Probability of two vowels from the word MATE if two letters are chosen at random, without a replacement? My textbook seems to have given me an answer without explaining. It states that the answer is 1/6 but it's a four-letter word, someone please explain. :)
 A: There are six ways to choose two letters from the word:
MAte MaTe MatE
mATe mAtE maTE

but only the choice of A and E results in two vowels. The probability is thus 1/6.
A: There $4$ choices  to draw the first letter because there are four possible letters.  Once you draw the first letter there will be three letters left and so there will by $3$ choices  to draw the second letter.
That means there are $4*3 = 12$ ways to draw two letters.
If you don't trust that, you can count them out.  The first letter can $M,A,T,E$.
If it's $M$ the second letter can be $A,T,$ or $E$. and you could draw $MA, MT$ or $ME$.  $3$ ways.
If the first letter is $A$ the second letter $M,T,$ or $E$.  $AM, AT,AE$.  $3$ more ways.
Then $TM, TA$ or $TE$.  $3$ more.
And finally $EM, EA,ET$.  $12$ ways total.
Now in order to draw two vowels there are $2$ choices for the first vowel: $A$ or $E$.  And for the second vowel there will only be one vowel left to be the second one.
So there are $2*1 =2$ ways to draw two vowels.  Namely:  $A$ then $E$.  And $E$ than $A$.
So of the $12$ total ways to draw two letters, there are $2$ ways to draw to vowels.
So the probability is  $\frac 2{12} = \frac 16$.
A: The answer is very simple. There are four letters in the word but actually there the sample space is 12 because 2 letters are choose . And event given is to be AE hence probability is 2/12=1/6
A: The word has 4 letters. Two of them are vowels so there is a 2/4 chance that the first pick will be a vowel.
That would leave three letters, containing one vowel, so there is a one in three chance that the second pick will be a vowel.
2/4 * 1/3 = 2/12 = 1/6

