Question

On a chessboard, small squares are either black or white, set alternately. Three pawns are placed at random on three squares of the chessboard. The probability that two are on the squares of the same color, is

A.\[\dfrac{8}{{21}}\]
B.\[\dfrac{5}{{21}}\]
C.\[\dfrac{{16}}{{21}}\]
D.None of these

Answer 1

Hint: At first we will find the total number of ways we can place 3 pawns on any box of the chessboard then in a similar way number of placement of the pawn such that two of the pawns are on one color and the third pawn on the other color, by making two different cases that 2 pawns are on white and one on black and the case with 2 pawns on black and third on white.
Now, using all these we can easily find the required probability with the formula of the probability of any event i.e. equal to $\dfrac{{favourable\,events}}{{total\,events}}$.

Complete step-by-step answer:
We know the combination of ‘r’ elements, i.e. the number of ways of selecting any ‘r’ elements out of total ‘n’ elements is given by the ${}^n{C_r}$ and this equal to $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ i.e.
$ \Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
It is well known that there are 64 small squares on a chessboard containing 32 black and 32 white colored squares.
Now, the number of ways of any spots for the three pawn at random $ = \,{}^{64}{C_3}$
Now let the pawns are on 2 black spots and 1 white, then the number of ways for such placements will be equal to $\,{}^{32}{C_2}\, \times \,{}^{32}{C_1}$
Similarly when pawns are on 2 white spots and 1 black, then the number of such placements will be $\,{}^{32}{C_2}\, \times \,{}^{32}{C_1}$
Now, the probability such that the 2 pawns are on the same color and 1 on another
$ = \,\dfrac{{\,{}^{32}{C_2}\, \times \,{}^{32}{C_1} + {}^{32}{C_2}\, \times \,{}^{32}{C_1}}}{{{}^{64}{C_3}\,}}$
Simplifying the numerator, we get
$ = \,\dfrac{{\,2{}^{32}{C_2}\, \times \,{}^{32}{C_1}}}{{{}^{64}{C_3}\,}}$
Now, using ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$ = \,\dfrac{{\,2\dfrac{{32!}}{{2!\left( {32 - 2} \right)!}}\, \times \,\dfrac{{32!}}{{1!\left( {32 - 1} \right)!}}}}{{\dfrac{{64!}}{{3!\left( {64 - 3} \right)!}}\,}}$
On simplification of the denominator of each term, we get
\[ = \,\dfrac{{\,2\dfrac{{32!}}{{2!\left( {30} \right)!}}\, \times \,\dfrac{{32!}}{{1!\left( {31} \right)!}}}}{{\dfrac{{64!}}{{3!\left( {61} \right)!}}\,}}\]
Now we’ll use $n! = n(n - 1)!$ , we get

\[ = \,\dfrac{{\,2\dfrac{{32 \times 31 \times 30!}}{{2!\left( {30} \right)!}}\, \times \,\dfrac{{32 \times 31!}}{{1!\left( {31} \right)!}}}}{{\dfrac{{64 \times 63 \times 62 \times 61!}}{{3!\left( {61} \right)!}}\,}}\]
On simplifying , we get
\[ = \,\dfrac{{\,32 \times 31 \times 32 \times 3 \times 2 \times 1}}{{64 \times 63 \times 62\,}}\]
Dividing the numerator and the denominator with \[64 \times 62\,\]
\[ = \,\dfrac{{\,16 \times 3}}{{63\,}}\]
Dividing the numerator and the denominator with \[3\], we get
\[ = \,\dfrac{{\,16}}{{21\,}}\]
Option(C) is correct.

Note: While finding the cases most of the students assume just one case of placing the pawns i.e. either 2 pawns are on white and one on black or the case with 2 pawns on black and third on white, so number their number of favorable events changes which result in the probability with error, so remember to count in all the cases to get a more accurate answer.