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lost822
11-14-2005, 05:51 PM
A recent study indicaed that 82% of US households speak English at home. If 4 households are randomlu selected then....
a. What is the probability that all 4 will speak English?

I am lost on this. I know that every probability has to be betwen 0 and 1. I keep wanting to say that .82/4 = .205 x 4 = .82 but that does not seem right. Can anyone help?

b. What is the probability that aleast one will speak English?

p(English)=.82 right? p(not English)=.18 right?
I am so lost. I would say for this that .82/4 =.205.

Thank you for your time!

JohnM
11-14-2005, 07:20 PM
Both of these problems use the binomial distribution.

p(English) = .82

(a.) if you select 4 at random, then all 4 need to be English-speaking, so it's:

English AND English AND English AND English = .82 * .82 * .82 *.82 = (.82)^4

(b.) P(at least 1) = 1 - P(none)

P(not English) = 1 - P(English) = 1 - .82 = .18

not English AND not English AND not English AND not English = .18 * .18 * .18 *.18 = (.18)^4

P(at least 1) = 1 - (.18)^4

Please check our Examples Section, which has a post that talks about binomial probabilities.

lost822
11-14-2005, 07:32 PM
It really made since when I saw it ALL laid out! Thank you so much!