Probability Question - three balls in a bag 
A ball contains three balls, one blue, one red, and one yellow. Every
  day, one ball is picked out of the bag and is then returned.
1) For how many days can you expect to not pick out a blue ball?
2) No blue balls have been drawn for one week. What is the probability
  one will not be drawn for at least one more week?
3) How many times on average will a blue ball be drawn in December?
4) What is the probability a blue ball will be drawn 1 to 30 times in
  that month?

Any help gratefully received... don't really get this :-( The only one I could try to get, but am not confident with, is the third one. If the chance of getting a blue ball is 1/3, and each drawing of a ball is independent, would it just be 31 (days in December) times 1/3 which would then be rounded to give 11 (rounded up) expected blue results?
Many thanks
Marc
 A: Hints:


*

*What is the expected value of the number of days you can go without drawing a blue ball? Recall what expected value is: If values $x_1,x_2\ldots x_n$ occur with respective probabilities $p_1,p_2\ldots p_n$, the expected value is 
$$x_1p_1+x_2p_2+\ldots+x_np_n=\sum_{j=0}^{n}x_jp_j$$
So how does that help us here? Our $x$ values are clearly $1,2,3\ldots$, the number of days since which we have not picked a blue ball. What are our $p$ values? $p_1=2/3$, but what about $p_2$ etc. ?

*The probability of drawing a blue ball is independent of other events. Thus the fact that a blue ball has not been drawn for a week is irrelevant towards the probability that a blue ball will continue to not be drawn. So this question becomes: What is the probability a blue ball will not be drawn for at least one week?

*The answer is correct, the method a little non-rigorous. In other words, it completely works, but most likely requires justification. Try using the definition of average, and hopefully you'll get something involving the binomial theorem.

*Consider the sample space. A blue ball can be drawn anywhere between $0$ and $31$ times, with probabilities $D_0\ldots D_{31}$. These probabilities must satisfy $D_0+\ldots+D_{31}=1$. We are trying to find $D_1+\ldots+D_{30}$. Thus our answer is equivalent to $1-D_0-D_{31}$, which is substantially easier to figure out.

