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?

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.

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.

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 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.