Originally Posted by

**GrigOrig99** 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.