- #1
Kumar8434
- 121
- 5
I thought of this problem:
The roads AB and CD have block E in common. There are 6 blocks in road AB and 13 blocks in road CD.
Someone has planted a mine in some of the blocks. He gives us this information:
1. There's only one mine in road AB in one of its blocks.
2. There's only one mine in road CD in one of its blocks.
Clearly, if there's a mine in block E then it satisfies both of the above conditions, hence there are no more mines in the roads. And, if there's no mine in E then there must be two mines, one in each road.
What is the probability of that there's a mine in E? If we consider only road AB, then the probability that E has the mine is ##\frac{1}{6}##. And, if we consider road CD, then it's ##\frac{1}{13}##. How do we combine these two probabilities? I don't think it should be the product of these probabilities, because the product is smaller than either of the probabilities, and intuition tells us that the mine is more likely to be in E than the rest of the blocks in road CD and less likely to be in E than the rest of the blocks in road AB. I think it should be the arithmetic mean of the probabilities. But how?
The roads AB and CD have block E in common. There are 6 blocks in road AB and 13 blocks in road CD.
Someone has planted a mine in some of the blocks. He gives us this information:
1. There's only one mine in road AB in one of its blocks.
2. There's only one mine in road CD in one of its blocks.
Clearly, if there's a mine in block E then it satisfies both of the above conditions, hence there are no more mines in the roads. And, if there's no mine in E then there must be two mines, one in each road.
What is the probability of that there's a mine in E? If we consider only road AB, then the probability that E has the mine is ##\frac{1}{6}##. And, if we consider road CD, then it's ##\frac{1}{13}##. How do we combine these two probabilities? I don't think it should be the product of these probabilities, because the product is smaller than either of the probabilities, and intuition tells us that the mine is more likely to be in E than the rest of the blocks in road CD and less likely to be in E than the rest of the blocks in road AB. I think it should be the arithmetic mean of the probabilities. But how?