In an extra-sensory-perception experiment, a blindfolded subject has two rows of blocks before him. Each row has blocks numbered 1 to 10 arranged in random order. The subject is to place one hand on a block in the first row and then try to place his other hand on the block having the same numerical in the second row. If the subject has no ESP, what is the probability of his making a match on the first try?
1:10
How did you get that?
I am thinking it should be 1/100, multiplying by the 1/10 x 1/10 (the probability of the identical numbers in each row)
To determine the probability of the subject making a match on the first try, let's break down the problem.
There are two key components affecting the probability: the number of blocks and the random order they are arranged in.
In the first row, there are 10 blocks (numbered 1 to 10). The subject randomly chooses one block in the first row.
In the second row, there are also 10 blocks (again numbered 1 to 10). The chance of the subject randomly choosing the block with the same number is 1 out of 10.
Therefore, the probability of making a match on the first try is 1 out of 10.
To summarize, if the subject has no extra-sensory perception (ESP) and is blindly selecting blocks from the first row, the probability of making a match on the first try is 1 out of 10.