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1) there is a square in the XY co-ordinate system with its [#permalink]
25 Sep 2003, 06:58

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1) there is a square in the XY co-ordinate system with its vertices
as {(1,1), (1,-1), (-1,-1), (-1,1)}. What is the probability that a
point picked at random within this square region will satisfy the
equation x^2+y^2<1?

2) if 2 numbers are chosen from 0-9 inclusive, what is the
probability that its product will be even?

1. Please can u explain how the equation is the origin of a circle. I do not understand.

2. I do not understand why 1-p(O,O) is used, as we need to select only one even number, the second number can be either odd or even, as we nedd the product ot be even.

(1) the equation of an origin-centered circumference is x^2+y^2=R^2, a direct use of the Pythagorean theorem.

(2) all the possible combinations for a product are EE, EO, OE, OO. The first three gives an even product; the last one gives an odd one. So, it is correct to calculate probabilities for each of the first three cases and add them up, but it is more concise and elegant to calculate the probability of the opposite case and subtract it from 1.

1. Because there is no term having x and y in the equation of circle. A general equation of circle is (X-x)**2 + (Y-y)**2=R**2. That's why it's circle having centre as origin and radius is 1.

2. There should be atleast one number should be even out of 2, if there multiplication is even. So if deduct probability of multiplication of (odd,odd) from total probability (i.e. 1), we can get desired probability.