Question

Containers A and B each hold several balls. Once a second, one of the balls is chosen at random and switched to the other container. After a long time has passed, you regularly record the number of balls in each container every second. In 8 x 105 seconds, you observe 3125 instances when container A held all the balls while container B was empty.
Required:
a. How many balls are there?
b. What is the most likely number of balls to be found in one of the containers?

Answer 1

Answer:

a) 8

b) 4

Step-by-step explanation:

Suppose N = the total number of balls.

The probability of determining an N ball in the box is $(0.5)^N$

However, the probability of finding all the balls in either of the one boxes is:

$= 2(0.5)^N$

If the number of favorable outcomes is 3125 out of 8 x 105 seconds

Then, the probability is:

$2(0.5)^N = \dfrac{3125}{8\times 10^5}$

$N = log_{0.5}(0.004)$

$N = \dfrac{log (0.004)}{log_{10}(0.5)}$

$\mathbf{N \simeq 8}$

The balance in the two boxes will be achieved when we have an equal number of balls in the same boxes and half the total no. of balls in the two boxes.

So;

$N_1 \sim N_2$

$N_1 \sim \dfrac{N}{2}$

where; N = no of balls

$N_1 \sim \dfrac{8}{2}$

$\mathbf{N_1 \simeq 4}$