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Sagot :
Answer:
Summary of Probabilities
[tex]1. P(\text{vowel or consonant}) = 1 [/tex]
[tex]2. P(\text{consonant or W}) = \frac{21}{26} [/tex]
[tex]3. P(\text{vowel or E}) = \frac{5}{26} [/tex]
[tex]4. P(\text{consonant or A}) = \frac{21}{26} [/tex]
[tex]5. P(\text{vowel or K}) = \frac{5}{26} [/tex]
Thus, the final answers are:
[tex]1. \: \boxed{1} [/tex]
[tex]2. \: \boxed{\frac{21}{26}}[/tex]
[tex]3. \: \boxed{\frac{5}{26}}[/tex]
[tex]4. \: \boxed{\frac{21}{26}} [/tex]
[tex]5. \: \boxed{\frac{5}{26}} [/tex]
Step-by-step explanation:
Total Letters in the English Alphabet
There are 26 letters in the English alphabet.
Vowels and Consonants
The vowels in the English alphabet are: A, E, I, O, U (5 vowels).
The consonants are the remaining letters: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Y, Z (21 consonants).
Probability Formula
The probability P of an event is given by:
[tex]P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}.[/tex]
Now, let's solve each part:
1. Probability of picking a vowel or a consonant
Since every letter is either a vowel or a consonant, the probability is:
[tex]{P(\text{vowel or consonant}) = \frac{26}{26} = 1.}[/tex]
2. Probability of picking a consonant or W
There are 21 consonants and 1 W, but W is already counted as a consonant. Thus, the number of favorable outcomes is 21.
[tex]P(\text{consonant or W}) = \frac{21}{26}.[/tex]
3. Probability of picking a vowel or E
There are 5 vowels, and E is one of them. Thus, the number of favorable outcomes is still 5 (since E is already counted).
[tex]P(\text{vowel or E}) = \frac{5}{26}.[/tex]
4. Probability of picking a consonant or A
There are 21 consonants, and A is a vowel. Thus, the number of favorable outcomes is still 21.
[tex]P(\text{consonant or A}) = \frac{21}{26}.[/tex]
5. Probability of picking a vowel or K
There are 5 vowels, and K is a consonant. Thus, the number of favorable outcomes is still 5.
[tex]P(\text{vowel or K}) = \frac{5}{26}.[/tex]
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