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.

_________________

hey ya......