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Until the letter is selected, it is a variable. Immediately after it is selected, the outcome is no longer a variable but a constant.
If the letters of computer are randomly arranged in all possible ways, the probability the word begins with a vowel in five out of 26, or 0.1923. You do not need to consider any other letters, or any permutations or combinations, because you only asked about the first letter.
Any string of two letters.
239,500,800 12!/2! * * * * * Actually, as the word "permutation" [not permutations] has 11 letters, the answer is 11!/2! = 19,958,400
360; I think.
3 out of 6
Word 1) 'math' has one vowel letter among a total of 4 letters. The probability of randomly selecting the vowel letter 'a' is P(v) = 1/4. Word 2) 'jokes' has two vowel letters among a total of 5 letters. The probability of randomly selecting a vowel letter is P(v) = 2/5. The probability of randomly selecting a vowel letter from the first word and a vowel letter from the second word is: P(v1,v2) = 1/4 (2/5) = 2/20 = 1/10 = 0.10 = 10.0%
There are 10 letters in the word "aspiration" and 5 of them are vowels. The probability of a randomly-selected letter being a vowel are 5/10 = 1/2 = 0.50.
1 in 5
Until the letter is selected, it is a variable. Immediately after it is selected, the outcome is no longer a variable but a constant.
There are five letters, and two of them are s's. The theoretical probability of choosing an s would be 2 out of 5.2/5 or 40%
If the letters of computer are randomly arranged in all possible ways, the probability the word begins with a vowel in five out of 26, or 0.1923. You do not need to consider any other letters, or any permutations or combinations, because you only asked about the first letter.
Any string of two letters.
239,500,800 12!/2! * * * * * Actually, as the word "permutation" [not permutations] has 11 letters, the answer is 11!/2! = 19,958,400
360; I think.
Randomly typed letters.
permutation without replacement