etf said:
Since there are 4 white and 6 red balls I should use formula for permutations with repetition to calculate all possible orders?
Orders of interest are where first three balls are white and other 7 balls (6 red and 1 white) could be in any order so m would be $$\binom{4}{3} \frac{7!}{6!1!}= 28$$ and $$P(A) = \frac{28}{210}$$?
You need to decide whether you want to use permutations or combinations. I think permutations are a more natural description of the actual, physical situation, and also would probably be what you would prefer to use if you were writing a Monte-Carlo simulation of the experiment for running on a computer. So: number the balls from 1--10; the first 4 are white and the remaining 6 are red. The sample space consists of all the permutations of the numbers (1,2,...,10), and we tacitly assume all are equally likely.
How many permutations are there altogether? How many permutations have numbers from 1--4 in the first three places? So, assuming we don't care what is in the 4th place, what would be the resulting probability? Next, if we assume we also want a non-white in position 4, how many permutations would there be now? What is the corresponding probability?
Another way to do the problem would be via conditional probability arguments. If the events E1, E2, E3 correspond to having a number from 1--4 in positions 1,2 and 3, respectively, then
P(E_1 \& E_2 \& E_3) = P(E_1) P(E_2 \& E_3 | E_1),\\<br />
P(E_2 \& E_3 | E_1) = P(E_3 | E_1 \& E_2) P(E_2|E_1)
What is ##P(E_1)?## What is ##P(E_2 | E_1),## etc.?
