Question 563780
You need to use the binomial formula. For A, the probability is (12!)/[(3!)(12-3)!](.45)^3(.55)^9=0.09233.

For B, calculate the probability of 4 or more via addition (i.e. P(4)+...+P(12)) using the same formula as in A.  The work for the latter would be (12!)/[(4!)(12-4)!](.45)^4(.55)^8+...+(12!)/[(12!)(12-12)!](.45)^12(.55)^0=0.865532.

Similarly, C should come to about 0.042142. Check those numbers yourself, since it's very easy to make typing errors on a calculator with so many numbers involved.