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?
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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.
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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?
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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
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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