As Mr. Reddly has shown (yes you are right), this problem is very easily done by looking at the failure situation and then subtracting that probability from one.
If you want a result where at least one ball is black, then the only scenario which fails to accommodate that is if all the balls are white.
Because we are looking at 5 independent events having all white balls is approximated by:
(27/35)*(27/35)*(27/35)*(27/35)*(27/35) or (27/35)^5
Since this is the DAT you should approximate in your head. Divide numerator and denominator by 9 (choose the highest common approximate denominator) and you will approximately have: (3/4)^5.
This can be converted mentally to 243/1024 or around 24% which gives you an approximate answer of 76% since 1-24% is 76%. The approximations are necessary if you don't have a calculator. Hopefully your answer choices aren't 74%, 75%, 76% etc.
You will never get a problem with such complex calculations on the real DAT since it is an easy test and they know you don't have a calculator.
The reason why you have to multiply when looking at successive independent events is because if you are looking for the same result on every trial you need it to be that result every time.
For example if you want all heads on five coin tosses, you need heads for the first toss. What is the probability of that? It is 0.5.
Then need heads on the second toss. The probability of that happening is also 0.5.
So you half a chance on the first toss and then half a chance on the second toss. Between the two throws you only have a quarter of a chance of getting all heads. If you took this to five successive tosses your chance of getting all heads is (0.5)^5 which is a very small number since it is pretty hard to get the same result 5 times in a row.
