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17 Mar 2020, 01:35
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A coin of diameter 'd' is tossed which lands on a square tile of side 'a'. What is the probability that the coin will fall completely within the tile
A. \(\frac{(a − d)^2}{a^2}\)
B. \(\frac{d(2a − d)}{a^2}\)
C. \(\frac{d(2a − d)}{(a − d)^2}\)
D. \(\frac{(a − d)^2}{d(2a − d)}\)
E. \(\frac{(a − d)}{d(2a − d)}\)
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17 Mar 2020, 04:09
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Let the square be ABCD with side '\(a\)'
Note that, coin will fall inside the square completely only if the center of the circle falls on or inside the square PQRS (as shown).
Side of square PQRS = \(a - (\frac{d}{2} + \frac{d}{2}) = a - d\)
--> So, Favorable area in which coin falls completely inside the square tile = Area of square PQRS = \((a - d)^2\)
Total area = \(a^2\)
--> Required Probability = \(\frac{(a - d)^2}{a^2}\)
Option A
Note that, coin will fall inside the square completely only if the center of the circle falls on or inside the square PQRS (as shown).
Side of square PQRS = \(a - (\frac{d}{2} + \frac{d}{2}) = a - d\)
--> So, Favorable area in which coin falls completely inside the square tile = Area of square PQRS = \((a - d)^2\)
Total area = \(a^2\)
--> Required Probability = \(\frac{(a - d)^2}{a^2}\)
Option A
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17 Mar 2020, 04:11
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If the coin will fall completely within the tile, center of the coin must lie inside the square ABCD.
Hence the favourable cases =\( [a- (2*\frac{d}{2})]^2 = (a-d)^2\)
The total area where the center of the coin can lie such that some part of the coin must land in the square tile = \((a+(2*\frac{d}{2}))^2= (a+d)^2\)
Total probability = \(\frac{(a-d)^2}{ (a+d)^2}\) (None of the options is correct imo)
But if we make another assumption - center of the coin must lie inside square tile.
Then total probability =\(\frac{ (a-d)^2}{ a^2}\)
Option A
A coin of diameter 'd' is tossed which lands on a square tile of side 'a'. What is the probability that the coin will fall completely within the tile
Hence the favourable cases =\( [a- (2*\frac{d}{2})]^2 = (a-d)^2\)
The total area where the center of the coin can lie such that some part of the coin must land in the square tile = \((a+(2*\frac{d}{2}))^2= (a+d)^2\)
Total probability = \(\frac{(a-d)^2}{ (a+d)^2}\) (None of the options is correct imo)
But if we make another assumption - center of the coin must lie inside square tile.
Then total probability =\(\frac{ (a-d)^2}{ a^2}\)
Option A
A coin of diameter 'd' is tossed which lands on a square tile of side 'a'. What is the probability that the coin will fall completely within the tile
Attachments
Untitled.png [ 8.52 KiB | Viewed 10252 times ]
