Probability
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Q.

Four unit squares are chosen at random on a chessboard. What is the probability that three of them are of one colour and fourth is of opposite colour?

 A.

80/427

 B.

160/427

 C.

320/1281

 D.

640/1281

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Solution:
Option(D) is correct

The number of ways of choosing 4 squares from 64 is \(^{64}C_4\)

The possible split ups as far as colour is concerned and the number of ways of selecting the square are listed below:

White = 0, Black = 4, No. of Ways = ${^{32}C_0} × {^{32}C_4}$

White = 1, Black = 3, No. of Ways = ${^{32}C_1} × {^{32}C_3}$

White = 2, Black = 2, No. of Ways = ${^{32}C_2} × {^{32}C_2}$

White = 3, Black = 1, No. of Ways = ${^{32}C_3} × {^{32}C_1}$

White = 4, Black = 0, No. of Ways = ${^{32}C_4} × {^{32}C_0}$

The number of ways in which 3 squares are of one colour and fourth is of opposite colour is: 

$P(A) =2\times {^{32}C_1} \times {^{32}C_3}$

\(=(2\times 32)\dfrac{32\times 31\times 30}{2\times 3}\)

The total number of ways of selecting 4 squares is:

\(= P(B)= {^{64}C_4}\)

\(=\dfrac{64\times 63\times62\times61}{2\times3\times4}\)

The required probability

\(=\dfrac{P(A)}{P(B)}\)

\(=\dfrac{640}{1281}\)

Edit: For an alternative solution, check comment by Laurence.


(8) Comment(s)


Siddharth
 ()

Let the combination, $WWWB$ be a solution of which there can be 4!/3! or 4 permutations. Similarly, there exists another such solution, $BBBW$ 

Now there are 32 White squares and 32 Black squares and the prob. of selecting either sequence is, 

$P=\dfrac{32*31*30*32}{64*63*62*61}$ 

Final answer would, 

$P*4*2 = 0.49960$



Laurence
 ()

Actually, there is an easy way.

If we want to have 3 squares having the same color and the other having the opposite color, then these 3 could be either white and the fourth is black(and knowing that a chess consists of 64 squares; 32 white and 32 black) then ${^{32}C_3} \times 32=158720$ or these 3 could be black and the fourth is white. 

So, we will do the same operation as proceeds and get the same answer (158720) . 

So, the number of combinations is $2×158720=317440$. 

The total number of combinations is ${^{64}C_4}=635376$. 

Hence, the probability is,

$=\dfrac{317440}{635376}$

$=\dfrac{640}{1281}$


Chinmay Katre
 ()

How is it different than the originally provided solution ?

Chinmay Katre
 ()

Sorry....I had too solved it by your method and assumed that this would be the solution provided by the site.....I better should have checked.


Edwin Gou
 ()

Dude... I just calculate the P(A)/P(B) by myself and I found out that the answer and I got the different answer. It's 620/79422 but it's not the simplest form yet but the numerator was already different.....

I'm so confused. Can anyone explain it to me why ?


Aarti
 ()

I also checked the calculation and got the answer as $\dfrac{640}{1281}$. Same as the solution.

I guess you have made some calculation mistake.


Sur
 ()

lol idk anything about permutation and combinations ;) ... isnt there an easier method or something?



Ankur
 ()

How about an alternate solution?

We just choose 3 random squares, 2 of which are definitely of 1 color. we just have to choose the fourth square appropriately.

If there are 3 black squares selected in the first 3, we select a white as the fourth one. if there were two white and one black in the first 3 we select a black for the fourth.

What would be the numeric solution then?