What is the probability that in at least 2 of the next 10 years less than 20 inches of rain are recorded in the region? Information:
The annual rainfall (in inches) in a certain region is normally distributed with $\mu=40$ and $\sigma=5$.
My attempt is:
probability of less than 20 inches of rain
$$P(x<20)=P(Z<-4)=1-\phi(4).$$
I'm not sure how to apply this to the part where at least 2 of the next ten years. 
 A: Its a binomial distribution with 10 years. (assume each year is independent of previous year.)
So, $X_i$~Norm(40,25) (this is the mean, and variance) represents the annual rainfall of a year.
P(X<20) which is aprx $\phi$($\frac{20-40}{5}$) which $\phi(-4)$
Y, composed of the $X_{i}$s, is ~Binomial(10,$\phi(-4)$). This represents the amount of years that have this rain or less.
So, we want P(Y>1) = 1 - P(Y=1) - P(Y=0) which you get by plugging in the binomial distribution here. 
$$P(X=x)= \begin{pmatrix} n\\x \end{pmatrix} p^x(1-p)^{n-x}$$
A: Let $X$ be the annual rainfall, then $X\sim N(40,5)$. Then you were right about
$$P(X<20)=P\left(Z<\frac{20-40}{5}\right)=P(Z<-4)=1-\Phi(4)$$
Let $Y$ be the number of years, in the next $10$ that will have less than 20 inches of rainfall. If we suppose that the annual rainfall is independant over year, then $Y$ follow a binomial distribution. $Y\sim \text{Binom}(10,1-\Phi(4))$
We are looking for $P(Y\geq 2)$. It will be simplier to look for the opposite.
$$P(Y\geq2)=1-P(Y<2)=1-P(Y=1)-P(Y=0)$$
$$P(Y\geq2)=1-{10\choose1}(1-\Phi(4))^1(\Phi(4))^9-{10\choose0}(1-\Phi(4))^0(\Phi(4))^{10}$$
We could simplify this last equation to
$$P(Y\geq2)=1-10(\Phi(4))^9+9(\Phi(4))^{10}$$
Hope it helps
