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yashar806
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3 are brown and so evidently 6 are not brown use the binomial \[\binom{9}{3}(.21)^3(.79)^6\]
i think you can use the formula (n x) . p^x . q ^(n-x)
what about the second one?
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.\[\mu=np=175\times 0.21\] \[\sigma=\sqrt{np(1-p)}=\sqrt{175\times 0.21\times 0.79}\] Then you find the required probability by means of a standard normal distribution table.
Basically you use the same method for the fourth one as for the third, but change the values of n and p accordingly.
You need to find one value of z-score for the third one and two values of z-score for the fourth.
because there two values 17% and 24%