Question

The word geometry has eight letters. Three letters are chosen at random. What is the probability that two consonants and one vowel are chosen?

Answer 1

Answer:  $\dfrac{15}{28}$

Step-by-step explanation:

The given word : geometry

Total letters in the above word = 8

Number of vowels (a e i o u) in word  =3

Number of consonants = 5

Number of ways to choose two consonant and one vowel is given by :-

$F=^5C_2\times^3C_1\\\\=\dfrac{5!}{2!(5-2)!}\times (3)\ [ \because ^nC_1=n]\\\\=5\times2\times3=30$

Number of ways to choose any three letters from the 8 alphabets:-

$T=^8C_{3}=\dfrac{8!}{3!(8-3)!}=8\times7=56$

Now, the probability that two consonants and one vowel are chosen :-

$=\dfrac{F}{T}\\\\=\dfrac{30}{56}=\dfrac{15}{28}$

Hence, the required probability = $\dfrac{15}{28}$

Answer 2

In geometry there are 3 vowels (e,e,o) and 5 consonants (g,m,t,r,y). 
There are 3 ways to pick two consonants and one vowel
(CCV)
(CVC)