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14 Jan 2021, 04:30
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C
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Difficulty:
95%
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Question Stats:
40% (02:30) correct
60%
(02:26)
wrong
based on 68
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What is the probability that two 1 by 1 squares selected randomly from a chess board will have only one common corner?
A. 7/288
B. 2/63
C. 7/144
D. 7/126
E. 7/72
A. 7/288
B. 2/63
C. 7/144
D. 7/126
E. 7/72
15 Jan 2021, 01:13
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Bunuel
Total ways to select 2 squares randomly = \(64C2=\frac{64*63}{2}=32*63\)
Ways to select two squares such that only one corner is common.
1) X - When the corner ones are picked as one of the two, so 4 of them : Only one other square has just one corner common = \(4*1=4\)
2) Y - When the square from the extreme edges less X are picked as one of the two, so 4*6=24 of them : Only 2 other squares have just one corner common = \(24*2=48\)
3) Z - When any other square less X and Y are picked as one of the two, so 8*8-4-24=36 of them : 4 other squares have just one corner common = \(36*4=144\)
Total = \(4+48+144=196\)
Look at point A : It is counted twice as common corner between P & S and S & P. So distinct ways = \(\frac{196}{2}=98\)
Probability = \(\frac{98}{32*63} = \frac{14}{32*9}=\frac{7}{16*9}=\frac{7}{144}\)
C
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07 Feb 2021, 00:27
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Bunuel
Got another way to solve the question. Hope it makes sense.
We need to choose two squares out of 64 so that they share only one corner.
On visualization, we see that this can happen if we move diagonally - either top left to bottom right or top right to bottom left or vice-a-versa i.e. we have a total of 4 ways to move for selection of two squares. However, the diagonal selection from top left to bottom right is same as selection from bottom right to top left. So, we either simply consider 2 ways of movement OR 4 ways of movement dividing by 2 for repetition.
Also. we have a total of 9*9 = 81 corner, out of which only some satisfy the condition. We can find the required number of squares in two ways:
1. Count individually the corners that are inside the chess board - ones which are shared by 4 squares i.e. 7*7 = 49
2. Subtract the outside corners that have only either 2 squares sharing one or the 4 corners of the big chess board square. 9*9 - 4*1 - 7*4 = 49
Now ways of selecting 2 squares in total = \(64_{C_2}\)
Required ways to select the 2 squares = \(49_{C_1}\) but this has to be multiplied by 2 to cover both the direction as mentioned above.
(1 represents a corner and thus selecting one would give us two squares)
P = \(\frac{49_{C_1}}{64_{C_2}} * 2 = \frac{7}{144}\)
Answer C.
) e.g. the diagonally opposite squares far separated by each other. Here no corner is counted twice or thrice.
