The Hawaiian language has only 12 letters: the vowels a, e, i, o and u and the consonants h, k, l, m,n,p and w what is the probability a randomly selected 3 letter "word" begins with a consonant and ends with 2 different vowels?
I did
$$\left(\frac 7{12}\right)\left(\frac 5{11}\right)\left(\frac 4{11}\right)= \frac {140}{1320}$$
did I make this too simple? I keep doubting everything I am doing!
Best Answer
Your answer would be correct if you were making selections of letters "without replacement" -- e.g., if you wrote each letter on a card, shuffled the cards, and then dealt out three of them in order. But this isn't how random words are usually thought of, as evidenced in the constraint that the word end in two different vowels. For the problem with replacement, the denominators on the left should all be $12$, for an answer of $(7/12)(5/12)(4/12)=140/1728=35/432$.