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Mathematics 7 Online
OpenStudy (anonymous):

According to a recent survey , about 1 in 3 new cars is leased rather than bought .What is the probability that 3 of 7 random select new cards are leased?

OpenStudy (anonymous):

Nancy your Looking Good

OpenStudy (anonymous):

ohhhh I read it wrong

OpenStudy (anonymous):

good what?

OpenStudy (amistre64):

this is binomial i believe

OpenStudy (anonymous):

1/3*3/7=2/21=1/7 I think that is right.

OpenStudy (anonymous):

*3/21*

OpenStudy (anonymous):

3/21

OpenStudy (anonymous):

binomial experiments

OpenStudy (amistre64):

we can use the ^7 coeffs to help out 14641 15101051 1 6 15 20 15 1 1 7 21 there definiantly a 21 in it

OpenStudy (amistre64):

.2561 perhaps?

OpenStudy (anonymous):

the key is ; 560/2187

OpenStudy (amistre64):

\[nCr(p)^r(1-p)^{n-r}\]

OpenStudy (anonymous):

you answer is correct,Please! explain how you get answer I don't get it

OpenStudy (amistre64):

7C3 = 35 (1/3)^3 (2/3)^4

OpenStudy (amistre64):

i have to remember why its right first lol

OpenStudy (anonymous):

n=? r=? p=?

OpenStudy (amistre64):

n = 7 in this case since there are 7 places to populate r = 3 since you want 3 of them to be 1/3 p = the proportion of success; 1 in 3 = 1/3

OpenStudy (anonymous):

TY

OpenStudy (anonymous):

I get it now

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