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?
The number of ways of choosing 4 squares from 64 is 64C4.
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 = 32C0×32C4.
White = 1, Black = 3, No. of Ways = 32C1×32C3.
White = 2, Black = 2, No. of Ways = 32C2×32C2.
White = 3, Black = 1, No. of Ways = 32C3×32C1.
White = 4, Black = 0, No. of Ways = 32C4×32C0.
The number of ways in which 3 squares are of one colour and fourth is of opposite colour is:
P(A)=2×32 C1×32 C3.
=>2×32x32×31×30/(32×3).
The total number of ways of selecting 4 squares is:
=>P(B)=64C4.
=>(64×63×62×61)/(2×3×4).
Hence, the required probability is :-
=>P(A)/P(B).
=>640/1281.
hope my answer is correct please give a smile
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 = 32C0×32C4.
White = 1, Black = 3, No. of Ways = 32C1×32C3.
White = 2, Black = 2, No. of Ways = 32C2×32C2.
White = 3, Black = 1, No. of Ways = 32C3×32C1.
White = 4, Black = 0, No. of Ways = 32C4×32C0.
The number of ways in which 3 squares are of one colour and fourth is of opposite colour is:
P(A)=2×32 C1×32 C3.
=>2×32x32×31×30/(32×3).
The total number of ways of selecting 4 squares is:
=>P(B)=64C4.
=>(64×63×62×61)/(2×3×4).
Hence, the required probability is :-
=>P(A)/P(B).
=>640/1281.
hope my answer is correct please give a smile