each second a particle x , which is constrained to move in a straight line , moves unit distance to the left or to the right with respective probabilities 1/3 and 2/3. another particle y, starting from the same initial positions and move independently of x, but in the same straight line, each second moves with unit distance to the right or to the left with probabilities both equal to 1/2. the probability that after 5 seconds, y will be in a position to the right of x ?

In this question , the probability of x moving to the right is more than what y moves to the right....so the deciding factor for the possibility of y being on the right of x is the no. of times x moves to the left

Let Lx and Ly be the probability of x and y to move to the left

Probability that x moves more towards left than y is given by binomial expansion

Prob. =⁵C₀Lx⁵+⁵C₁Lx⁴Ly + ⁵C₂Lx³Ly²

Therefore by putting Lx=1/3 and Ly=1/2

We get prob. = 31/243

 

If u have any doubt or any other answer....plz reply

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

You ask for expert and i am here. Sorry for the delay friends, but this question is a case of binomial probability distribution. What we are suppose to do is to find probabilities for general cases and then add them for all the cases. Instead of finding all the cases first and then finding there probabilities.

Consider right side displacement as +1 nad left side as -1

After five moves x will be at +5,+3,+1,-1,-3,-5 depending upon how many right or left moves he makes.

 

In case he makes i right moves and 5-i left moves his position will be at 2i-1 with a probability of C(5,i)(2/3)ⁱ(1/3)^5-i  [C(5,i) ways to make i right and 5-i left moves]

I can write Px(2i-5)=C(5,i)(2/3)ⁱ(1/3)^5-i

Simillarly Py(2j-5)=C(5,j)(1/2)ⁱ(1/2)^5-i

 

I want y to be on right of x

Therefore j>i

So required probability is

_{[i=0 ]}^[5 (Px(2i-5)*_{[j=i+1 ]}^{[5 ]} Py(2j-5))

 

Add this to get final answer