If a child's knowledge of the alphabet is limited to the letters a, b, c, i, and e, and if the child writes two letters at random (assume the child may write the same letter twice), what is the probability that one is a vowel and the other is a consonant?
Accepted Solution
A:
Answer:[tex]\frac{6}{25}[/tex]Step-by-step explanation:In probability"AND" means "MULTIPLICATION""OR" means "ADDITION"Here, we want P(vowel) "AND" P(consonant). So we find individual probabilities and "MULTIPLY" them.Firstly, P(vowel):From the letters a,b,c,i,e ------ we know 2 are vowel and 3 are consonants (total 5)Hence, P(vowel) = 2/5Second, P(consonant):There are 3 consonants, henceP(consonant) = 3/5Now, we multiply:P(vowel and consonant) = P(vowel) * P(consonant) = 2/5 * 3/5 = 6/25The probability that one is vowel and the other is a consonant = [tex]\frac{6}{25}[/tex]