Two containers 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 record the number of balls in each container every second. In 10,000 s, you find 80 times when all the balls were in one container (either one) and the other container 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?

Question

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MrRoyal MrRoyal · Answer 1 · 2021-05-24T20:36:29+02:00

Answer:

(a) 8 balls

(b) 4 balls

Step-by-step explanation:

Let

[tex]N \to[/tex] Number of balls

For a box, the probability that there are N balls in it is:

[tex]Pr = 0.5^N[/tex]

For 2 boxes, it is:

[tex]Pr = 2 * 0.5^N[/tex]

From the question, we have:

[tex]\frac{80}{10000} \to[/tex] Favorable outcome

To solve for N, we have:

[tex]2 * 0.5^N = \frac{80}{10000}[/tex]

[tex]2 * 0.5^N = 0.008[/tex]

Divide both sides by 2

[tex]0.5^N = 0.004[/tex]

Take log of both sides

[tex]\log(0.5^N) = \log(0.004)[/tex]

Apply law of logarithm

[tex]N\log(0.5) = \log(0.004)[/tex]

Make N the subject

[tex]N = \frac{\log(0.004)}{\log(0.5)}[/tex]

[tex]N =7.966[/tex]

Approximate

[tex]N = 8[/tex]

Solving (b): Balls in one of the two boxes.

Here, we assume that each ball will have almost the same number of balls at a given instance;

Hence, we have:

[tex]Balls \approx \frac{N}{2}[/tex]

[tex]Balls \approx \frac{8}{2}[/tex]

[tex]Balls \approx 4[/tex]

4 balls in each box