each letter of the alphabet is placed in a bag. One letter is drawn and not replaced. If three letters are drawn without replacement, what is the probability that all three letters chosen are NOT any of the five vowels, a, e, i, o, u?

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  1. surely you can see that the probability of three consonants is

    21/26 * 20/25 * 19/24

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  2. 21C3 / 26C3

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  3. 82%

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  4. Im stuck on the same question. Studying for the asvab. they set it up to 21C3/26C3 and apparently that equals 54,264/65,780 which equals 0.82 which turns into 82% which is the answer. But how in the world they did the work, im trying to figure that out.I tried using the combinations formula and i got different answers from 54,264/65,780. Im trying everything to figure out what they did.

    Question: A bag Contains 26 tiles representing the 26 letters of the English Alphabet. If 3 tiles are drawn from the bag without replacement, what is the probability that all 3 will be consonants?

    Consonants are the other letters that aren't vowels. And sometimes Y apparently.

    If we use steves method, it doesnt make sense.

    I dont really have an answer on how they did the work, maybe this is just a mistake. im not sure, ive been working on this question for about 4 hours now.

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