What are the correct interpretations for these basic probability questions? I have this homework on basic probability for my engineering mathematics course, but I don't understand these two sub-questions.
Here is the text.

The probabilities of the monthly snowfall exceeding 10 cm at a
  particular location in the months of December, January and February
  are 0.20, 0.40 and 0.60, respectively.

Here are the questions that I do not know how to interpret.

(For a particular winter:)
  
  
*
  
*What is the probability of receiving at least 10 cm snowfall in a
  month, in at least two of the three months of that winter?
  
*Find the expected number of months in which monthly snowfall does not exceed 10 cm.
  

Given and my assumptions:


*

*There are three consecutive months, each with a corresponding probability of exceeding 10 cm snowfall.

*A month of exceeding snowfall does not affect the amount of snowfall for the next month (not to be confused with the months' relationship in a tree diagram, they are still related)


Sorry to ask about this. English is not my main language.
 A: More assumptions are needed.  If $X_1$, $X_2$ and $X_3$ are the amounts of snowfall in each month, (2) says $X_1$ and $X_2$ are independent and $X_2$ and $X_3$ are independent.
But that does not imply $X_1, X_2, X_3$ are independent, which is what you need.
In that case, if $p_i = P(X_i \ge 10)$, the answer to (1) would be
$P(X_1 \ge 10, X_2 \ge 10) + P(X_1 \ge 10, X_3 \ge 10) + P(X_2 \ge 10, X_3 \ge 10) - 2 P(X_1 \ge 10, X_2 \ge 10, X_3 \ge 10) = p_1 p_2 + p_1 p_3 + p_2 p_3 - 2 p_1 p_2 p_3$.
A: a)  The probability of the monthly snowfall NOT exceeding $10 $ cm is $0.8, 0.6$, and $0.4$. Since these are assumed to be independent (not really a safe assumption), the probability of all three being less than $10$ cm is:
\begin{align*}
0.8 \times 0.6 \times 0.4 = 0.19200  
\end{align*}
b)
All $3 = .2 \times .4 \times .6 = 0.04800$
Dec & Jan & not Feb  $ =0.2 \times 0.4 \times 0.4 = 0.03200$
Dec, not Jan, Feb $= 0.2 \times 0.6 \times 0.6 = 0.07200$
not Dec, Jan, Feb $= 0.8 \times 0.4 \times 0.6 = 0.19200 $
Now add up these four numbers: $0.34400 $
c)
Consecutive
Dec-Jan-not Feb $= 0.03200$
Jan-Feb-not Dec $= 0.19200$
$0.19200 + 0.03200 = 0.22400$
Not Consecutive
Dec - Feb-not Jan $= 0.07200$
$0.07200 + 0.19200 + 0.03200 = 0.29600$
probability of consecutive $= 0.244/0.296 = 0.756$
