Question 588172
ONE WAY:
I would calculate the probability of no X and subtract from 1.
Not counting Y as a vowel, there are 5 vowels in the 26 alphabet letters, leaving 21 consonants to choose from.
The probability of not having an X as the first letter would be {{{20/21}}}.
The probability of not having an X as the second letter would be {{{19/20}}}, because whatever letter was chosen first would not be available.
The probability of not having an X as the third letter would be {{{18/19}}}.
The probability of not having an X as the fourth letter would be {{{17/18}}}.
The probability of not having an X would be
{{{(20/21)(19/20)(18/19)(17/18)=20*19*18*17/(21*20*19*18)=17/21}}}
The probability of having one X would be
{{{1-17/21=4/21}}}
ANOTHER WAY:
There are {{{20*19*18}}} ways to choose a 3-letter combination with no repeated letters from the set of consonants that are not X.
Inserting an X in any of the 4 possible positions, you get the
{{{4*20*19*18}}} combinations of four different consonants including one X.
There is a total of {{{21*20*19*18}}} combinations of four different consonants.
Among those, we had found that only {{{4*20*19*18}}} could include an X.
The fraction including an X is
{{{4*20*19*18/(21*20*19*18)=4/21}}}