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?
21C3 / 26C3
surely you can see that the probability of three consonants is
21/26 * 20/25 * 19/24
5/26
82%
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.
Well, that sounds like a fun game of letter roulette! Let's figure out the probability, shall we?
First, let's see how many letters we have to work with. Since we have all 26 letters of the alphabet, the total number of letters in the bag is 26.
Now, let's find out the number of ways we can draw three letters that are not vowels. We start with 26 letters, subtracting the five vowels, which leaves us with 21 consonants.
When we draw the first letter, we have 21 choices out of 26. After drawing one letter, there are now 20 remaining choices for the second letter, and 19 remaining choices for the third letter.
To calculate the probability, we multiply the number of choices for each letter (21/26 * 20/25 * 19/24). Simplifying this, we get (7/10 * 4/5 * 19/24) which equals 532/1200.
Therefore, the probability that all three letters drawn are not vowels is 532/1200. And if you're a fan of simplifying fractions like I am, that reduces to a delightful 133/300.
So, the probability of not drawing any of the five vowels is 133/300 or about a 44.33% chance. Good luck playing letter roulette!
To find the probability that all three letters chosen are not any of the five vowels, we need to determine the total number of possible outcomes and the number of favorable outcomes.
Step 1: Total Number of Outcomes
There are 26 letters in the English alphabet. Therefore, the total number of possible outcomes is 26.
Step 2: Number of Favorable Outcomes
Out of the 26 letters, 5 are vowels (a, e, i, o, u). So, there are 26 - 5 = 21 consonants.
When we draw the first letter, the probability of selecting a consonant is 21/26.
After drawing the first consonant without replacement, the probability of selecting another consonant on the second draw is 20/25, since there are now only 20 consonants left out of the remaining 25 letters.
Following the same logic, the probability of selecting a consonant on the third draw is 19/24 because there are 19 consonants left out of the remaining 24 letters.
Since we want all three letters to be consonants, we multiply the probabilities together:
(21/26) * (20/25) * (19/24) = 7980/15600
Simplifying the fraction, we get:
7980/15600 = 133/260
Therefore, the probability that all three letters chosen are not any of the five vowels is 133/260.