



 
Josh, (A) "All white" and "at least one red" are complementary events: one, and only one of them, must be true. In such cases the probabilities always add to 1. This formula is useful because it is usually easier to compute the probability that all of a set of things happen than to compute directly the probability that at least one happens. (B) As you said, the [conditional] probability changes each time you remove a marble. The formula that you should use for a problem like this is
So the probability of a second white marble given that the first was white is 99/199, because IF your first draw was white there are now 99 white and 100 red marbles left. The probability of a third white marble, given that the first two were white, is 98/198, and so on. So the probability of three white marbles in a row is $(100 \times 99 \times 98)/(200 \times 199 \times 198)$ which is just a shade less than 1/8. and the probability of the complementary event (at least one red) is $1  (100 \times 99 \times 98)/(200 \times 199 \times 198)$ which is just over 7/8. That should be enough of a hint for Part A. Part B is easier and requires no arithmetic. Good Hunting!
 


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 