I think I have worked this out correctly but wanted to have someone else take a look to confirm it.
Q.: A box contains letters used in a word game. At a certain stage in the game, the 8 letters in the box are A, A, C, E, L, P, P, P. 1 player draws, at random, 3 letters in succession without replacing them. Calculate the probability that the 3 letters drawn do not include E or P.
Attempt: P(3 letters without E or P) = (2/8 x 1/7 x 1/6) x 3! = 12/336 = 1/28, P(3 letters without P or E) = (2/8 x 1/7 x 1/6) x 3! = 1/28. Thus, P(3 letters without E/ P) = 1/28 + 1/28 = 1/14.
This matches the text book answer. I have separated the probabilities as 'P or E' and 'E or P' due to appearance of the 'or' in the question which, according to the book, means the answer totals are to be added.