Mathematics, 23.06.2020 23:01 alyo31500

A recent study conducted by a health statistics center found that 25% of households in a certain country had no landline service. This raises concerns about the accuracy of certain surveys, as they depend on random-digit dialing to houeholds via landlines. Pick five households form this country at random. Complete parts a) through c). a) what is the probability that all five of them have a landline?
b) what is the probability that at least one of them does not have a landline?
c) what is the probability that at least one of them does have a landline?

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Answer from: shamarmosley

a) 23.73% probability that all five of them have a landline

b) 76.27% probability that at least one of them does not have a landline

c) 99.90% probability that at least one of them does have a landline

Step-by-step explanation:

For each household, there are only two possible outcomes. Either it has landline service, or it does not. The probability of a household having landline service is independent of other households. 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.

25% of households in a certain country had no landline service.

This means that 100-25 = 75% have, so p = 0.75

Pick five households form this country at random.

This means that n = 5

a) what is the probability that all five of them have a landline?

This is P(X = 5).

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

$P(X = 5) = C_{5,5}.(0.75)^{5}.(0.25)^{0} = 0.2373$

23.73% probability that all five of them have a landline

b) what is the probability that at least one of them does not have a landline?

Either all have, or at least one does not have. The sum of the probabilities of these events is 100%.

From a), 23.73% probability that all five of them have a landline

100 - 23.73 = 76.27

76.27% probability that at least one of them does not have a landline

c) what is the probability that at least one of them does have a landline?

Either none have a landline, or at least one has. The sum of the probabilities of these events is 1. So

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

We want $P(X \geq 1)$

Then

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

In which

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

$P(X = 0) = C_{5,0}.(0.75)^{0}.(0.25)^{5} = 0.0010$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0010 = 0.9990$

99.90% probability that at least one of them does have a landline

Answer from: Quest

y-y1=4(x-x1)

step-by-step explanation:

this is point-slope form since you only have 4 as the slope it goes in place of m

hope this : )

Answer from: Quest

step-by-step explanation:

4227473658779875563886538705958641728563470543621287463824132573847969p05749365273847906970680574936214

Iam stuck on this problem and am having trouble solving it. dies anyone know the steps to solving it

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