i did it in a different way.
it is a binomial distribution.
so let probability of rain on any single day is x and probability of not rainint is 1-x.
so from statement 1
we get, The probability of no rain throughout the fist two days is 36%
that means nP0 = (1-x)^2 = .36
so, 1-x = .6 so x = .4
therefore probability that it will rain on each of the next 3 days = x^3= .4^3 = .064
so it is suff...
from statement 2) we get The probability of rain on the 3rd day is 40%
that means 3c1 * x. ((1-x)^2 = .4
or, x. (1-x) ^2= .4/3
we can solve for x from here as well and find what is x^3.
so this one is also suff....
Hence, D is the answer.