A box contains 10 marbles: 6 white ones and 4 red ones. A person picks 3 marbles at once. Question: A box contains 10 marbles: 6 white ones and 4 red ones. A person picks 3 marbles at once. What is the probability that the person picks at least one white marble?
So what confuses me here is picking the marbles at once. Isn't this the same as picking one-by-one without replacement? I came up with ((6/10) + (6/9) + (6/8)) / 3 but am not sure this is the right answer.
 A: Probability that the person picks only red marbles (no white marbles),
$P(3R) =  \displaystyle \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{1}{30}$
This is same as saying $P(3R) = \displaystyle  \frac{4 \choose 3}{10 \choose 3}$
The desired probability $ = 1 -  P(3R) = \frac{29}{30}$
A: It's easiest to consider the complement: the probability that the person draws no white marble, so only red ones. There are $\binom{10}{3}$ ways to draw three marbles form $10$ in one go and $\binom{4}{3}$ ways to draw $3$ red ones (out of the four available ones).
So the answer is $$1- \frac{\binom{4}{3}}{\binom{10}{3}}$$
It indeed is the same as picking one-by-one without replacement. But in your answer you compute the chance of $3$ white ones, not at least one white one, so much smaller.
A: P(at least one white) = 1 - P(all red)
For the probability of all red you should count the number of ways you can choose three red marbles and divide that by the number of ways you can choose three marbles:
$P(AllRed) = (4*3*2)/(10*9*8)$
And the answer is:
$P(At Least One White) = 1 - (4*3*2)/(10*9*8) = 0.9666667$
