Question

see attachment

Answer 1

3 are brown and so evidently 6 are not brown use the binomial
\[\binom{9}{3}(.21)^3(.79)^6\]

Answer 2

i think you can use the formula (n x) . p^x . q ^(n-x)

Answer 3

what about the second one?

Answer 4

sorry third one

Answer 5

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.

Answer 6

oh, fourth one?

Answer 7

Basically you use the same method for the fourth one as for the third, but change the values of n and p accordingly.

Answer 8

You need to find one value of z-score for the third one and two values of z-score for the fourth.

Answer 9

because there two values 17% and 24%