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?

(a) 8 balls

(b) 4 balls

Step-by-step explanation:

Let

Number of balls

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

For 2 boxes, it is:

From the question, we have:

Favorable outcome

To solve for N, we have:

Divide both sides by 2

Take log of both sides

Apply law of logarithm

Make N the subject

Approximate

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:

4 balls in each box

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

none cuz there are no "following statements" regarding the question

step-by-step explanation: