## anonymous 3 years ago see attachment

1. anonymous

2. anonymous

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

3. anonymous

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

4. anonymous

5. anonymous

sorry third one

6. kropot72

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.

7. anonymous

oh, fourth one?

8. kropot72

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

9. kropot72

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

10. anonymous

because there two values 17% and 24%