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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
3 are brown and so evidently 6 are not brown use the binomial \[\binom{9}{3}(.21)^3(.79)^6\]
 one year ago

dhruv_khatkar Group TitleBest ResponseYou've already chosen the best response.0
i think you can use the formula (n x) . p^x . q ^(nx)
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
what about the second one?
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
sorry third one
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
For the third one you use the fact that the binomial distribution can be approximated by the normal distribution when n is large and p is neither small nor near 1.\[\mu=np=175\times 0.21\] \[\sigma=\sqrt{np(1p)}=\sqrt{175\times 0.21\times 0.79}\] Then you find the required probability by means of a standard normal distribution table.
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
oh, fourth one?
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
Basically you use the same method for the fourth one as for the third, but change the values of n and p accordingly.
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
You need to find one value of zscore for the third one and two values of zscore for the fourth.
 one year ago

yashar806 Group TitleBest ResponseYou've already chosen the best response.0
because there two values 17% and 24%
 one year ago
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