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

Answer: 1) Third option is correct.

2) First Option is correct.

Explanation:

Answer 2

Answer: 1) Third option is correct.

2) First Option is correct.

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=(4!)/(3!* 1!)=(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
  `(2)/(4)`
• For the second letter to be vowel is
  `(2)/(3)`
• For the third letter to be vowel is
  `(1)/(2)`

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

`(2)/(4)* (2)/(3)* (1)/(2)=(4)/(24)=(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
  `(2)/(4)`
• For the second letter to be consonant is
  `(2)/(3)`
• For the third letter to be consonant is
  `(2)/(2)`

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

`(2)/(4)* (2)/(3)* (2)/(2)=(8)/(24)=(1)/(3)`

Hence, First option is correct.