A 2010 study conducted by the National Center for Health Statistics found that 25% of U.S. households had no landline service. This raises concerns about the accuracy of certain surveys, as they depend on random-digit dialing to households via landlines. We are going to pick five U.S. household s at random: 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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bayosanuade bayosanuade · Answer 1 · 2019-10-25T16:18:01+02:00

Answer: (a) 0.02373

(b) 0.9854

(c) 0.9990

Step-by-step explanation:

[tex]\\[/tex]This is a binomial experiment , because the events are independent, the probability of each trial remain constant.

[tex]\\[/tex]Let P be the probability that they have land lines and q be the probability that they do not have land line, that is

[tex]\\[/tex]P = 0.75

[tex]\\[/tex]q = 0.25

[tex]\\[/tex]From binomial distribution formula,

[tex]\\[/tex]P(X=r) = nCr [tex]p^{r}[/tex] [tex]q^{n-r}[/tex]

[tex]\\[/tex](a) p ( all the 5 have land lines) = 5C5 [tex]0.75^{5}[/tex][tex]0.25^{0}[/tex]

[tex]\\[/tex]= 1 x0.02373 x1

[tex]\\[/tex]= 0.02373

[tex]\\[/tex](b) p ( at least one of them has land line) = 1 – p( 0ne of them has)

[tex]\\[/tex]P( one of them has ) = 5C1 [tex]0.75^{1}[/tex][tex]0.25^{4}[/tex]

[tex]\\[/tex] = 5 x 0.75 x 0.00390625‬

[tex]\\[/tex]=0.01465

[tex]\\[/tex]Therefore p ( at least one of them has land line) = 1-0.01465

[tex]\\[/tex]= 0.9854

[tex]\\[/tex](c) p ( at leats one of them does have ) = 1 – p( 5 of them does not)

[tex]\\[/tex]P( 5 0f them does not) = 5C5 [tex]0.25^{5}[/tex][tex]0.75^{0}[/tex]

[tex]\\[/tex] = 1 X 0.0009765625‬

[tex]\\[/tex] = 0.0009765625

[tex]\\[/tex]Therefore , p ( at least one of them does have ) = 1 - 0.0009765625‬

[tex]\\[/tex] = 0.9990