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Monica [59]
3 years ago
13

Two containers hold several balls. Once a second, one of the balls is chosen at random and switched to the other container. Afte

r 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?
Mathematics
1 answer:
natka813 [3]3 years ago
6 0

Answer:

(a) 8 balls

(b) 4 balls

Step-by-step explanation:

Let

N \to Number of balls

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

Pr = 0.5^N

For 2 boxes, it is:

Pr = 2 * 0.5^N

From the question, we have:

\frac{80}{10000} \to Favorable outcome

To solve for N, we have:

2 * 0.5^N = \frac{80}{10000}

2 * 0.5^N = 0.008

Divide both sides by 2

0.5^N = 0.004

Take log of both sides

\log(0.5^N) = \log(0.004)

Apply law of logarithm

N\log(0.5) = \log(0.004)

Make N the subject

N = \frac{\log(0.004)}{\log(0.5)}

N =7.966

Approximate

N = 8

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:

Balls \approx \frac{N}{2}

Balls \approx \frac{8}{2}

Balls \approx 4

<em>4 balls in each box</em>

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8 0
3 years ago
Read 2 more answers
20.There is an 80% chance David will eat a healthy breakfast and a 25% chance that it will rain. If these events are independent
Mariulka [41]

Answer:

A. 20%

Step-by-step explanation:

These events are independent, because if David eats a healthy breakfast cannot influence on would be rain or not.

"And" for probabilities  of independent events means "x" (times).

80% = 0.8

25% = 0.25

0.8*0.25 = 0.2 = 20%

4 0
3 years ago
Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can
Thepotemich [5.8K]

Answer:

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)}

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)}

Step-by-step explanation:

S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)} (second die is even)

B = {(1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)} (sum of the two numbers is even)

C = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,3) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,3) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,3) (6,5)} (at least one in the pair is odd i.e one of the pair is odd or both are odd)

A bar = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (the pairs that are not in A)

B bar = {(1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (3,6) (4,1) (4,3) (4,5) (5,2) (5,4) (5,6) (6,1) (6,3) (6,5)} (the pairs that are not in B)

C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (the pairs that are not in C)

▪A = {(1,2) (1,4) (1,6) (2,2) (2,4) (2,6) (3,2) (3,4) (3,6) (4,2) (4,4) (4,6) (5,2) (5,4) (5,6) (6,2) (6,4)(6,6)}

▪C bar = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)}

▪A∩B = {(2,2) (2,4) (2,6) (4,2) (4,4) (4,6) (6,2) (6,4) (6,6)} (intersection: the pairs that are common to both A and B)

▪A∩B bar = {(1,2) (1,4) (1,6) (3,2) (3,4) (3,6) (5,2) (5,4) (5,6)} (intersection: the pairs that are common to both A and B bar)

▪A bar∪B = {(1,1) (1,3) (1,5) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,3) (3,5) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,3) (5,5) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)} (union: all the pairs in A bar and B )

▪A bar∩C = {(1,1) (1,3) (1,5) (2,1) (2,3) (2,5) (3,1) (3,3) (3,5) (4,1) (4,3) (4,5) (5,1) (5,3) (5,5) (6,1) (6,3) (6,5)} (intersection: the pairs that are common to both A bar and C)

6 0
3 years ago
Your housing and fixed expenses are $702.38 per month do you want to have a four month emergency fund and save it over a nine mo
MrMuchimi

Answer:

For Review: Planning Ahead and Contracts Quick Check... these are the answers...

1) B. Yes, you want to begin saving at least 25 years before you plan to retire

2) D. all of the above

3) $2,341.27 per month realized income

4) Your housing and fixed expenses are $702.38 per month do you want to have a four month emergency fund and save it over a nine month period of time how much do you need to save per month?

Answer #4) $312.17 /month

Step-by-step explanation:

702.38 * 4 = 2,809.52 How much you will need to save for 4 months worth or expenses.

Now divide by 9 because that is how many months you will take to save up...

2,809.52 / 9 = 312.17

Answer) $312.17 per month, for 9 months to save 4 months worth or expenses.

8 0
3 years ago
Factorise 6+ 9x show working please
Stolb23 [73]
6 + 9x.....a common factor in both terms is 3

3(2 + 3x) <==
5 0
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