Aptitude Discussion

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 |

**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.**

**Siddharth**

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**Laurence**

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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,

$=\dfrac{317440}{635376}$

$=\dfrac{640}{1281}$

How is it different than the originally provided solution ?

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**

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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 ?

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**

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

**Ankur**

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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?

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,

$P*4*2 = 0.49960$