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%
3 out of 6
Since the word "probability" contains only letters, then the probability of choosing a letter from the word "probability" is 1, i.e. it is certain to happen.
There are eight different letters in 'mathematics' - a, c, e, h, i, m, s and t. Since there are 26 letters in the alphabet, the chance of a randomly chosen one being a letter that's in mathematics is 8/26, or 4/13.
I think it depends on how many students are thinking of alphabet letters at the same time :) The first student will think of a letter from the English alphabet. Then, the second student will think of a letter. The probability that the second student will select what the first student selected is 1/26.
The answer depends on which letters are in the bag!
3 out of 6
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.
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.
To determine the probability of randomly selecting the permutation "abacus" from the letters AABCSU, we first calculate the total number of distinct permutations of the letters. The total permutations are given by the formula ( \frac{n!}{n_1! \cdot n_2! \cdots n_k!} ), where ( n ) is the total number of letters and ( n_i ) are the counts of each distinct letter. Here, we have 6 letters (2 A's, 1 B, 1 C, 1 S, 1 U), resulting in ( \frac{6!}{2!} = 360 ) distinct permutations. Since "abacus" is one specific permutation, the probability is ( \frac{1}{360} ).
Since the word "probability" contains only letters, then the probability of choosing a letter from the word "probability" is 1, i.e. it is certain to happen.
2/7
There are eight different letters in 'mathematics' - a, c, e, h, i, m, s and t. Since there are 26 letters in the alphabet, the chance of a randomly chosen one being a letter that's in mathematics is 8/26, or 4/13.
I think it depends on how many students are thinking of alphabet letters at the same time :) The first student will think of a letter from the English alphabet. Then, the second student will think of a letter. The probability that the second student will select what the first student selected is 1/26.
The statement about the probability of selecting the letter 'z' from the alphabet being 126 is incorrect. The probability of selecting any one specific letter from the 26 letters of the English alphabet is 1/26, not 126. Therefore, the probability of selecting 'z' is approximately 0.0385, or about 3.85%.
5/24, or five out of twenty four, or with numbers, five out of infinity