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8 red card number 1-8 put in one box， 8 blue card number 1

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8 red card number 1-8 put in one box， 8 blue card number 1

by Max@Math Revolution » Fri Jun 10, 2016 10:14 pm
8 red card number 1-8 put in one box， 8 blue card number 1-8，put in another box, take one card from each of the box，what is the probability that the number match each other?
A. 1/2
B. 1/4
C. 1/8
D. 1/16
E. 1/64

*An answer will be posted in 2 days.
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by Gmatasap » Sat Jun 11, 2016 7:59 am
The probability of getting any of the number 1-8 card is 1. Since any specific number has not be provided, any number will do.
The probability of getting the same number in the draw of blue card is 1/8

The required probability is
[spoiler]1*1/8=1/8[/spoiler]

Actually modifying the question, can you help with 2 cards were drawn from each box, what is the probability that we get at least one pair of numbers?
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by Max@Math Revolution » Mon Jun 13, 2016 8:52 am
From (1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8), the probability of the number match each other is 8. Hence, 8C1/(8C1)(8C1)=1/8. So, the correct answer is C.
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by 800_or_bust » Mon Jun 13, 2016 9:23 am
Max@Math Revolution wrote:8 red card number 1-8 put in one box， 8 blue card number 1-8，put in another box, take one card from each of the box，what is the probability that the number match each other?
A. 1/2
B. 1/4
C. 1/8
D. 1/16
E. 1/64

*An answer will be posted in 2 days.
The probability of selecting any card from box A is 8/8 = 1. The probability of selecting the same card from Box B is 1/8. Thus, the probability of selecting two matching cards is just 1/8.
800 or bust!
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by 800_or_bust » Mon Jun 13, 2016 9:24 am
Gmatasap wrote:The probability of getting any of the number 1-8 card is 1. Since any specific number has not be provided, any number will do.
The probability of getting the same number in the draw of blue card is 1/8

The required probability is
[spoiler]1*1/8=1/8[/spoiler]

Actually modifying the question, can you help with 2 cards were drawn from each box, what is the probability that we get at least one pair of numbers?
Wouldn't it just be 1/56 (assuming no replacement) or 1/64 (with replacement) using the same logic?

Edit: Nevermind, I misread your question. Thought you meant probability that both were the same. Let me think about that.

Edit 2: Hmmm, this is tricky. If the cards are replaced each time, I believe it would just be 15/64. The probability of not getting a match either time would be 7/8 x 7/8 = 49/64. So, the probability of at least one match would be 1 - 49/64 = 15/64. Without replacement, it gets tough. The first draw from Box B could have removed the card that was drawn from Box A on the second draw, in which case the probability of a match is 0. Otherwise, the probability would of a match would be 1/7. I'm guessing you would need to determine the relative probabilities of both of these cases.
800 or bust!
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by 800_or_bust » Mon Jun 13, 2016 9:41 am
800_or_bust wrote:
Gmatasap wrote:The probability of getting any of the number 1-8 card is 1. Since any specific number has not be provided, any number will do.
The probability of getting the same number in the draw of blue card is 1/8

The required probability is
[spoiler]1*1/8=1/8[/spoiler]

Actually modifying the question, can you help with 2 cards were drawn from each box, what is the probability that we get at least one pair of numbers?
Wouldn't it just be 1/56 (assuming no replacement) or 1/64 (with replacement) using the same logic?

Edit: Nevermind, I misread your question. Thought you meant probability that both were the same. Let me think about that.

Edit 2: Hmmm, this is tricky. If the cards are replaced each time, I believe it would just be 15/64. The probability of not getting a match either time would be 7/8 x 7/8 = 49/64. So, the probability of at least one match would be 1 - 49/64 = 15/64. Without replacement, it gets tough. The first draw from Box B could have removed the card that was drawn from Box A on the second draw, in which case the probability of a match is 0. Otherwise, the probability would of a match would be 1/7. I'm guessing you would need to determine the relative probabilities of both of these cases.
My best guess at it...

The probability that the second card drawn from Box A matches the first card drawn from Box B equals the probability that the first card drawn from Box A did not match the first card drawn from Box B, times the probability that the second card drawn from Box A does match the first card drawn from Box B given that the first card drawn from Box A did not match the first card drawn from Box B. Thus, the probability of this would be 7/8 x 1/7 = 1/8. And the probability that the second card drawn from Box A does not match the first card drawn from Box B would be 1 - 1/8 = 7/8.

Thus, the overall probability that the second cards match would be (1/8)(0) + (7/8)(1/7) = 1/8. And the probability that at least one of the two cards drawn match would again be 15/64.
800 or bust!
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