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MissTica
3 years ago
12

100÷50000 it's has to be step by step aka show work ​

Mathematics
1 answer:
lys-0071 [83]3 years ago
4 0

100÷50000 can be written as 100/50000, then cancel out 2 of the zeroes, and you will get 1/500 or 0.002 as your final answer.

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Answer:

a) 0.7412 = 74.12% probability that all the three orders will be filled correctly.

b) 0.0009 = 0.09% probability that none of the three will be filled correctly

c) 0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d) 0.9991 = 99.91% probability that at least one of the three will be filled correctly

e) 0.0082 = 0.82% probability that only your order will be filled correctly

Step-by-step explanation:

For each order, there are only two possible outcomes. Either it is filled correctly, or it is not. Orders are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The percentage of orders filled correctly at Burger King was approximately 90.5%.

This means that p = 0.905

You and 2 friends:

So 3 people in total, which means that n = 3

a. What is the probability that all the three orders will be filled correctly?

This is P(X = 3).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.905)^{3}.(0.095)^{0} = 0.7412

0.7412 = 74.12% probability that all the three orders will be filled correctly.

b. What is the probability that none of the three will be filled correctly?

This is P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{3,0}.(0.905)^{0}.(0.095)^{3} = 0.0009

0.0009 = 0.09% probability that none of the three will be filled correctly.

c. What is the probability that one of the three will be filled correctly?

This is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{3,1}.(0.905)^{1}.(0.095)^{2} = 0.0245

0.0245 = 2.45% probability that at least one of the three will be filled correctly.

d. What is the probability that at least one of the three will be filled correctly?

This is

P(X \geq 1) = 1 - P(X = 0)

With what we found in b:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0009 = 0.9991

0.9991 = 99.91% probability that at least one of the three will be filled correctly.

e. What is the probability that only your order will be filled correctly?

Yours correctly with 90.5% probability, the other 2 wrong, each with 9.5% probability. So

p = 0.905*0.095*0.095 = 0.0082

0.0082 = 0.82% probability that only your order will be filled correctly

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