Dear MALAY,

ur first case x is left of y is the same case as y is right of x

So the answer remains same

in ur second case x is right of y ,

the other three expressions of the binomial expression gives the answer ,

so the answer is same as (Lx+Ly)⁵ - previous answer

Putting Lx,Ly as 1/3 and 1/2

Answer = 79/288

 

in ur third case , i feel it is not at all possible for x and y to be at the same place if there r an odd number of moves........it is because if x moves 3 paces to right or left , it means that acc to the probability , 1 should be to its left and two to its right but in 2 paces of y , one should be to the left then the other should be to the right.......so for an odd number of cases , it is not possible for x and y to be there at the same place.....

 

As far as binomial expansion is concerned ,

Let us take the example of a toss of coin with probability of heads p and probability of tails q

Possible cases for

One toss = H,T

Two tosses = (H,H),(H,T),(T,H),(T,T)

Three tosses = similar but third toss will also be considered

 

 

For n tosses , a similar possibility cant be easily derived

So in the expansion ,

(p+q)ⁿ = nC0 pⁿ+nC1 p^n-1q+ nC2 p^n-2q².........so on till n+1 th term

 

In this expansion , we take into consideration the powers of p or q as per the condition

In this expansion , powers of p and q represent the no. of cases in which p is true and no.of cases in which q is true in the given n number of chances.....

So u may accordingly solve the question by considering that x moves to left more than y for y to remain to right of x and no. of moves of y to the left must be more for x to be on the right of left.....

If u have any doubt ....plz reply