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 rando

Question

3 years ago

8

m (assume the child may write the same letter twice), what is the probability that one is a vowel and the other is a consonant?

2 answers:

lina2011 [118] · Answer 1 · 2022-06-19T01:47:16+03:00

The probability that one is a vowel and the other is a consonant is $P = \frac{12}{25}$ .

Step-by-step explanation:

As given,

The child's knowledge of the alphabet is limited to the letters a, b, c, i, and e.

If child writes two letter at random, we have to find the probability that one is a vowel and the other is a consonant.

So, total letters (L) = 5

Vowels (V) = 3 (a,i,e)

Consonants (C) = 2 (b,c)

Probability that one is a vowel and the other is a consonant (P) = ?

P = Desired possibilities (D) /Total possibilities (T)

Total possibilities of writing any two letters assuming he can also repeat the letter is = L * L = 5 * 5 = 25

There can be two scenarios of a writing letter with a constant and vowel.

1) V.C (vowel followed by consonant)

=> 3*2 = 6

1) C.V (consonant followed by vowel)

=> 2*3 = 6

So, Desired possibilities (D) = 6+6 = 12

Hence.

$P = \frac{12}{25}$

Therefore, the probability that one is a vowel and the other is a consonant is $P = \frac{12}{25}$ .

Keywords: probability, alphabets

Learn more about probability at:

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DiKsa [7] · Answer 2 · 2022-06-19T00:25:27+03:00

Answer:

$\frac{6}{25}$

Step-by-step explanation:

In probability

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) = 2/5

Second, P(consonant):

There are 3 consonants, hence

P(consonant) = 3/5

Now, we multiply:

P(vowel and consonant) = P(vowel) * P(consonant) = 2/5 * 3/5 = 6/25

The probability that one is vowel and the other is a consonant = $\frac{6}{25}$