Question

Higgins was asked to make a unique three-letter arrangement using only letters from the word home. He can use each letter only once. (His arrangement doesn’t need to be a valid word.)
:The probability that Higgins forms a three-letter arrangement with vowels as the second and third letters is
1/3 1/4 1/6 1/8

The probability that Higgins forms a three-letter arrangement with two consecutive consonants is
1/3 1/4 1/6 1/8

Answer 1

What is the question.?

Answer 2

Answer: 1) Third option is correct.

2) First Option is correct.

Step-by-step explanation:

Since we have given that

There is unique three- letter arrangement using only letters from the word home.

so, The total possible arrangement would be :

$^4C_3=\dfrac{4!}{3!\times 1!}=\frac{24}{6}=4$

1) We need to find the probability that Higgins forms a three letter arrangement with vowels as the second and third letters is as follows:

• For the first letter to be not vowels is $\dfrac{2}{4}$
• For the second letter to be vowel is $\dfrac{2}{3}$
• For the third letter to be vowel is $\dfrac{1}{2}$

So, the total probability that a three letter arrangement with vowels as the second and third letters is

$\dfrac{2}{4}\times \dfrac{2}{3}\times \dfrac{1}{2}=\dfrac{4}{24}=\dfrac{1}{6}$

Therefore, Third option is correct.

Now, we need to find the probability that HIggins forms a three letter arrangement with two consecutive consonants is as follows:

• For the first letter to be consonants is $\dfrac{2}{4}$
• For the second letter to be consonant is $\dfrac{2}{3}$
• For the third letter to be consonant is $\dfrac{2}{2}$

So, the Probability that Higgins forms a three letter arrangement with two consecutive consonants is given by

$\dfrac{2}{4}\times \dfrac{2}{3}\times \dfrac{2}{2}=\dfrac{8}{24}=\dfrac{1}{3}$

Hence, First option is correct.