Consider a room divided in half. Which is more likely to have an imbalance of ex

ID: 3027569 • Letter: C

Question

Consider a room divided in half. Which is more likely to have an imbalance of exactly five particles (i.e., N_R = N/2 + 5, where N_R is the number in the right half of the room): a room with N = 20 particles, or a room with N = 60 particles? Which is more likely to have an imbalance of 5% (i.e., N_R = N/2 + 0.05N)? Show that the most probable value to observe is N_R = N/2. If numbers get too big to handle easily, you may want to use Stirling's approximation for the factorial function, ln N! N ln N - N

Explanation / Answer

The binomial distribution is appropriate for counting the number of a given type of outcomes from n independent trials, which is prespecified and each with 2 probable outcomes with the same probability of p.

The probability mass function is P{Y = j} = {n! / j!(n j)!} p j (1 p) nj

a.N= 20 particles, we have to compute the probability of having N/2 + 5 = 15 particles in right half of room ie. 5 particles in left half of room.

Each particle has same probability = 0.5 for being in left or right half of room.

P(Y=5) = {20! / 5!(15)!} 0.5 5 (1-0.5) 15

= 0.015

N= 60 particles, we have to compute the probability of having N/2 + 5 = 35 particles in right half of room ie. 25 particles in left half of room.

P(Y=25) = {60! / 25!(35)!} 0.5 25 (1-0.5) 35

= 0.045

Hence N= 60 particles is likely to have more imbalance of having Nr = N/2 + 5

b)N= 20 particles, we have to compute the probability of having N/2 + 0.05N = 11 particles in right half of room ie. 9 particles in left half of room.

Each particle has same probability = 0.5 for being in left or right half of room.

P(Y=9) = {20! / 9!(11)!} 0.5 9 (1-0.5) 11

= 0.16

N= 60 particles, we have to compute the probability of having N/2 + 0.05N = 33 particles in right half of room ie. 27 particles in left half of room.

P(Y=27) = {60! / 27!(33)!} 0.5 25 (1-0.5) 35

= 0.076

Hence N= 20 particles is likely to have more imbalance of having Nr = N/2 + 0.05N

c)The probability for Nr = N/2

P(Y=N/2) = N!/(N/2)!( N/2)!*(0.5)N which will be the largest for any N

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Consider a room divided in half. Which is more likely to have an imbalance of ex

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