1. Consider eight particles, four of which are black and four white. Four partic

Question

1. Consider eight particles, four of which are black and four white. Four particles can fit left of a permeable membrane and four can fit right of the membrane. Imagine that due to random motion of the particles every arrangement of the eight particles is equally likely Some possible arrangements are BBBBWwWW. BBBWBWww. WBWB)WBWB; the membrane position is denoted by | (20 points). (a) How many different arrangements are there (5 points)? (b) Calculate the probability of having all four black particles on the left of the permeable membrane (2 points). What is the probability of having one white particle and three black particles on the left of the membrane (2 points)? Finally, calculate the probability that two white and two black particles are left of the membrane (2 points). Compare these three probabilities (2 points). Which arrangement is most likely (2 points)? (c) Imagine that, in one time instant, a random particle from the left-hand side exchanges places with a random particle on the right-hand side. Starting with three black particles and one white particle on the left of the membrane, compute the probability that after one time instant there are four black particles on the left. What is the probability that there are two black and two white particles on the left, after that same time instant (4 points)? Which is the more likely scenario of the two (1 point)? 9 0 9

Explanation / Answer

a) If we have 8 distinguishable particles to fit into 8 slots, the total of distinct arrangement is explained by:

The number of particle available for filling the first slot will be 8
The number of particle available for filling the second slot will be 7
The number of particle available for filling the third slot will be 6
The number of particle available for filling the fourth slot will be 5……and so on

So the total of distinct arrangement would be 8 x 7 x 6 x …….x 1 = 8! = 40320

This is the probability of arrangements if the particles are distinct, like labeled B1, B2 or W1 or W2. But if we consider all black particles are indistinguishable and all white particles are also indistinguishable we over counter by a factor of 4! for black as well as 4! for white. So the result will be 8! / 4!x 4! = 70

b) To compute the probability of various arrangement, lets figure out how many arrangements are favorable and divide be total number of arrangements.

Probability of 4 black particles on left.

For the first slot on left there are 4 black particles to occupy, then 3 left for next slot on left and 2 for next and so on.

So, the number of arrangements will be (4 x 3 x 2 x 1) x (4 x 3 x 2 x 1) = 4! 4!
There are 8! total number of arrangements as we calculated above.

So, the probability of all black particles on left will be 4! 4! / 8! = 1/70

Probability of one black particle on left

The number of arrangements with one black particle in one slot and white particle on the other 3 slots on the same side is:

4 x ( 4 X 3 x 2) x (4 x 3 x 2 x 1) = 4 x 4! 4!

So, the probability will be 4 x 4! 4! / 8! = 4/70

But we should also remember that we have 4 spots in the left and the probability of one black particle on the left is spot 2,3 or 4 is equal to slot 1.
So, the final probability will be 4 x 4/70 = 16/70
The probability for one white particle is also the same i.e. 16/70

Probability of two black and two white on left side of membrane

number of black particle available to occupy first place 4
number of black particle available to occupy second place 3
number of white particle available to occupy first place 4
number of white particle available to occupy second place 3

The number of such arrangements = (4 x 3 x 4 x 3) x (4 x 3 x 2 x 1)
So, probability will be (4 x 3 x 4 x 3) 4! / 8! = 6/70
But,

B B W W
B W B W
B W W B
W B B W
W B W B
W W B B
all these are equally suitable arrangement, i.e. there are 6 of them with all equal probability to occur. So final probability will be 6 x 6 /70 = 36/70

c) Starting with three black particle on the left, if 1 particle is allowed to move at random
The possible arrangement before the movement will be
B B B W / B W W W

The particle at any slot can swap so there are total 4 swapping p[possibilities but the probability of swapping only the one black to the other white is ¼ x ¼ = 1/16

Probability to have 2 black and 2 white on left
similar to above there are 9/16 possibilities
So the this event is more likely to occur than the one above

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