on the red chip).
(1/2)*(1)(6/11)(1/6)=1/22
