The x-and y- coordinates of point P are each to be chosen at random from the set of integers 1 through 10. What is the probability that P will be in quadrant II?
The probability that point P will be in quadrant II is 0.
In quadrant II, the x-coordinate must be negative, while the y-coordinate remains positive. Since the x- and y-coordinates of point P can only be chosen from the set of integers 1 through 10, it is impossible for the x-coordinate to be negative, thereby making the probability of point P being in quadrant II equal to 0.
This choice is correct because, as mentioned, the x-coordinate cannot be negative when chosen from the set of integers 1 through 10. Consequently, point P cannot fall into quadrant II at all.
This choice implies a probability of 1/8, which suggests that there is a chance for point P to be in quadrant II. However, since the x-coordinate must be negative for quadrant II and is only selected from positive integers, this probability is incorrect.
This choice suggests a probability of 1/4, again implying that there is a possibility for point P to be in quadrant II. As established, the x-coordinate's restriction to positive integers means there is no chance for point P to occupy this quadrant, rendering this option incorrect.
This choice indicates a probability of 1/2, which also incorrectly assumes that point P could potentially be in quadrant II. Given the constraints of the coordinate selections, the probability is definitively 0, making this option invalid.
In conclusion, the coordinates of point P can only fall within the positive range of integers 1 through 10. Since the criteria for being in quadrant II cannot be met due to the impossibility of a negative x-coordinate, the probability is unequivocally 0. Thus, understanding the definition of quadrants and the restrictions of coordinate selection leads to the recognition that point P cannot be in quadrant II.
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