Mathematics, 23.07.2021 03:20 tyaslovell

it is found that 4% of watches produced at a particular factory are defective. If 20 watches made at this factory are randomly selected, what is the probbility that at mpst 1 watch in the same sample is found to be defective

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Answer from: claudiniweenie

0.81 = 81% probability that at most 1 watch in the sample is defective.

Step-by-step explanation:

For each watch, there are only two possible outcomes. Either it is defective, or it is not. The probability of a watch being defective is independent of any other watch, which means that the binomial probability distribution is used 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.

4% of watches produced at a particular factory are defective.

This means that p = 0.04

20 watches made at this factory are randomly selected

This means that n = 20

What is the probability that at most 1 watch in the same sample is found to be defective?

This is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

In which

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

$P(X = 0) = C_{20,0}.(0.04)^{0}.(0.96)^{20} = 0.442$

$P(X = 1) = C_{20,1}.(0.04)^{1}.(0.96)^{19} = 0.368$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.442 + 0.368 = 0.81$

0.81 = 81% probability that at most 1 watch in the sample is defective.

Answer from: Quest

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

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

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

g(x)

g(2)

=2x

=2⋅2

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!

Answer from: Quest

step-by-step explanation:

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