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Evgesh-ka [11]
2 years ago
15

A particular library has 58 visitors. When the visitors were asked how many books they read that year, 20 said they read one boo

k, 22 said they read three, and 16 said they read five. Assuming the visitors are telling the truth, what is the empirical probability that a visitor read five books? Write your answer as an exact fraction which is reduced as much as possible
Mathematics
1 answer:
irina1246 [14]2 years ago
7 0

Answer:

16/58 = 8/29

Step-by-step explanation:

p = number of required outcome/number of total out come

p = 16/58 ie number of people that read five book divided by total number of people

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A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st
Alex

Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So 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.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that n = 15, p = \frac{1}{5} = 0.2

What is the probability that he answers at least 10 questions correctly?

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

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

P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001

P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011

P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0

P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0

P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0

P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0

P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111

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Answer:  3.50x + 4.00y ≤ 45

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<u>Step-by-step explanation:</u>

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