There is a square board and i'm shooting at it. It is equally likely to hit any point on the borad. What is the probability that i hit a point that is closer to the centre than the boundary??
2006-12-01 15:14:54 · 11 answers · asked by geord 1 in Science & Mathematics ➔ Mathematics
The probability to hit a given part of the board is
(area of that part)/(area of the board)
Now what does the set of points that are closer to the center than to the boundary look like ? That's not such an easy question actually, which makes me wonder at what level that problem was given to you it is an area with four "sides" which are parabolas.
If you assume your board is the square centered at 0,0 with side lengths equal to 4, then for example the equation of the parabola bounding your domain "to the right" is 4*(1-x)=y^2.
Using symmetries, with a little integration work you'll find the area of the set of points discussed above is
4*(8^(1/2)-2)^2+32/3 *(3-8^(1/2))^(3/2)
and the area of the square is 4*4=16 so the probability you're looking for is
1/4 *(8^(1/2)-2)^2+2/3 *(3-8^(1/2))^(3/2)
which is approximately 21.9%
2006-12-01 23:50:46 · answer #1 · answered by frank m 2 · 0⤊ 0⤋
Well, if the board were 1 foot square, then the parts closest to the center all lie in a box 6 inches square centered in the middle of the board. Six inches square contains 36 square inches. One foot square contains 144 square inches, so the part of the box outside the center square is 144-36 or 108 square inches. Thus the area of the box closest to the center is 36 square inches, the part closest to the edges is 108 sqare inches. The odds are 36/144 that you will hit a part closest to the center or 25%, assuming you cannot miss the board at all. The odds are 108/144 that you will hit the area farthest from the center, or 75%.
So you have a 25% chance of hitting in the area closest to the center, or 1 in 4.
2006-12-01 23:22:34 · answer #2 · answered by Kokopelli 7 · 2⤊ 0⤋
The problem is to be solved as geometrical probability. Let the length of the square be 'l'. From the property which is taught in the geometry at school level, we know that all the points lying within the circle of radius 'l/4' having centre as the centre of the square will be closure to the centre than the boundaries of the square. Therefore the required probability will be the ratio of the area of the circle to the area of the square i.e.(pie) (l/4)**2/ (l)**2= (pie)/(16).
2006-12-02 01:16:34 · answer #3 · answered by meshu 1 · 0⤊ 1⤋
say the given square board is of length x. make a square of side x/2 having the same center as the given square board. if u hit at any point inside this smaller square, it is closer to the center than the boundary.
Hence, probability = (area of smaller square) / ( area of square board)
= (x/2)^2 / x^2
= 1/4
2006-12-02 00:04:55 · answer #4 · answered by placebo 2 · 1⤊ 0⤋
Your center vs. boundary will be 1/4 from each of the edges. The center will be 1/2 by 1/2 of the total side. 1/4*s^2 is the center. 3/4*s^2 is left for the outer portion.
25% center
75% outer
2006-12-01 23:46:12 · answer #5 · answered by J G 4 · 1⤊ 0⤋
25%. or 1/4.
if you draw another square inside the first square that the length of its rib (boundary) is half the first square,the area of that is 1/4 first square.
2006-12-02 00:26:46 · answer #6 · answered by Melika 3 · 1⤊ 0⤋
i think it would be more likely to hit the outside
2006-12-01 23:16:51 · answer #7 · answered by Secret Agent Man 3 · 0⤊ 0⤋
25% center
75% outer
2006-12-02 00:55:42 · answer #8 · answered by arpita 5 · 0⤊ 0⤋
You have not indicated the number of columns, without which calculations may not be possible
2006-12-01 23:19:00 · answer #9 · answered by cvrk3 4 · 0⤊ 2⤋
not enough information is given in your problem to solve the solution/probability.
2006-12-01 23:22:08 · answer #10 · answered by immageek 1 · 0⤊ 2⤋