Latest Elitmus Aptitude Question SOLUTION: Two squares are chosen on a chessboard at random. What is the probability that they have a side in common? a) 1/18 b) 64/4032 c) 63/64 d) 1/9

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(#M40155908) ELITMUS Numerical Ability question Keep an EYE

Two squares are chosen on a chessboard at random. What is the probability that they have a side in common?
a) 1/18
b) 64/4032
c) 63/64
d) 1/9

Asked In Elitmus Rizul (8 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

tejaswi  (8 years ago)

Ans: a) 1/18

In 64 squares, there are:

(1) 4 at-corner squares, each has ONLY 2 squares each having a side in common with...
(2) 6*4 = 24 side squares, each has ONLY 3 squares such that each has a side in common with...
(3) 6*6 = 36 inner squares, each has 4 squares such that each has a side in common with...
So we have the calculation:

P = (4/64)*(2/63) + (24/64)*(3/63)+ (36/64)*(4/63)
P = 1/18

Is this solution Helpfull? Yes (20) | No (3)  

vivek singh  (8 years ago)

Solution:

Sample space will be total ways of selecting 2 squares (order doesn't matter) out of available 64 squares.
So sample space =64C2

Now there are 7 unique adjacent square sets in each row and each column.
i.e. favourable cases will be 7×(8 rows + 8 columns) = 112.
Hence Required Probability=(
Favourable cases
Sample Space
)
=
112
64C2

= 1/18

Is this solution Helpfull? Yes (13) | No (2)  

himani1  (8 years ago)

Ana is 1/9
Solution:
4 squares are corner ones which will have 2 adjacent squares each. So, no of ways of choosing corner square and an adjacent square are 4C1*2C1
Similarly, ways of choosing an edge square and it's adjacent square are 24C1*3C1

Remaining squares=64-4-24=36
Ways of choosing middle square and it's adjacent squares are 36C1*4C1

Total no of ways of choosing any two squares is 64C2

Therefore probability= (8+72+144)/64C2
=1/9

Is this solution Helpfull? Yes (1) | No (4)  

diti  (8 years ago)

Answer is 1/9.
4 squares at corner share 2 squares. So 4*2=8
Remaining 24 sidemost squares share 3 squares 24*3=72
Remaining 36 squares share 4 squares so 36*4=144
total no. of ways = 8+72+144=224
total no of ways selecting 2 from 64 squares is 64C2
so probability=224/64c2=(1/9)

Is this solution Helpfull? Yes (1) | No (4)  

Mallikarjuna Reddy  (8 years ago)

I think its 2/9
sample space is 64c2
consider single row having 8 squares let number'em as 1,2,3..8
for a single row we have selecting two squares is 8*7 ways
but,(1,2)(2,1) are same bcz order is not important so total ways is (8*7)/2=>28
for 8 rows we have 28*8 selections
similarily
for columns we have 28*8 selections
so total=28*8+28*8=>448
probability is 448/64c2=>2/9

Is this solution Helpfull? Yes | No (5)  

sapna sharma  (7 years ago)

Ans-(a)1/18

Is this solution Helpfull? Yes | No (1)  

Ria Vohra  (7 years ago)

Sample space will be total ways of selecting 2 squares (order doesn't matter) out of available 64 squares.
So sample space =64C2

Now there are 7 unique adjacent square sets in each row and each column.
i.e. favourable cases will be 7×(8 rows + 8 columns) = 112.
Hence Required Probability=(
Favourable cases
Sample Space
)
=
112
64C2

= 1/18

Is this solution Helpfull? Yes | No   

shiv  (4 years ago)

if we choose a particular square then the number of ways for choosing another square on the same side are : 7 + 7 = 14
choosing such a pair in whole chessboard = 8 * 14 = 112
now probability = 112/64c2 = 1/8.

Is this solution Helpfull? Yes | No   

tarun rajpoot  (4 years ago)

Total number of squares = 64

Two squares can be selected in 64C2 ways

In each column, there are 7 pairs of adjacent squares where each pair share 1 side in common.
Total such pairs = 8×7 = 56

In each row, there are 7 pairs of adjacent squares where each pair share 1 side in common.
Total such pairs = 8×7 = 56

Therefore, favourable cases = 56 + 56 = 112

Required probability
=
112/
64
C
2
=
112/
2016
=
1/
18

Is this solution Helpfull? Yes | No   

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